The smallest such prime...
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- เผยแพร่เมื่อ 21 ต.ค. 2024
- We look at a nice number theory problem.
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For subcase 1 you didn't need to involve x at all because (p-1)^2
That's actually quite elegant.
Yeah, that makes sense.
If anyone is wondering: a=6083, b=78, p=23
I wish every math YTer would go back and show that their answer actually satisfies the initial problem, at least where applicable.
how did you find that out?
@@KiLLJoYTH-cam I was curious if it would take me long to write a shitty way of finding it in C, so I found it that way.
Nice!
Also, a=b^2-1
4:58 "See what I did there?" mp = Michael Penn?
🅱️ = Michael Penn
I thought it meant math professor
I always teach my students to use their names in the selection of their variables. I struggled to guess why this gymnast 🤸♂️ kept using m. It all makes sense to me now.
@@nelsblair2667 LMAO Gymnast 😂
Your presentation style is unbeatable. So enthralling, It’s usually you find a good solution but bad presentation, or the inverse, but rarely both.
That was a fun one, but the hint totally gave the game away. :-) So the problem statement is equivalent to p³ = b⁴ - a² = (b² - a)(b² + a). Since p is prime, p³ can only be split up two ways, p * p² and 1 * p³. And since a and b are positive integers, b² - a < b² + a, so the factors can be mapped to the terms in only two ways.
Case 1: b² - a = p, b² + a = p²
Add up both equations yields 2b² = p(1+p). The case p=2 can be discarded quickly, so p must be odd. Therefore 1+p is even, so we can divide the equation by 2 and get
b² = p (1+p)/2 (where the last term is an integer). Therefore b² is divisible by p. Since p is an integer, it follows that b is divisible by p. So there is some integer n with b = np, so
n² p² = p (1+p)/2
n² p = (1+p)/2.
So (1+p)/2 is divisible by p. But (1+p)/2 < p for all p >= 3, so that can't be right. So there can be no solutions in this case.
Case 2: b² - a = 1, b² + a = p³
Add up both equations yields 2b² = 1 + p³ --> b² = (1 + p³)/2. This again discounts p=2 as possibility, since in that case the RHS is not natural. I had no idea how to proceed from here, so I just wrote down a table of prime numbers, their cubes, the RHS of the equation, and their square roots. By checking every known prime number, I was able to find that p=23 leads to b=78, and a=6083. A quick check later confirmed this to be a solution. And since the other case found none, and this case found none before this one, the smallest solution is:
6083² + 23³ = 78⁴
Hvjvb
G6f
I did it this way as well, but before I watched the video, so I didn't have the hint to spoil it, although these sort of problems almost always involve unique factorisations of expressions.
Fortunately, I tried the b² - a = 1, b² + a = p³ case first and used an Excel spreadsheet to calculate b = SQRT((1+p^3)/2) against a column of primes, showing p=23 and b=78 as a solution. For the second case, I was then able to stop calculating p=(SQRT(8*b^2+1)+1)/2 once p grew larger than 23. I often "cheat" and use Excel for trial-and-error solutions as it's trivial to "knock up" a quick series of calculations, especially if you keep a text file with a column of primes to paste as your starting point.
Once one has p³ = 2b² − 1, a quick analysis modulo 8 yields p ≡ ±1 mod 8. Then one just needs to test p = 7, 17, then 23 works. That's much faster, but Michael's solution could have been faster if the minimum p were much larger.
Writing a short program and test primes till it finds a valid answer would be much faster.
@@happygimp0 One doesn't always have a computer or even a calculator, in particular for math Olympiads.
p = 2 doesn't work for case 2 because if b^2 + a = 8, and b^2 - a = 1, then 2a = 7. But a is an integer: contradiction.
Yes, it was premature to carry over this "fact" from the first case. It had to be proved separately for the second case.
@@davidblauyoutube he technically assumed it outside of both cases, left as an exercise to the viewer from the given equation at the start. tho he didnt mention it until it became relevant in case 1
If you just take those 2 equations as they are, and then eliminate a, then you get 2*b^2 = 9, which is a contradiction, since we are given that b is an integer.
The relatively prime case is easier. You get (p-1)^2
Yep!
National mathematics day. 22 dec.🇮🇳🇮🇳🇮🇳
Happy national mathematics day! :)
Nice problem!
So I made a spreadsheet of the possibilities as we found in the only subcase that works.
x, p=6x^2-1, y=sqrt((p^2-p+1)/3)
As it turns out, x=2 is the only x
In subcase 1, we need not consider x at all to see that y^2=p^2-p+1 implies (p-1)^2 < y^2 < p^2
14:30 *Michael Penn:* Tony, there was no other way. We're in the endgame now...
37,002,889+12,167=37,015,056 or 6,083^2+23^3=78^4
Thanks to the guy who invented internet. I get to watch and learn such cool stuff.
You eliminated p=2 only in respect to the first factorization, cannot bring that result into another case I think
..amazing ..so many layers of concepts revealed as Michael Penn went on..
Case 1 can be made easier by: since p | b, p² | b², but p² ∤ p(p+1), so we have a contradiction.
Also, I thought of finding all values of p that satisfy this equation and there may actually not be so many solutions. I tried to narrow down the possibilities of p and got this:
p + 1 = 6x²
p = 6x² - 1
3y² = p² - p + 1 = (36x⁴ - 12x² + 1) - (6x² - 1) + 1 = 36x⁴ - 18x² + 3
y^2 = 12x⁴ - 6x² + 1
Since y² is odd, y ≡ 1 (mod 8), thus x is even.
Let 2w = x, then p = 24w² - 1 (this means that p ≡ 23 (mod 24))
y^2 = 192w⁴ - 24w² + 1 (this polynomial cannot be factorised without going into complex numbers)
Additionally, since for any perfect square k^2 ≡ -1, 0, 1 (mod 5), w ≡ -1, 0, 1 (mod 5). (And in particular, y ≡ 1 (mod 100) if w ≡ 0 (mod 5) and y ≡ 69 (mod 100) if w ≡ ±1 (mod 5), though I'm not sure how helpful this is.) In fact, I plugged in values for w = 4, 5, 6, 10, 11 (9 gives a composite value for 24w² - 1) and none of them gave a perfect square for 192w⁴ - 24w² + 1.
National mathematics Day :)
Can mathematics be national?
@@Uni-Coder Yes it is in India
One of the best videos I have found!
Find all m,n natural number such that
(3^m) +23=2^n
Penn has become a Teller - magical problem!
13:29 I find it sort of confusing bounding y^2 in terms of x. Could have just written as (p-1)^2
I know right? Me too.
The way I discarded the first case where 2b² = p(p+1) you have b² = p(p+1)/2. Geometrically that can't work since a stick of length p that can't be broken down (because p can't be factored) to fit into a square of size b², because square root of b is necessarily smaller than p (geometric mean lies between p and (p+1)/2 ). There you go. Not as elegant as how you gave it, but it's what I came up with.
My first thought is p^3 = b^4 - a^2 = (b^2)^2 - a^2 = (b^2 - a)(b^2 + a). Examine small primes in order to find a solution. (b^2 -a) is obviously less than (b^2 + a) so it has to be either 1 or p. If we assume p is 2, we get b^2 - a = 1 or 2 and b^2 + a = 2 or 4. This gives 2 b^2 = 3 or 6, which has no solutions in Z. So p must be an odd prime. Just check all of the small ones until you get a hit.
p|b^2 => p|b, this is only true for prