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5:24 I didn't even notice they were fibonacci numbers. I was seeing that the 3rd entry was twice the 2nd plus the 1st; the 4th was twice the 3rd plus the 2nd and the 1st; the 5th was twice the 4th plus the 3rd, the 2nd, and the 1st; etc.
1 2 |1 2 | 1 2 3 4|3 4 | 3 4 2 1 |2 1 | 2 1 4 3| 4 3| 4 3 I think this permutation would be easier to see, because then you are just mirroring the square above it…and since your are just mirroring the first square it is obvious that each corner must be 1 of 4 colors. The question I have is how many different ways you can arrange this tiling to still hold this property? Aside from just swapping a single color for another color? In some way this seems like a permutation of roots kinda question…
You can do the first problem without Fibonacci numbers or cassini’s identity. The thing you have to notice is that 2*13=5^2+1 and 5*34=13^2+1. From there, conjecture that the next pair after (n,m) looks like (m, (m^2+1)/n). This turns out to be easy to prove inductively. Cassini’s identity basically just gives that this pattern will generate Fibonacci numbers, but the connection is unnecessary.
@@shimonschlessinger7262 that follows immediately from the induction, no? if you say that (n,m) is a valid pair, it must mean that n divides (m^2 + 1), since that is what it means for the pair to be valid
I was trying to prove this by writing the pairs as a sequence a_n where a_n+1 = 3a_n - a_n-1 and proving this works by induction. Not sure if it's even the correct approach though
The question about colouring the positive integers can be solved using Ramsey's theorem, which (I assume) is taught to most Olympiad students. Assume the positive integers are equipped with some 3-colouring. Connect each pair of distinct positive integers m < n by an edge, and colour this edge with the colour of n-m (also a positive integer as m and n are distinct). Ramsey's theorem tells us that in any complete graph with at least R(4,4,4) vertices (where R(4,4,4) is some particular Ramsey number, i.e. the least positive integer large enough for this property to hold), any 3-colouring of the edges will yield a monochromatic K_4 (a set of 4 vertices where all edges between them are monochromatic). Take some finite subset of at least R(4,4,4) positive integers with the edges coloured according to the rule above. Then there exist elements of this subset n_1 < n_2 < n_3 < n_4 such that n_2 - n_1, n_3 - n_1, n_4 - n_1, n_3 - n_2, n_4 - n_2 and n_4 - n_3 are all monochromatic. Then label these integers as a, a+b, a+b+c, b, b+c and c respectively, and double check everything :).
Here's my (arguably) simpler proof of the first problem. It's a proof by induction. Base case: the pair 1, 2 works (easy to find). Induction step: assume we have found a pair a, b such that a < b and a | b^2 +1 and b | a^2 +1. Then we can write b^2 + 1 = a * c for some positive integer c. Then it is clear that c | b^2 + 1. Thus it suffices to show that b | c^2 + 1. But c = (b^2 + 1) / a, so c^2 +1 = (b^4 + 2b^2 + 1) / a^2 + 1 = (b^4 + 2b^2 + a^2 + 1) / a^2. We know b | a^2 + 1, so b divides the numerator. But if b | a^2 + 1, then b is relatively prime to a^2 (the denominator). It follows that b divides the quotient, and we are done. Thus, starting from any pair (a, b) we can find, we can generate an infinite sequence as follows. Start with (a, b). b^2 + 1 = a * c gives us the pair (b, c). c^2 + 1 = b * d gives us the pair (c, d). And so on. This proof is nice because it does not require any knowledge about properties of Fibonacci numbers. And it follows easily from the pattern of the first few examples that you found (and they were the first few examples I found as well).
I have no idea what you're doing in that first paragraph, but my brain did naturally jump to your second paragraph. Because each pair will be related to the next with these simple a^2+1 and b^2+1 relations, you should be able to just take that relation and apply it to a working pair. 1^2+1 = 2, and 2^2+1 = 5, which is divisible by 1, the requirement of this question. Then 5^2+1=26, which now just has to be divisible by the previous number in the sequence (2), to meet the requirement of the question, and you have your value, 13. I guess the last question is, to prove the formula always yields an integer and not a fraction, or maybe that's just obvious, I'm not sure at all.
I should also add that c > b, otherwise there is no guarantee that my construction would go in a circle. That follows from the fact that b > a, which implies c = (b^2 + 1)/a > (a^2 + 1)/a > a^2/a = a
It can also be described via Vieta jumping, where we have a, c as the two solutions to a^2 + (kb)a + (b^2+1)=0. Since c is the other solution, we have c^2 + (kb)c + (b^2+1)=0.
This video means so much to me, thank you! I last watched X+Y before my diagnosis and that scene with Luke crushed me. I need to watch it again, but your summary and perspective was really valuable.
Another Roof in some Q&A video a while back: "Don't get me started on films" Another Roof today: basically a film critic in a niche genre (no complaints here)
I wonder if you can do problem 1 by noticing that in the sequence 1, 2, 5, 13, 34 ... each term is three times the previous one minus the one before that: 5 = 3x2-1, 13 = 3x5-2, 34 = 3x13-5 and so on. Then the result can be proved by induction. Suppose c, d, e and f are four terms of this series so we know that e = 3d-c and f = 3e - d, and suppose inductively we already know that ce = d^2+1. Then we have to show that df = e^2+1, so we evaluate e^2+1-df. First we replace f, and then we factorise out the e: so e^2+1-df = e^2+1-3de+d^2 = e^2-3de+ec (by the inductive assumption) = e(e-3d+c) = 0 because the bit in the brackets is zero by the definition of e. I suppose I only "noticed" the formula for the sequence because the video said it was every other term of the Fibonacci sequence, but really I should have looked more closely earlier on!
In Problem 1, we can generate a sequence that follows Catalan's identity without knowing about the Fibonacci sequence. We simply start with F_1=1, F_2=2 (because we know (1,2) is a solution) and we define F_n=[F_(n-1)^2+1]/F_(n-2). By induction, we can show that all F_n are natural numbers and F_n>F_(n-1) for all n, so we have infinite solutions.
Great video! X+Y is actually one of my most favorite movies since it presents the atmosphere at the Olympiad training camp so well (and since I'm easily influenced by cheap sentimental music I guess). I don't know if you chose this movie for MAFS because I suggested it in a comment under a previous video or it came to you in some other way, but I am really glad you watched it and made this episode with your take on it. I will definitely become a patreon to check out the extra video!
I didn't see the whole movie but having seen only the "card problem" scene in the mathologger video, I have to agree with the mushiness and the critic of the music! I can't imagine the whole class applauding for someone solving their first problem at the blackboard, even if the class is aware of the difficulty of the student and sympathetic. Actually, that would probably make Nathan feel even worse in this situation (I'm assuming they don't do it each time a student solves a problem).
The trick is to slowly reword the rule into more powerful rules.You can color the numbering, because there isn't a rule saying a and b can't be equal and play basketball. Assume you can color the numbering in such a way. What other axiom can we construct? Consider a=b=1, and c=n. What we get is that 1, n, n+1, and n+2 can't the same color. no three consecutive numbers can match the color of 1. Next, consider thw special case, a=b=2, and C is N We get 2, N, N+2, N+4. In other words no alternating numbers can share the same color as 2. I think you can see where we're going with this. Speaking generally, no collinear numbers can simultaneously share a color with the number which is the distance between them. Or, you can refresh this another way. Given any 3 numbers that match in color, (something that is guarenteed past 4, and gets ridiculously high at small values) the sum of the highest and the lowest must be different to all three of them, independent of the number between them. Call this number, the cap. But the middle number dosen’t matter for determining what the cap number is, as long as it exists somewhere between the largest and the smallest. Given that 2 colors with an identical color somewhere in between them, their sum must nessesaralu be a diffrent color. You are also guaranteed you will encounter at least 3 of 1 color every 7 units along the numberline. So, we can pick one of them where this property is true for every 7-segmant of the numberline can be associated with 1 or multiple cap numbers. We've stripped out half the fat of the problem. the six matching colors we needed were whittled down to basically 3.and generalized it to basically A, B and A+B. Haven't proven it yet. But that's pretty damn restrictive.
I like how you solved the similar-colour issue in your depiction of the train problem. It's generally a good idea to couple colour-coding with at least one other form of coding to ensure colour-blind people can follow along also. I feel this would also have been useful for the 72-gon problem. You could have marked the vertices as red dots, blue diamonds and green crosses for example. Perhaps you can keep this in mind for future videos. Not colour-blind myself, it's just something I picked up in a profession that also involves graphic design. All in all great video. As someone with both Dyscalculia and a deep love for maths, I always get a kick out of seeing someone demonstrate solutions for problems I would have an extremely hard time solving myself. Good stuff!
The solution to the three color problem relies on the van der warden theorem (if someone who doesn't know what that is is reading this just look it up on wikipedia, the proof is there too. It doesn't require background and it's rather beautiful). First aacording to the van der waerden theorem there is an arithmetic progression with one color, let's say green without loss of generality. Let's assume the jump of the arithmetic progression is d (the series being all the numbers that are some r mod d), let's look at all of the numbers that are divisible by d. If we find two green numbers there such that a,b,a+b are all green then if we take any number c from the arithmetic progression with jump d we found earlier then a,b,a+b are green by assumption and a+c, b+c and a+b+c are all green because they are part of the green arithmetic progression. Now when we restricted our view to a subset which also has the same properties as all the integers namely all the integers divisible by d (in the sense that a coloring of all of those is equivalent to the coloring of the integers) we only need to find two numbers a and b with the above property which are green or a b c with the property desired by the question which are blue or red where as before it had to be red blue or green so effectively we managed to reduce the numbers needed in one color, keep doing this process until the numbers needed for one color reaches one, at that point that color can't appear anymore. Keep doing the process with the rest of the two colors until both of them are eliminated meaning there is no color that can appear after a certain point but that is a contradiction since the numbers after that point have to be colored something, i think this approach also proves the theorem stated in the video. The van der waerden theorem is something that is used quite a bit in olympiad combinatorics questions, i wonder if that was an actual question once, also it's placement as the first problem in a team selection test seems fitting difficulty-wise
It looks like a beautiful argument. There is a flaw though that I don't see a direct solution for: The van der Waerden theorem does not state there is an infinite arithmetic progression, only that you can find one of any finite length. This means that the subproblem is finite - it cannot use numbers larger than the last number in the chosen progression - and then we cannot apply induction.
that would be the case if the induction was infinite but it isn't in this case. at most it ends after 9 steps so if we take the arithmetic progression to be big enough for these 9 steps it should work as an example here is a possible sequence for the induction to take: at first we need to find a b and c satisfying the question that are blue red or green, next we need to find a,b,c red or blue or just a,b that are green. next we would need a,b,c red or a,b that are red or blue. this continues until all three colors terminate @@spaanse
Thanks for the very engaging and interesting video! I very much like your style and delivery, and exceptional quality. Very much looking forward to your videos in 2024! Happy New Year!
I'd love to see your reaction to "The Man Who Knew Infinity" about Ramanujan! I honestly can't remember how deeply they delved into the math topics because I was so invested in the human story 😅
Problem 1 gave me the strangest jumpscare. I'm 80% sure proving Casini's identity (although it wasn't labelled as such ) by strong induction was a question on one of the midterms for an undergraduate elementary number theory course I took this semester. I messed up the proof during the midterm but I'm tempted to attempt it again after this video
I remember the problem about trains as pretty basic induction problem. Like in the first or second class on induction. But I haven't heard the solution from the video before, it's beautiful
5:22 I saw a different pattern. The next number is created by doubling the last and adding sum of all before. x[ i ] = x[ i -1 ]*2 + sum (from n=1 to i-1) It can be also written as x[ i ] = x[ i - 1 ] + sum (from n=1 to i) And this function really gives out every second fibbonachi number haha
This is a fun continuing series, thanks! Not sure exactly what film to suggest but it would be good to see one that you are more negative on or that is much less accurate, so we can calibrate the scale a bit! There must be some sci-fi action flick with terrible orbital mechanics or something 😅
The 3 color positive integer problem feels like it could be solved using some tools from graph coloring and/or hypergraphs. I will try finding a solution using these tomorrow, when I am with my math books.
I think it has something to do with hasse diagrams consider the integer a,b,c and all colorings can be represented by the power set of a,b,c and by transitivity
Pi (not Life of): I haven't watched it in a while, and even though the movie doesn't have a huge amount of mathematics in it, it does exploit the motif of Go and the titular constant throughout, and therefore ultimately I do think it probably fits the mould you are after. The golden ratio is in there a lot too actually (maybe more than pi?) but is treated a bit mystically rather than with mathematical rigour.
Wow, this video is great! As a maths major I found the maths interesting, and as someone interested in media critique and analysis you're review and comments were fascinating, perfect combo!
for problem 3. A. there is at least 1 color that forms a set with infinite elements card(S)=card(N) since the 3 colors form a partition of the naturals and the naturals are infinite, card(S1)+card(S2)+card(S3)=card(N) and it is clear that there is Sn such that card(Sn)=card(N). say the set of integers colored in green. B. it is clearly possible to pick 3 numbers from one of the infinite sets. since the sets are infinite, it is possible to pick at least 3 elements (a,b,c) in S^3. we need to somehow show that a+b,a+c,b+c and a+b+c are also in S for carefully picked (a,b,c) and S. C. We can proof the analogue case for 2 elements lets show that it is possible to pick a,b such that a+b is also green. assume the contrary, a+b is always red. we must also assume that the sum of 2 red is a green(or red is the set we are looking for). pick a,b such that a red and b green. Wlog assume a+b is green. we know that a+2b is red, 2(a+b) is green. this implies that 2a+b is red. 3(a+b) = (a+2b)+(2a+b) is green but at the same time 3(a+b)=(a+b)+2(a+b) is red. contradiction, it is possible to choose a set S and 2 elements (a,b) in it such that a+b is in S. We can say more, there are infinite pairs of (a,b) with such property. assume that there are finitely many pair (a,b) with such property and consider the subset of the naturals larger than all such a and such b. then identical reasoning shows that it is impossible. That is, there are an infinite amount of pairs(a,b) in S^2 such that a+b is also in S for a set S. not sure if there is a simple way to directly generalize this to the 3 variables case but that is an extremely elementary way of solving a similar problem.
I believe induction on the number of colors and numbers should do the trick, if it does work congats on solving the problem. Edit: it does work but i think you need the van der waerden theorem for it but i am not sure
A solution to the 3 color positive integer problem: As an overview: We are going to find a lot of rectangles in a lattice that all have the vertices same color, then show that the distances between all of those rectangles form their own lattice, and there are enough of them that we are guaranteed at least one with the same coloring. First, make a rectangular lattice of 13 points 3^4*3+1 points (numbers explained in next paragraph). Choose the spacing in the grid so that all the distances are distinct; this is important in the last paragraph. One way of doing this is to have the nth point in a column and mth point in a row have coordinates (2^n+n-2, 2^(m+13)+m-2^14-1). This will make all of our distances distinct later. Color each vertices the same as the number with its coordinate sum. This means that the point in the second column and 3 row (i.e. in point (2,3) in the lattice) would have coordinates (9,2^14+1) and be colored the same as the number 2^14+10 was colored. Notice that in a column of 13 points, at least 5 of them have the same coloring by the pigeon hole principle (PHP). Likewise, across 3^4*3+1 columns, at least 5 of them have the exact same colorings. Thus, if we take the 5 columns that have the same coloring, and choose the 5 rows in those columns that are colored the same, we have a 5x5 lattice where all the vertices are colored the same. Choose any rectangle in the 5x5 lattice. Notice that we could call the lower left vertex in our chosen rectangle a, we can call the upper left vertex a+b, the lower right vertex a+c, and the upper right vertex a+b+c. So, we're halfway done; we still have to show that b, c, and b+c have the same color. Notice that in a 5x5 lattice there are (5 choose 2)=10 distances along one edge. Create a new lattice with where the x and y coordinates correspond to each of the distances in our large 5 x5 lattice with all the vertices that are colored the same; this is a 10x10 lattice. We only need the lower left 4x7 of this lattice. By the pigeon hole principle, the are at least 2 points of 4 with the same color, and as there are only (4 choose 2) ways of choosing which 2 points in the row have the same color, in our column of 25 four of the rows chose the same two to color, and by the pigeon hole principle at least 2 of those 4 columns choose the same color to color their points. Thus, we have found a rectangle in our smaller lattice whereas 3 points are colored the same. Let the upper left one be b, the lower right one be c, and then their sum, b+c, is exactly the upper right corner. Thus, we found a+b, a+c, a+b+c, all matching color, and b, c, and b+c all matching color, although I realize now that is a possibly distinct color. So.... this doesn't work, but I'm going to publish this answer anyway.
My solution to the one with colored trains: Let's generalise this problem to 2nx2m grid. Now, let's use transfinite induction based on any good ordering implementing the natural partial ordering. For 2x2, that's by the problem statement. For 2x4: Each row has different colors because it participates in a 2x2 square. Let's track a value that says if a color in a cell is also in the first row. In each row, it's the same iff it was the same in the previous one. Therefore, it's the same for each row. For each row, it's different from the previous one. Therefore, it's different in the first and last ones. Therefore, the first and last rows don't have the same colors. Therefore, the corners all have different colors. For 4x2: Same except with columns instead of rows For 2nx2m: Choose any way to arrange 2n-1 and 2m-1 as a sum of odd numbers other than the trivial two. The trivial two are each as itself as the only one and each as itself of ones. Select the rows and columns starting with the first one in steps equal to the summands. The values in the chosen rows and columns obey the condition. Therefore, the corners are all different Another solution, also by me: Let's consider the first two columns. Apply the reasoning used for the 2x4 case in the above solution. Now, consider two cases: There are different odd-numbered rows and/or different even-numbered rows in the first two columns. In this case, pick any two consecutive of them. The third element in the row between them must be the same as in the first column. If one is the same as in the first column, the ones adjacent to it are, too. Therefore, the entire column is the same. By the same reasoning, all odd-numbered columns are the same, and all even-numbered columns are the same. Therefore, the colors at the corners are the same as the colors at the corners of the first two columns. Therefore, they're all different. Otherwise: Apply the same reasoning with rows and columns switched. In doing so, instead of picking consecutive different same-parity columns as the seed, use the first column
I feel like the red/yellow/green integers can be solved with the pigeon hole principle. To start: Pick any 7 integers at random, you know there has to be a color that is represented 3 or more times.
Damn to think question 2 is just a wonderful example of the power of the pigeonhole principle is just amazing, would have never thought of it in that way, but that is just amazing.
Once you notice that the numbers are Fibonacci sequence, solution is trivial. Pretty much everyone familiar with Fibonacci sequence knows that it can be expressed in a closed form using the golden ratio, thus as F_n = ((1+√5)^n - (1-√5)^n) / √5. Now, using this closed form, it's very easy to show that F_n-2*F_n+2 = F_n^2 + 1.
I actually saw the pattern in problem 1 and was able to tell it was fibonacci related without recognizing 13 and 34 as fibonacci numbers. Because I noticed that each number was nearly 3 times the previous, then that the difference was exactly the number before that. So Xn = Xn-1 * 3 - Xn-2 I hadn't realized this was the same as the odd fibonaccis until I unpaused and then checked to make sure it was
I also saw the pattern, but just because one of the first things you check is addition/subtraction/multiplication/division of adjacent, and you immediately strike gold as you calc a bunch of Fibonacci numbers when you try subtraction.
When you asked for the pattern in 1,2,5,13,34 i got f(n) = 2(n-1) +sum from i=0 to i=n-2 E.g. 2*1=2, 2*2+1=5, 2*5+1+2=13, 2*13+5+2+1=34 Is that true for all n
@@Joe_PayneYeah, let’s take F(10), the 10th fibonacci number. By definition, F(10)=F(9)+F(8). But then we repeat with F(8)=F(7)+F(6), so F(10)=F(9)+F(7)+F(6). Repeat with F(6), then F(4), then... until F(2) which is F(1)=1 + F(0)=0, so we can remove F(0) to get F(10)=F(8)+F(6)+F(4)+F(2). Note that this also works with odd fibonacci numbers, except that when you try to do F(3)=F(2)+F(1), you can’t make F(1) disappear. But since F(1)=1, then every odd fibonacci numbers are the sum of every previous even fibonacci numbers +1 To go a bit further, let’s combine both results : if you take F(10), F(10)=F(9)+F(8). F(8) is the sum of every odd entries (starting at F(7)), and F(9) is the sum of every even entries before it (starting at F(8)) +1. So F(n) is the sum of every fibonacci numbers up to n-2, then add 1
As quentind points out, that observation is essentially the definition of the Fibonacci sequence, so it makes sense. Back in the day I noticed an interesting correspondence between adjacent numbers squared, which after playing around with it a bit, actually just reduced down to the arithmetic series. Kind of disappointing, but also cool to rediscover a common formula from a totally different way.
The first problem leaves me with a question... are the odd-term Fibonacci sequence pairs the ONLY pairs possible like that? It certainly answers the question and establishes there are infinitely many of them... but are they the only such pairs? Also, when you're discussing what the father tells the very young character in the film in framing autism as like 'having special powers like a wizard', do you really think it would be helpful to confront a very young child who is recognizing their difference from their peers with all of the real challenges people with autism face? The father wasn't talking to the audience, he was talking to his (again) very young child. It also contributes to making the self harm motivations of the other character more relatable later in the film. In addition it also shows that parents don't always do the 'right thing' in dealing with their children, and that the 'wrong thing' is often done with the best intentions.
Brilliant video! Do you have a Letterboxd? You have such nuanced thoughts on the films you discuss, in 'addition' to the math, and I would love to read more of them!
My solution to the problem Mathologer also covered: Let's assume that goes on forever. Pick the leftmost pair that is flipped infinitely many times. After last flipping the previous pair, you can only flip it once You can flip each pair at most 1 more time than the previous one. This allows one to easily find the maximum total number of flips. The minimum is 0 because you can start from all face down
I sat and pondered the colored squares problem, and got that we can roll up the square into a cylinder and then into a torus, and the property still holds. The corners then form a single 2x2 square, and so they all have different colors.
@@AnotherRoof Pick any two adjacent columns. Then looking at the top three rows, we can form two 2x2 squares enforcing local color neutrality. Then because the two colors in the second row is common to both squares, the first row and third row have the same 2 colors (not necesarily in the same order tho). This extends to the fact that all even rows have same two colors and all odd rows have the other two colors (for the chosen two-wide strip). Now suppose we roll it up into a cylinder by joining the top and bottom rows, since the side length is even, the top row is an odd row and the bottom row is an even row. Thus the new 2x2 square formed in this 2-wide strip using the top and bottom row had all four colors, since the rows contain 2 different sets of 2 different colors. Thus the property holds on a cylinder, but the color commonality holds along the columns too, so the property holds after we form a torus too by the same logic. QED Actually, we can show that either all even rows are identical to each other (same for odd), or all rows contain only 2 colors each in an alternating pattern, and successive even(or odd) rows can only differ by a 180 phase shift. Same for columns
After reading Varad's idea, I can see how it can work. if you take two rows together they have an equal amount of each color because they can be tiled by squares. if we look at the pair of rows without the leftmost and rightmost column we still have an even length so still an equal number of each color, thus, rightmost and leftmost columns together have all colors. which is a square in the cylinder. this means we can fold it into a cylinder, and from the same argument for columns we can roll it into a torus.@@AnotherRoof
@@razdvora709 How do you know the question is tile-able by squares? And how does that mean that you must have an equal amount? I wish the full question had been written somewhere, I feel like this is almost in the question itself, but I'm not sure how; maybe knowing what a train is, or how that has to be in a 2x2 grid, would help.
for question 3 this is where i got to divide the positive integers into groups of an arbitrary a. It must repeat somewhere So i guess proving that there exist a,b,a+b with the same colour given a 3 colouring for a arbitrary a is a way but i am totally stuck
@@octosaurinvasion by repeat i mean like divide it into segments every n numbers. so the first n will be one segments, n+1 to 2n one segments. at some point the segments will repeat?
@@boiii2148 They won't necessarily repeat. If you look at my example there's no way you could divide it into repeating segments. The GRRG string near the beginning will never happen again, and then the GRRRG after it will also never appear, and so on.
I must rewatch this film. I saw it some time ago, ironically my non mathematical flat mate brought it to my attention, I’d never heard of it. As someone who did maths at uni for 4 years, this is possibly my favourite film with significant maths content, though I’m not sure how long that list is.
Im thinking tne dad was killed off maybe to shift some of tne autism traits along the spectrum. Personally I have found when I have had some trauma or difficulty some of my traits slide up and down the spectrum. They did try to do this by linking synasthesia in. The weakness of the film is the love plot line and I think the parallel is more related to that personallyI think a director like Mike Leigh would have handled this aspect better. I think porcupine trees "fear of a blank planet may have been a better fit for some of the score. Also another interesting thing I found was dealing with "non autistic" competitors I have often found that I myself can present as non autistic or less autistic depending on a percieved threat level but then I grew up in a time where it wasnt recognised nor accepted.
Interesting. I didn't spot the Fibonacci and instead went with term n+1 is 3xn - term n-1. Now I have no idea if that's just a coincidence or a property of Fibonacci series. If I cared more I would find out I guess.
I think this is related to Vieta jumping. The two divisibility relations can be condensed into one Diophantine equation: a^2 + kab + b^2 + 1 = 0 We notice this is a quadratic in a, and we can set k=-3. The sum of roots is -kb = 3b. Swapping a with the other root would give a new solution.
Problem 3: I'm going by the assumption that a!=b!=c bc otherwise it's trivial. Note that the problem from the movie allows negative integers. So, let a = -b, b != 0, c = 0. All of these addition terms will become either a, b, c; so now we just need to prove that there is a case in which a = b = c (color wise). Of course, that is not always the case. Let's analyze the cases in which this does not work: 1. if the colour of 0 is unique to 0. For instance if 0 is red and every other number isn't. 2. If the colours are NEVER symmetric about 0 (excluding 0). I.E if Color(-n) is NEVER = to Color(n). I think case 2 is the easier one to solve for. This is because the set of possible colors is now restricted to the fact that Color(-n)!=Color(n). This is where I got after a couple mins of thinking, I will post replies as(if) I progress
In case I explained this poorly, I'm hoping to take a "two-pronged" approach to this, where if the color conditions are satisfied, then we can use the method I've listed, and if they aren't, we can use a different algorithm
For the following, I'm not confident this logic is sound, so somebody please correct me if this is incorrect! For the first case: let a=a, b=-2a,c=3a. therefore, a+b=-a, a+c=4a,b+c=a,a+b+c=2a. So, we need to show that there is always a case s.t -2a, -a, a, 2a, 4a are all the same of 2 colors (since 0 is a unique 3rd color), where a is a non-zero integer. Observe that if you have a=2^k, it will share exactly 2 values with a=2^(k+1) and 1 value with a=2^(k+2). However, since we have 2 possible colours, via the pigeonhole principle, we can see that these colours will become saturated. Thus there is some a=2^(k), keN s.t the condition is satisfied.
I'm actually still just trying to understand the question. Is it assuming that the integers are all randomly colored? Or can you even try and adversarially color it to make it false? Couldn't you just make 1=red, 2=green and 3->infinity = yellow, so it will never work?
I think there's a much, much simpler solution for problem 4. Let's suppose we use your diagram and the upper left corner is Red, Blue Green, Yellow We know that the next square to the right (that shares a column with this one) must contain all four colors. And so there's only the possibility of Blue, Red/Green Yellow, Green/Red Note this is a square in an odd position. By going to the next square, we reason again that the column we add has to have Blue and Yellow, so it must be of the form Red/Green, Blue/Yellow Green/Red, Yellow/Blue This argument repeats for each "move" to the right. Since the last square is in even position, it must be of that latter form. Notably, the upper right square is Blue or Yellow, *not* Red. And since none of this argument depends on orientation, the same argument can be repeated for the other corners by symmetry. No counting, no algebra! Just a very elegant problem reasoning about parity. If you wanted to formalize it could probably make some inductive argument on n. It's immediately true for n=1, and we can reason about swaps for the k+1 case.
@@mranonymous5268 that's the problem I encountered when I tried this argument as well! For example we use this to reason that the upper right corner must be blue or yellow, and that the lower left corner must be green or yellow. Who's to say they aren't both yellow?
I think the trick to #3 is to take 0 as one of the values. If you have (0,a,b) is a solution, then the problem reduces to find a and b st 0, a, b and a+b are all the same colour. Should be doable by assuming it doesn't exists and getting some contradiction. like the set with 0 has to be a singleton or something.
I tried something like this but got bored and thought there must be a cleaner way -- maybe the nice way will become apparent once it's all worked out though!
The problem statement specifies positive integers, so 0 is not included - starting with 0 was my first thought. My next thought is to investigate the possibility that for some n, all multiples of n are the same colour, which is a much stronger result than required, and trivially implies the desired result. If there is no n such that all multiples of n are the same colour, then that gives you a lot of structure to work with.
I think that allowing a character to express a sentiment that many parents of autistic people share seems like it isn't that bad, at least unless the film presents it as actually true
I understand why you took issue with the comments from Nathan's father, but I do think those words served an interesting purpose in representating the somewhat misguided views many hold
Damn, it's too bad I already knew I had autism, otherwise I would've gone to some psychologist and show them tbose clips and say "that's me". /half-serious
Because I couldn't find the Mathologer video in the description or pinned comment, here is the link for y'all's convenience: th-cam.com/video/04x4ZdLpN-0/w-d-xo.html&ab_channel=Mathologer
I mean, you need to prove Catalans Identity to properly proove problem 1, no? Seems a bit "Yes, because Yes: look at this Truth here:" The only other thing you are really adding is the fact there are infinite Fib numbers
Results can be stated without proof in the IMO. In fact, it is expected that students learn many theorems as part of their preparation (common ones are Fermat's Little Theorem and Muirhead's Inequality).
I've never understood why anybody would expect a character in a book or movie, etc., to represent *all* people with whatever trait that have. They're one person. No matter how they are depicted, they can't possibly pull off such a representation. Unless a work has at least several characters with that trait, representation of the spectra within that trait is never going to happen. And frankly, a movie about a dozen autistic people would itself be a weird representation, since that is uncommon in almost any real life situation.
I don't think anyone expects a character to represent all people with that trait. What matters to me is that they are a fair representation and not an offensive stereotype or anything.
What a great case for having all types of people in most media you just made! You are correct, we can't expect a single movie to show us loads of different characters that fit a minority, so we should instead have loads of movies with one or two minority characters, and in all of those, give a different - yet realistic and not offensive - representation. this way there will be a character for everyone somewhere, it just doesn't have to be in a single one.
I knew someone would say this -- in any case, 0 is deemed the 0th term and 1 the 1st term, and so on. Thus the odd-positioned terms are still those to which I referred in the video. Hope that helps!
Is there a source for your spectral breakdown of autistic traits or is it your own idea? Either way it seems like a pretty convenient way to discuss autism, and much better than the "high vs low functioning" dialect 🌟
No there isn't and like I said in the video, I must stress that this isn't a psychologically rigourous way to look at autism, just an informal way to think about it. I talk about it more in the Patreon video but essentially if you look at the diagnostic criteria (see the ICD-11 or DSM-V), you can more or less slot them into these categories. They also diagnose with three levels of autism, where level 1 is someone who requires minimal support and 3 is someone who requires substantial support -- this is another way of avoiding the "high functioning / low functioning" phrasing. Hope that helps and glad you found it useful!
Problem 1 is exactly the type of problems which made me abandon competitive math in favour of more fruitful uses of my time. They always boil down to guess it's this thing, pull this specific trick out of a hat, try a bunch of random😊 things and tadaa you solved a useless trivia you'll never touch again. And in the rare case math is actually needed in a practical application you get approximations and iterative algorithms which invalidate 80% of math and produce a more accurate solution because the purely-theoretical one had to make N simplifying assumptions.
You point out a spesific problem that need a theorem or formula but in fact most of olympiad problem didn’t require a single theorem, even when it is, there is not many theorem to memorize for you to use. And math competition itself benefit you by making your logic and critical thinking sharper, its like solving infinite puzzle with infinite rules
I know what you mean, but once you spot the pattern in the Fibonnaci numbers, the result can be proved algebraically. I mentioned this off-hand but should have spoken more about it in the video I think!
I (think) someone could suppose that a or b is a prime, then with some number theory you could get to the answer. (Im trying but the problem is pretty hard)
If you have a high res (1 mm^3 voxels) MRI of your brain, look at the ratio of grey matter volume over the white matter volume for both left and right hemispheres. Most people have a higher GM/WM ratio in the left hemisphere. I personally am even more left of the mean. Also look at the size of the cerebellum. Did you know you only use 42% of your brain? That's because (yes this is true) the so-called "small" brain, the cerebellum, has 58% of the neurons in the entire brain. They are granule cells, and they are very small, creating a crossbar that helps coordinate muscles. Or not, if it's smaller than normal. The new Mac M3 Max has more transistors (90 billion) than the human brain has neurons (86 billion) and you only use 36 billion neurons for thinking. Normals have an IQ of 100
9:40 They don't mention this in the film, because that's wrong. It's not random, it's arbitrary. And they don't mention that because they don't have to. They say how many of each colour there are, and nothing more. So you have no more information to go on. The use of the word "always" should remove any unclarity that's left.
@@WeReNeR-k2f Neither the screenshot in the corner nor the audio of this video specified “positive” or even “nonnegative”. Maybe the movie audio did but I have not seen the movie so don’t know.
@@danielrhouck the problem is stated on the desk exactly one second after the timestamp in your comment, but ig it can just be its wrong interpretation
@@WeReNeR-k2f Hmm, I see that, but it still does not match what was in the movie or what he said, so I would go with the screenshot instead of the blackboard
For problem 3, couldn't you just trivially pick a=b=c=0? The problem doesn't *technically* say that they have to be different. Pointing out stupid edge cases like this is why I never got very far in competitive math
I've worked out that each term in 1, 2, 5, 13, 34 … is
n = round(m * (ϕ + 1))
I don't know if it will hold true forever, but as the rounding amount is getting less each time, I can't see why it shouldn't. (Proof needed, though). If it does hold, then there will be an infinite number of pairs. (I'm only at 8'00", so this might be addressed later.) Does not work for the pair (1,1), but I don't know if that's even being considered, as it would require m = n.
Its because fibonacci number scales approximately with phi, the more term you calculate, the better the approximations is. So every odd fibonacci number scales with phi^2, but we know that phi^2=phi+1 because phi is the solution to x^2-x-1=0. That is why you get that result, but i really doesn’t see why is this significant, because this is definitely not needed for the proof, you just have to proof every successive odd fibonacci sequence is a pair of solution.
@@rafiihsanalfathin9479 Yes, I soon as I heard Fibonacci sequence, I though there would be a relationship to ϕ. It's been literally decades since I've looked at Fibonacci and related sequences in any depth. As for proving, I meant that I didn't have a proof, not that it should have to be done. I agree that the proof offered is sufficient.
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Check out Mathologer's video on the cards problem here:
th-cam.com/video/04x4ZdLpN-0/w-d-xo.htmlsi=XrMdD4KJVZyhpaX2
My farts are better than Another Roof's farts
5:24 I didn't even notice they were fibonacci numbers. I was seeing that the 3rd entry was twice the 2nd plus the 1st; the 4th was twice the 3rd plus the 2nd and the 1st; the 5th was twice the 4th plus the 3rd, the 2nd, and the 1st; etc.
1 2 |1 2 | 1 2
3 4|3 4 | 3 4
2 1 |2 1 | 2 1
4 3| 4 3| 4 3
I think this permutation would be easier to see, because then you are just mirroring the square above it…and since your are just mirroring the first square it is obvious that each corner must be 1 of 4 colors.
The question I have is how many different ways you can arrange this tiling to still hold this property? Aside from just swapping a single color for another color?
In some way this seems like a permutation of roots kinda question…
You can do the first problem without Fibonacci numbers or cassini’s identity. The thing you have to notice is that 2*13=5^2+1 and 5*34=13^2+1. From there, conjecture that the next pair after (n,m) looks like (m, (m^2+1)/n). This turns out to be easy to prove inductively. Cassini’s identity basically just gives that this pattern will generate Fibonacci numbers, but the connection is unnecessary.
I was about to say finding that connected is almost impossible in contest setting. There has to be a simpler solution using induction
@@shimonschlessinger7262 that follows immediately from the induction, no? if you say that (n,m) is a valid pair, it must mean that n divides (m^2 + 1), since that is what it means for the pair to be valid
I was trying to prove this by writing the pairs as a sequence a_n where a_n+1 = 3a_n - a_n-1 and proving this works by induction. Not sure if it's even the correct approach though
@@Nnm26 You can use viete jumping
The question about colouring the positive integers can be solved using Ramsey's theorem, which (I assume) is taught to most Olympiad students. Assume the positive integers are equipped with some 3-colouring. Connect each pair of distinct positive integers m < n by an edge, and colour this edge with the colour of n-m (also a positive integer as m and n are distinct). Ramsey's theorem tells us that in any complete graph with at least R(4,4,4) vertices (where R(4,4,4) is some particular Ramsey number, i.e. the least positive integer large enough for this property to hold), any 3-colouring of the edges will yield a monochromatic K_4 (a set of 4 vertices where all edges between them are monochromatic). Take some finite subset of at least R(4,4,4) positive integers with the edges coloured according to the rule above. Then there exist elements of this subset n_1 < n_2 < n_3 < n_4 such that n_2 - n_1, n_3 - n_1, n_4 - n_1, n_3 - n_2, n_4 - n_2 and n_4 - n_3 are all monochromatic. Then label these integers as a, a+b, a+b+c, b, b+c and c respectively, and double check everything :).
Here's my (arguably) simpler proof of the first problem. It's a proof by induction. Base case: the pair 1, 2 works (easy to find). Induction step: assume we have found a pair a, b such that a < b and a | b^2 +1 and b | a^2 +1. Then we can write b^2 + 1 = a * c for some positive integer c. Then it is clear that c | b^2 + 1. Thus it suffices to show that b | c^2 + 1. But c = (b^2 + 1) / a, so c^2 +1 = (b^4 + 2b^2 + 1) / a^2 + 1 = (b^4 + 2b^2 + a^2 + 1) / a^2. We know b | a^2 + 1, so b divides the numerator. But if b | a^2 + 1, then b is relatively prime to a^2 (the denominator). It follows that b divides the quotient, and we are done.
Thus, starting from any pair (a, b) we can find, we can generate an infinite sequence as follows. Start with (a, b). b^2 + 1 = a * c gives us the pair (b, c). c^2 + 1 = b * d gives us the pair (c, d). And so on. This proof is nice because it does not require any knowledge about properties of Fibonacci numbers. And it follows easily from the pattern of the first few examples that you found (and they were the first few examples I found as well).
Amazing, thanks for sharing!
I have no idea what you're doing in that first paragraph, but my brain did naturally jump to your second paragraph.
Because each pair will be related to the next with these simple a^2+1 and b^2+1 relations, you should be able to just take that relation and apply it to a working pair.
1^2+1 = 2, and 2^2+1 = 5, which is divisible by 1, the requirement of this question.
Then 5^2+1=26, which now just has to be divisible by the previous number in the sequence (2), to meet the requirement of the question, and you have your value, 13.
I guess the last question is, to prove the formula always yields an integer and not a fraction, or maybe that's just obvious, I'm not sure at all.
I should also add that c > b, otherwise there is no guarantee that my construction would go in a circle. That follows from the fact that b > a, which implies c = (b^2 + 1)/a > (a^2 + 1)/a > a^2/a = a
It can also be described via Vieta jumping, where we have a, c as the two solutions to a^2 + (kb)a + (b^2+1)=0. Since c is the other solution, we have c^2 + (kb)c + (b^2+1)=0.
This video means so much to me, thank you! I last watched X+Y before my diagnosis and that scene with Luke crushed me. I need to watch it again, but your summary and perspective was really valuable.
I really appreciate the care you put before showing the self-harming part.
Another Roof in some Q&A video a while back: "Don't get me started on films"
Another Roof today: basically a film critic in a niche genre
(no complaints here)
Hahaha! I've been waiting for someone to make this exact point 😅 Glad you're enjoying though!
I mean, someone got him started, and now he can't contain himself. I love it, I'm adding X+Y to my watch list lol
I wonder if you can do problem 1 by noticing that in the sequence 1, 2, 5, 13, 34 ... each term is three times the previous one minus the one before that: 5 = 3x2-1, 13 = 3x5-2, 34 = 3x13-5 and so on. Then the result can be proved by induction. Suppose c, d, e and f are four terms of this series so we know that e = 3d-c and f = 3e - d, and suppose inductively we already know that ce = d^2+1. Then we have to show that df = e^2+1, so we evaluate e^2+1-df. First we replace f, and then we factorise out the e: so e^2+1-df = e^2+1-3de+d^2 = e^2-3de+ec (by the inductive assumption) = e(e-3d+c) = 0 because the bit in the brackets is zero by the definition of e. I suppose I only "noticed" the formula for the sequence because the video said it was every other term of the Fibonacci sequence, but really I should have looked more closely earlier on!
In Problem 1, we can generate a sequence that follows Catalan's identity without knowing about the Fibonacci sequence. We simply start with F_1=1, F_2=2 (because we know (1,2) is a solution) and we define F_n=[F_(n-1)^2+1]/F_(n-2). By induction, we can show that all F_n are natural numbers and F_n>F_(n-1) for all n, so we have infinite solutions.
Great video! X+Y is actually one of my most favorite movies since it presents the atmosphere at the Olympiad training camp so well (and since I'm easily influenced by cheap sentimental music I guess). I don't know if you chose this movie for MAFS because I suggested it in a comment under a previous video or it came to you in some other way, but I am really glad you watched it and made this episode with your take on it. I will definitely become a patreon to check out the extra video!
Thanks! It was already on my shortlist but comments like yours made me want to make this one next, so thanks!
I didn't see the whole movie but having seen only the "card problem" scene in the mathologger video, I have to agree with the mushiness and the critic of the music! I can't imagine the whole class applauding for someone solving their first problem at the blackboard, even if the class is aware of the difficulty of the student and sympathetic. Actually, that would probably make Nathan feel even worse in this situation (I'm assuming they don't do it each time a student solves a problem).
The trick is to slowly reword the rule into more powerful rules.You can color the numbering, because there isn't a rule saying a and b can't be equal and play basketball.
Assume you can color the numbering in such a way. What other axiom can we construct?
Consider a=b=1, and c=n. What we get is that 1, n, n+1, and n+2 can't the same color. no three consecutive numbers can match the color of 1.
Next, consider thw special case, a=b=2, and C is N We get 2, N, N+2, N+4. In other words no alternating numbers can share the same color as 2.
I think you can see where we're going with this. Speaking generally, no collinear numbers can simultaneously share a color with the number which is the distance between them.
Or, you can refresh this another way. Given any 3 numbers that match in color, (something that is guarenteed past 4, and gets ridiculously high at small values) the sum of the highest and the lowest must be different to all three of them, independent of the number between them. Call this number, the cap.
But the middle number dosen’t matter for determining what the cap number is, as long as it exists somewhere between the largest and the smallest.
Given that 2 colors with an identical color somewhere in between them, their sum must nessesaralu be a diffrent color.
You are also guaranteed you will encounter at least 3 of 1 color every 7 units along the numberline. So, we can pick one of them where this property is true for every 7-segmant of the numberline can be associated with 1 or multiple cap numbers.
We've stripped out half the fat of the problem. the six matching colors we needed were whittled down to basically 3.and generalized it to basically A, B and A+B.
Haven't proven it yet. But that's pretty damn restrictive.
I like how you solved the similar-colour issue in your depiction of the train problem. It's generally a good idea to couple colour-coding with at least one other form of coding to ensure colour-blind people can follow along also. I feel this would also have been useful for the 72-gon problem. You could have marked the vertices as red dots, blue diamonds and green crosses for example. Perhaps you can keep this in mind for future videos.
Not colour-blind myself, it's just something I picked up in a profession that also involves graphic design.
All in all great video. As someone with both Dyscalculia and a deep love for maths, I always get a kick out of seeing someone demonstrate solutions for problems I would have an extremely hard time solving myself. Good stuff!
The solution to the three color problem relies on the van der warden theorem (if someone who doesn't know what that is is reading this just look it up on wikipedia, the proof is there too. It doesn't require background and it's rather beautiful). First aacording to the van der waerden theorem there is an arithmetic progression with one color, let's say green without loss of generality. Let's assume the jump of the arithmetic progression is d (the series being all the numbers that are some r mod d), let's look at all of the numbers that are divisible by d. If we find two green numbers there such that a,b,a+b are all green then if we take any number c from the arithmetic progression with jump d we found earlier then a,b,a+b are green by assumption and a+c, b+c and a+b+c are all green because they are part of the green arithmetic progression. Now when we restricted our view to a subset which also has the same properties as all the integers namely all the integers divisible by d (in the sense that a coloring of all of those is equivalent to the coloring of the integers) we only need to find two numbers a and b with the above property which are green or a b c with the property desired by the question which are blue or red where as before it had to be red blue or green so effectively we managed to reduce the numbers needed in one color, keep doing this process until the numbers needed for one color reaches one, at that point that color can't appear anymore. Keep doing the process with the rest of the two colors until both of them are eliminated meaning there is no color that can appear after a certain point but that is a contradiction since the numbers after that point have to be colored something, i think this approach also proves the theorem stated in the video. The van der waerden theorem is something that is used quite a bit in olympiad combinatorics questions, i wonder if that was an actual question once, also it's placement as the first problem in a team selection test seems fitting difficulty-wise
It looks like a beautiful argument. There is a flaw though that I don't see a direct solution for: The van der Waerden theorem does not state there is an infinite arithmetic progression, only that you can find one of any finite length. This means that the subproblem is finite - it cannot use numbers larger than the last number in the chosen progression - and then we cannot apply induction.
that would be the case if the induction was infinite but it isn't in this case. at most it ends after 9 steps so if we take the arithmetic progression to be big enough for these 9 steps it should work
as an example here is a possible sequence for the induction to take: at first we need to find a b and c satisfying the question that are blue red or green, next we need to find a,b,c red or blue or just a,b that are green. next we would need a,b,c red or a,b that are red or blue. this continues until all three colors terminate
@@spaanse
Thanks for the very engaging and interesting video! I very much like your style and delivery, and exceptional quality. Very much looking forward to your videos in 2024! Happy New Year!
I'd love to see your reaction to "The Man Who Knew Infinity" about Ramanujan! I honestly can't remember how deeply they delved into the math topics because I was so invested in the human story 😅
I watched it for the math and was very disappointed, they barely mention a fraction of his work
Problem 1 gave me the strangest jumpscare. I'm 80% sure proving Casini's identity (although it wasn't labelled as such ) by strong induction was a question on one of the midterms for an undergraduate elementary number theory course I took this semester. I messed up the proof during the midterm but I'm tempted to attempt it again after this video
Casini's identity jumpscare 🤨🧐🤨🧐🤨🧐😳🥵🥶🥵🥶🥵🥶🍷🗿
I remember the problem about trains as pretty basic induction problem. Like in the first or second class on induction. But I haven't heard the solution from the video before, it's beautiful
I just noticed the brilliant transition from the "another roof" logo to the "mafs" logo! That was awesomly done!
5:22 I saw a different pattern. The next number is created by doubling the last and adding sum of all before.
x[ i ] = x[ i -1 ]*2 + sum (from n=1 to i-1)
It can be also written as
x[ i ] = x[ i - 1 ] + sum (from n=1 to i)
And this function really gives out every second fibbonachi number haha
I would love to see your take on 21 the film about the card counting students or good will hunting
This is a fun continuing series, thanks! Not sure exactly what film to suggest but it would be good to see one that you are more negative on or that is much less accurate, so we can calibrate the scale a bit! There must be some sci-fi action flick with terrible orbital mechanics or something 😅
The 3 color positive integer problem feels like it could be solved using some tools from graph coloring and/or hypergraphs. I will try finding a solution using these tomorrow, when I am with my math books.
Did you find any?
I think it has something to do with hasse diagrams consider the integer a,b,c and all colorings can be represented by the power set of a,b,c and by transitivity
Was the solution for problem 4 referring to my comment 2 years ago? So glad that this solution is mentioned here🤩. Great video as always.
Pi (not Life of): I haven't watched it in a while, and even though the movie doesn't have a huge amount of mathematics in it, it does exploit the motif of Go and the titular constant throughout, and therefore ultimately I do think it probably fits the mould you are after. The golden ratio is in there a lot too actually (maybe more than pi?) but is treated a bit mystically rather than with mathematical rigour.
Wow, this video is great! As a maths major I found the maths interesting, and as someone interested in media critique and analysis you're review and comments were fascinating, perfect combo!
for problem 3.
A. there is at least 1 color that forms a set with infinite elements card(S)=card(N)
since the 3 colors form a partition of the naturals and the naturals are infinite, card(S1)+card(S2)+card(S3)=card(N) and it is clear that there is Sn such that card(Sn)=card(N). say the set of integers colored in green.
B. it is clearly possible to pick 3 numbers from one of the infinite sets.
since the sets are infinite, it is possible to pick at least 3 elements (a,b,c) in S^3. we need to somehow show that a+b,a+c,b+c and a+b+c are also in S for carefully picked (a,b,c) and S.
C. We can proof the analogue case for 2 elements
lets show that it is possible to pick a,b such that a+b is also green. assume the contrary, a+b is always red. we must also assume that the sum of 2 red is a green(or red is the set we are looking for). pick a,b such that a red and b green. Wlog assume a+b is green. we know that a+2b is red, 2(a+b) is green. this implies that 2a+b is red. 3(a+b) = (a+2b)+(2a+b) is green but at the same time 3(a+b)=(a+b)+2(a+b) is red. contradiction, it is possible to choose a set S and 2 elements (a,b) in it such that a+b is in S. We can say more, there are infinite pairs of (a,b) with such property. assume that there are finitely many pair (a,b) with such property and consider the subset of the naturals larger than all such a and such b. then identical reasoning shows that it is impossible. That is, there are an infinite amount of pairs(a,b) in S^2 such that a+b is also in S for a set S.
not sure if there is a simple way to directly generalize this to the 3 variables case but that is an extremely elementary way of solving a similar problem.
I believe induction on the number of colors and numbers should do the trick, if it does work congats on solving the problem.
Edit: it does work but i think you need the van der waerden theorem for it but i am not sure
A solution to the 3 color positive integer problem:
As an overview: We are going to find a lot of rectangles in a lattice that all have the vertices same color, then show that the distances between all of those rectangles form their own lattice, and there are enough of them that we are guaranteed at least one with the same coloring.
First, make a rectangular lattice of 13 points 3^4*3+1 points (numbers explained in next paragraph). Choose the spacing in the grid so that all the distances are distinct; this is important in the last paragraph. One way of doing this is to have the nth point in a column and mth point in a row have coordinates (2^n+n-2, 2^(m+13)+m-2^14-1). This will make all of our distances distinct later.
Color each vertices the same as the number with its coordinate sum. This means that the point in the second column and 3 row (i.e. in point (2,3) in the lattice) would have coordinates (9,2^14+1) and be colored the same as the number 2^14+10 was colored.
Notice that in a column of 13 points, at least 5 of them have the same coloring by the pigeon hole principle (PHP). Likewise, across 3^4*3+1 columns, at least 5 of them have the exact same colorings. Thus, if we take the 5 columns that have the same coloring, and choose the 5 rows in those columns that are colored the same, we have a 5x5 lattice where all the vertices are colored the same.
Choose any rectangle in the 5x5 lattice. Notice that we could call the lower left vertex in our chosen rectangle a, we can call the upper left vertex a+b, the lower right vertex a+c, and the upper right vertex a+b+c. So, we're halfway done; we still have to show that b, c, and b+c have the same color.
Notice that in a 5x5 lattice there are (5 choose 2)=10 distances along one edge. Create a new lattice with where the x and y coordinates correspond to each of the distances in our large 5 x5 lattice with all the vertices that are colored the same; this is a 10x10 lattice. We only need the lower left 4x7 of this lattice. By the pigeon hole principle, the are at least 2 points of 4 with the same color, and as there are only (4 choose 2) ways of choosing which 2 points in the row have the same color, in our column of 25 four of the rows chose the same two to color, and by the pigeon hole principle at least 2 of those 4 columns choose the same color to color their points. Thus, we have found a rectangle in our smaller lattice whereas 3 points are colored the same.
Let the upper left one be b, the lower right one be c, and then their sum, b+c, is exactly the upper right corner.
Thus, we found a+b, a+c, a+b+c, all matching color, and b, c, and b+c all matching color, although I realize now that is a possibly distinct color. So.... this doesn't work, but I'm going to publish this answer anyway.
My solution to the one with colored trains:
Let's generalise this problem to 2nx2m grid. Now, let's use transfinite induction based on any good ordering implementing the natural partial ordering. For 2x2, that's by the problem statement.
For 2x4: Each row has different colors because it participates in a 2x2 square. Let's track a value that says if a color in a cell is also in the first row. In each row, it's the same iff it was the same in the previous one. Therefore, it's the same for each row. For each row, it's different from the previous one. Therefore, it's different in the first and last ones. Therefore, the first and last rows don't have the same colors. Therefore, the corners all have different colors.
For 4x2: Same except with columns instead of rows
For 2nx2m: Choose any way to arrange 2n-1 and 2m-1 as a sum of odd numbers other than the trivial two. The trivial two are each as itself as the only one and each as itself of ones. Select the rows and columns starting with the first one in steps equal to the summands. The values in the chosen rows and columns obey the condition. Therefore, the corners are all different
Another solution, also by me:
Let's consider the first two columns. Apply the reasoning used for the 2x4 case in the above solution. Now, consider two cases:
There are different odd-numbered rows and/or different even-numbered rows in the first two columns. In this case, pick any two consecutive of them. The third element in the row between them must be the same as in the first column. If one is the same as in the first column, the ones adjacent to it are, too. Therefore, the entire column is the same. By the same reasoning, all odd-numbered columns are the same, and all even-numbered columns are the same. Therefore, the colors at the corners are the same as the colors at the corners of the first two columns. Therefore, they're all different.
Otherwise: Apply the same reasoning with rows and columns switched. In doing so, instead of picking consecutive different same-parity columns as the seed, use the first column
I feel like the red/yellow/green integers can be solved with the pigeon hole principle. To start: Pick any 7 integers at random, you know there has to be a color that is represented 3 or more times.
perhaps "Travelling Salesman" would be appropriate for a future episode
Damn to think question 2 is just a wonderful example of the power of the pigeonhole principle is just amazing, would have never thought of it in that way, but that is just amazing.
Once you notice that the numbers are Fibonacci sequence, solution is trivial. Pretty much everyone familiar with Fibonacci sequence knows that it can be expressed in a closed form using the golden ratio, thus as F_n = ((1+√5)^n - (1-√5)^n) / √5. Now, using this closed form, it's very easy to show that F_n-2*F_n+2 = F_n^2 + 1.
I actually saw the pattern in problem 1 and was able to tell it was fibonacci related without recognizing 13 and 34 as fibonacci numbers.
Because I noticed that each number was nearly 3 times the previous, then that the difference was exactly the number before that.
So Xn = Xn-1 * 3 - Xn-2
I hadn't realized this was the same as the odd fibonaccis until I unpaused and then checked to make sure it was
I also saw the pattern, but just because one of the first things you check is addition/subtraction/multiplication/division of adjacent, and you immediately strike gold as you calc a bunch of Fibonacci numbers when you try subtraction.
i know you do a lot of movies for this series but i would love if you would review some futurama episodes, ive been told they're very accurate math
My instinct is to tackle the three-colour problem in trinary
Why do I have a very visual depiction of what this means.
When you asked for the pattern in 1,2,5,13,34 i got f(n) = 2(n-1) +sum from i=0 to i=n-2
E.g. 2*1=2, 2*2+1=5, 2*5+1+2=13, 2*13+5+2+1=34
Is that true for all n
After looking into it more, it seems like every evenly positioned number = the sum of all previous odd numbers in the sequence.
@@Joe_PayneYeah, let’s take F(10), the 10th fibonacci number. By definition, F(10)=F(9)+F(8). But then we repeat with F(8)=F(7)+F(6), so F(10)=F(9)+F(7)+F(6). Repeat with F(6), then F(4), then... until F(2) which is F(1)=1 + F(0)=0, so we can remove F(0) to get F(10)=F(8)+F(6)+F(4)+F(2).
Note that this also works with odd fibonacci numbers, except that when you try to do F(3)=F(2)+F(1), you can’t make F(1) disappear. But since F(1)=1, then every odd fibonacci numbers are the sum of every previous even fibonacci numbers +1
To go a bit further, let’s combine both results : if you take F(10), F(10)=F(9)+F(8). F(8) is the sum of every odd entries (starting at F(7)), and F(9) is the sum of every even entries before it (starting at F(8)) +1. So F(n) is the sum of every fibonacci numbers up to n-2, then add 1
As quentind points out, that observation is essentially the definition of the Fibonacci sequence, so it makes sense.
Back in the day I noticed an interesting correspondence between adjacent numbers squared, which after playing around with it a bit, actually just reduced down to the arithmetic series. Kind of disappointing, but also cool to rediscover a common formula from a totally different way.
I like the solution for problem 3 is just “oh yeah that’s been proved more generally, just use that.”
The first problem leaves me with a question... are the odd-term Fibonacci sequence pairs the ONLY pairs possible like that? It certainly answers the question and establishes there are infinitely many of them... but are they the only such pairs? Also, when you're discussing what the father tells the very young character in the film in framing autism as like 'having special powers like a wizard', do you really think it would be helpful to confront a very young child who is recognizing their difference from their peers with all of the real challenges people with autism face? The father wasn't talking to the audience, he was talking to his (again) very young child. It also contributes to making the self harm motivations of the other character more relatable later in the film. In addition it also shows that parents don't always do the 'right thing' in dealing with their children, and that the 'wrong thing' is often done with the best intentions.
The man who knew infinity movie has it as well
Another great video in the MAFS series! Happy to get here early
You're making math entertaining 🔥💪
Brilliant video! Do you have a Letterboxd? You have such nuanced thoughts on the films you discuss, in 'addition' to the math, and I would love to read more of them!
My solution to the problem Mathologer also covered:
Let's assume that goes on forever. Pick the leftmost pair that is flipped infinitely many times. After last flipping the previous pair, you can only flip it once
You can flip each pair at most 1 more time than the previous one. This allows one to easily find the maximum total number of flips. The minimum is 0 because you can start from all face down
9:30 The teacher is the same actor from "The Expert" sketch!
Haven't worked it out, but I assume problem 4 can be solved with something similar to Stokes' theorem...
You should make a playlist with all the vids from this series so we can easily find old ones
I highly recommend Suugaku Golden too
Its a manga, but has the same topic with a bunch of interesting questions
I sat and pondered the colored squares problem, and got that we can roll up the square into a cylinder and then into a torus, and the property still holds. The corners then form a single 2x2 square, and so they all have different colors.
This is a great idea, but how do you know that the property still necessarily holds after rolling it up?
@@AnotherRoof
Pick any two adjacent columns. Then looking at the top three rows, we can form two 2x2 squares enforcing local color neutrality. Then because the two colors in the second row is common to both squares, the first row and third row have the same 2 colors (not necesarily in the same order tho).
This extends to the fact that all even rows have same two colors and all odd rows have the other two colors (for the chosen two-wide strip).
Now suppose we roll it up into a cylinder by joining the top and bottom rows, since the side length is even, the top row is an odd row and the bottom row is an even row. Thus the new 2x2 square formed in this 2-wide strip using the top and bottom row had all four colors, since the rows contain 2 different sets of 2 different colors.
Thus the property holds on a cylinder, but the color commonality holds along the columns too, so the property holds after we form a torus too by the same logic. QED
Actually, we can show that either all even rows are identical to each other (same for odd), or all rows contain only 2 colors each in an alternating pattern, and successive even(or odd) rows can only differ by a 180 phase shift. Same for columns
After reading Varad's idea, I can see how it can work.
if you take two rows together they have an equal amount of each color because they can be tiled by squares.
if we look at the pair of rows without the leftmost and rightmost column we still have an even length so still an equal number of each color, thus, rightmost and leftmost columns together have all colors. which is a square in the cylinder.
this means we can fold it into a cylinder, and from the same argument for columns we can roll it into a torus.@@AnotherRoof
@@razdvora709 Lovely stuff! Thank you both for putting this argument together -- kinda wish I'd used this one in the video!
@@razdvora709 How do you know the question is tile-able by squares? And how does that mean that you must have an equal amount? I wish the full question had been written somewhere, I feel like this is almost in the question itself, but I'm not sure how; maybe knowing what a train is, or how that has to be in a 2x2 grid, would help.
for question 3 this is where i got to
divide the positive integers into groups of an arbitrary a. It must repeat somewhere
So i guess proving that there exist a,b,a+b with the same colour given a 3 colouring for a arbitrary a is a way
but i am totally stuck
it does not have to repeat. consider the sequence of colors R = red G = green RGRRGRRRGRRRRG...
@@octosaurinvasion by repeat i mean like divide it into segments every n numbers. so the first n will be one segments, n+1 to 2n one segments. at some point the segments will repeat?
@@boiii2148 They won't necessarily repeat. If you look at my example there's no way you could divide it into repeating segments. The GRRG string near the beginning will never happen again, and then the GRRRG after it will also never appear, and so on.
For problem 3 there is a result just for 2 numbers a and b called schur's theorem
Thanks for the video! ❤️❤️❤️🙏
Thanks for the interesting video, I'd love to hear your opinion on the film Super 30 about the maths teacher Anand Kumar
Pure gold. Definitely going to check out the film now!
Is there enough math in The Accountant to discuss it? Or Arrival?
Sir can you drop a video on calculas
I must rewatch this film. I saw it some time ago, ironically my non mathematical flat mate brought it to my attention, I’d never heard of it.
As someone who did maths at uni for 4 years, this is possibly my favourite film with significant maths content, though I’m not sure how long that list is.
my friend actually won silver on IMO, but I am two steps short from even getting to it 😢
Same 😢. I was his rival and training partner too😢
Im thinking tne dad was killed off maybe to shift some of tne autism traits along the spectrum. Personally I have found when I have had some trauma or difficulty some of my traits slide up and down the spectrum. They did try to do this by linking synasthesia in.
The weakness of the film is the love plot line and I think the parallel is more related to that personallyI think a director like Mike Leigh would have handled this aspect better. I think porcupine trees "fear of a blank planet may have been a better fit for some of the score.
Also another interesting thing I found was dealing with "non autistic" competitors I have often found that I myself can present as non autistic or less autistic depending on a percieved threat level but then I grew up in a time where it wasnt recognised nor accepted.
Interesting. I didn't spot the Fibonacci and instead went with term n+1 is 3xn - term n-1. Now I have no idea if that's just a coincidence or a property of Fibonacci series. If I cared more I would find out I guess.
I think this is related to Vieta jumping. The two divisibility relations can be condensed into one Diophantine equation:
a^2 + kab + b^2 + 1 = 0
We notice this is a quadratic in a, and we can set k=-3. The sum of roots is -kb = 3b. Swapping a with the other root would give a new solution.
Problem 3: I'm going by the assumption that a!=b!=c bc otherwise it's trivial.
Note that the problem from the movie allows negative integers. So, let a = -b, b != 0, c = 0. All of these addition terms will become either a, b, c; so now we just need to prove that there is a case in which a = b = c (color wise).
Of course, that is not always the case. Let's analyze the cases in which this does not work:
1. if the colour of 0 is unique to 0. For instance if 0 is red and every other number isn't.
2. If the colours are NEVER symmetric about 0 (excluding 0). I.E if Color(-n) is NEVER = to Color(n).
I think case 2 is the easier one to solve for. This is because the set of possible colors is now restricted to the fact that Color(-n)!=Color(n). This is where I got after a couple mins of thinking, I will post replies as(if) I progress
In case I explained this poorly, I'm hoping to take a "two-pronged" approach to this, where if the color conditions are satisfied, then we can use the method I've listed, and if they aren't, we can use a different algorithm
For the following, I'm not confident this logic is sound, so somebody please correct me if this is incorrect!
For the first case: let a=a, b=-2a,c=3a. therefore, a+b=-a, a+c=4a,b+c=a,a+b+c=2a. So, we need to show that there is always a case s.t -2a, -a, a, 2a, 4a are all the same of 2 colors (since 0 is a unique 3rd color), where a is a non-zero integer.
Observe that if you have a=2^k, it will share exactly 2 values with a=2^(k+1) and 1 value with a=2^(k+2). However, since we have 2 possible colours, via the pigeonhole principle, we can see that these colours will become saturated. Thus there is some a=2^(k), keN s.t the condition is satisfied.
I'm actually still just trying to understand the question. Is it assuming that the integers are all randomly colored? Or can you even try and adversarially color it to make it false? Couldn't you just make 1=red, 2=green and 3->infinity = yellow, so it will never work?
I believe that it is asking to prove it for any arbitrary coloration. @@kindlin
@@SpinDip42069 So, does my 1/2/3-∞ example show a counter example and thus, it's not true? As anything added to 3 just yellow still, so....
Can't wait to see The Man Who Knew Infinity.
how about reviewing The Man Who Knew Infinity? I haven't seen it, but I hear it's good :O
15:05 am i color blind or is the green near identical to the yellow?
13:14
Mmmm tastes like C
we all love some tasty bit manipulation :)
Do Fermat's room
man im just waiting for more math movies like x+y+z. come on already.
14:00 why would you say positive integers when you can just say natural numbers tho?
how are they different?
@@user-xy5yg6se1k Some people include 0 as a natural number, which isn't positive.
I think there's a much, much simpler solution for problem 4. Let's suppose we use your diagram and the upper left corner is
Red, Blue
Green, Yellow
We know that the next square to the right (that shares a column with this one) must contain all four colors. And so there's only the possibility of
Blue, Red/Green
Yellow, Green/Red
Note this is a square in an odd position. By going to the next square, we reason again that the column we add has to have Blue and Yellow, so it must be of the form
Red/Green, Blue/Yellow
Green/Red, Yellow/Blue
This argument repeats for each "move" to the right. Since the last square is in even position, it must be of that latter form. Notably, the upper right square is Blue or Yellow, *not* Red. And since none of this argument depends on orientation, the same argument can be repeated for the other corners by symmetry.
No counting, no algebra! Just a very elegant problem reasoning about parity. If you wanted to formalize it could probably make some inductive argument on n. It's immediately true for n=1, and we can reason about swaps for the k+1 case.
Doesn't this only prove that the top-right and bottom-left corners are not red, but not that they must be different?
@@mranonymous5268 that's the problem I encountered when I tried this argument as well! For example we use this to reason that the upper right corner must be blue or yellow, and that the lower left corner must be green or yellow. Who's to say they aren't both yellow?
I think the trick to #3 is to take 0 as one of the values.
If you have (0,a,b) is a solution, then the problem reduces to find a and b st 0, a, b and a+b are all the same colour.
Should be doable by assuming it doesn't exists and getting some contradiction. like the set with 0 has to be a singleton or something.
I tried something like this but got bored and thought there must be a cleaner way -- maybe the nice way will become apparent once it's all worked out though!
The problem statement specifies positive integers, so 0 is not included - starting with 0 was my first thought.
My next thought is to investigate the possibility that for some n, all multiples of n are the same colour, which is a much stronger result than required, and trivially implies the desired result. If there is no n such that all multiples of n are the same colour, then that gives you a lot of structure to work with.
I think that allowing a character to express a sentiment that many parents of autistic people share seems like it isn't that bad, at least unless the film presents it as actually true
the 72-gon problem is stated so badly
I understand why you took issue with the comments from Nathan's father, but I do think those words served an interesting purpose in representating the somewhat misguided views many hold
Damn, it's too bad I already knew I had autism, otherwise I would've gone to some psychologist and show them tbose clips and say "that's me".
/half-serious
Because I couldn't find the Mathologer video in the description or pinned comment, here is the link for y'all's convenience: th-cam.com/video/04x4ZdLpN-0/w-d-xo.html&ab_channel=Mathologer
Whoops! Totally my fault sorry, I'm fixing it now but thanks for reminding me!
I mean, you need to prove Catalans Identity to properly proove problem 1, no? Seems a bit "Yes, because Yes: look at this Truth here:" The only other thing you are really adding is the fact there are infinite Fib numbers
Results can be stated without proof in the IMO. In fact, it is expected that students learn many theorems as part of their preparation (common ones are Fermat's Little Theorem and Muirhead's Inequality).
I've never understood why anybody would expect a character in a book or movie, etc., to represent *all* people with whatever trait that have. They're one person. No matter how they are depicted, they can't possibly pull off such a representation. Unless a work has at least several characters with that trait, representation of the spectra within that trait is never going to happen. And frankly, a movie about a dozen autistic people would itself be a weird representation, since that is uncommon in almost any real life situation.
I don't think anyone expects a character to represent all people with that trait. What matters to me is that they are a fair representation and not an offensive stereotype or anything.
What a great case for having all types of people in most media you just made! You are correct, we can't expect a single movie to show us loads of different characters that fit a minority, so we should instead have loads of movies with one or two minority characters, and in all of those, give a different - yet realistic and not offensive - representation. this way there will be a character for everyone somewhere, it just doesn't have to be in a single one.
Cube
My last MAFS video was on Cube, check it out!
@@AnotherRoof Thanks, will do !
Have you checked Melancholia already?
The Fibonacci sequence starts with zero.
I knew someone would say this -- in any case, 0 is deemed the 0th term and 1 the 1st term, and so on. Thus the odd-positioned terms are still those to which I referred in the video. Hope that helps!
Is there a source for your spectral breakdown of autistic traits or is it your own idea? Either way it seems like a pretty convenient way to discuss autism, and much better than the "high vs low functioning" dialect 🌟
No there isn't and like I said in the video, I must stress that this isn't a psychologically rigourous way to look at autism, just an informal way to think about it.
I talk about it more in the Patreon video but essentially if you look at the diagnostic criteria (see the ICD-11 or DSM-V), you can more or less slot them into these categories. They also diagnose with three levels of autism, where level 1 is someone who requires minimal support and 3 is someone who requires substantial support -- this is another way of avoiding the "high functioning / low functioning" phrasing.
Hope that helps and glad you found it useful!
Problem 1 is exactly the type of problems which made me abandon competitive math in favour of more fruitful uses of my time.
They always boil down to guess it's this thing, pull this specific trick out of a hat, try a bunch of random😊 things and tadaa you solved a useless trivia you'll never touch again.
And in the rare case math is actually needed in a practical application you get approximations and iterative algorithms which invalidate 80% of math and produce a more accurate solution because the purely-theoretical one had to make N simplifying assumptions.
You point out a spesific problem that need a theorem or formula but in fact most of olympiad problem didn’t require a single theorem, even when it is, there is not many theorem to memorize for you to use. And math competition itself benefit you by making your logic and critical thinking sharper, its like solving infinite puzzle with infinite rules
I know what you mean, but once you spot the pattern in the Fibonnaci numbers, the result can be proved algebraically. I mentioned this off-hand but should have spoken more about it in the video I think!
I (think) someone could suppose that a or b is a prime, then with some number theory you could get to the answer.
(Im trying but the problem is pretty hard)
A guy in a group chat Im in just solved using basic number theory and Newton identities, so yes you dont need a really profound knowledge
Just once I'd like to see a movie with an autistic main character who doesn't talk like you fed ChatGPT the script for every Big Bang Theory episode
@@johnnyjoestar4473 Then maybe watch x+y :)
If you have a high res (1 mm^3 voxels) MRI of your brain, look at the ratio of grey matter volume over the white matter volume for both left and right hemispheres. Most people have a higher GM/WM ratio in the left hemisphere. I personally am even more left of the mean. Also look at the size of the cerebellum. Did you know you only use 42% of your brain? That's because (yes this is true) the so-called "small" brain, the cerebellum, has 58% of the neurons in the entire brain. They are granule cells, and they are very small, creating a crossbar that helps coordinate muscles. Or not, if it's smaller than normal. The new Mac M3 Max has more transistors (90 billion) than the human brain has neurons (86 billion) and you only use 36 billion neurons for thinking. Normals have an IQ of 100
9:40 They don't mention this in the film, because that's wrong. It's not random, it's arbitrary. And they don't mention that because they don't have to. They say how many of each colour there are, and nothing more. So you have no more information to go on. The use of the word "always" should remove any unclarity that's left.
Yeah wasn't really correcting the film just adding additional clarity. Hope that helps!
14:12 That’s trivial: a=b=c=0. Assuming a, b, and c have to be distinct, it’s harder.
a,b,c are POSITIVE integers. Positive means "greater than zero", so a,b,c > 0
@@WeReNeR-k2f Neither the screenshot in the corner nor the audio of this video specified “positive” or even “nonnegative”. Maybe the movie audio did but I have not seen the movie so don’t know.
@@danielrhouck the problem is stated on the desk exactly one second after the timestamp in your comment, but ig it can just be its wrong interpretation
@@WeReNeR-k2f Hmm, I see that, but it still does not match what was in the movie or what he said, so I would go with the screenshot instead of the blackboard
@@danielrhouck yeah, you're probably right and, according to the problem, presented in the movie, it's just "integers", so negative numbers aswell
❤❤❤🙏🏽🙏🏽🙏🏽
For problem 3, couldn't you just trivially pick a=b=c=0? The problem doesn't *technically* say that they have to be different.
Pointing out stupid edge cases like this is why I never got very far in competitive math
0 isn't a positive integer. Generally, I'd expect that a, b and c have to be different.
im sorry but i hate math so much
what are you even doing here
@@NStripleseven i slipped and fell into a manhole and broke a homeless man's neck and have been using his tv's TH-cam account since march 21st
Memo to TH-cam ... I boycott all products you advertise. Including Brilliant. Because knowledge is for free.
grammar alert! math is already plural and does not require an S.
I'm British.
Do I need calculator for the movie or not ?
I've worked out that each term in 1, 2, 5, 13, 34 … is
n = round(m * (ϕ + 1))
I don't know if it will hold true forever, but as the rounding amount is getting less each time, I can't see why it shouldn't. (Proof needed, though).
If it does hold, then there will be an infinite number of pairs.
(I'm only at 8'00", so this might be addressed later.)
Does not work for the pair (1,1), but I don't know if that's even being considered, as it would require m = n.
Its because fibonacci number scales approximately with phi, the more term you calculate, the better the approximations is. So every odd fibonacci number scales with phi^2, but we know that phi^2=phi+1 because phi is the solution to x^2-x-1=0. That is why you get that result, but i really doesn’t see why is this significant, because this is definitely not needed for the proof, you just have to proof every successive odd fibonacci sequence is a pair of solution.
@@rafiihsanalfathin9479 Yes, I soon as I heard Fibonacci sequence, I though there would be a relationship to ϕ. It's been literally decades since I've looked at Fibonacci and related sequences in any depth.
As for proving, I meant that I didn't have a proof, not that it should have to be done. I agree that the proof offered is sufficient.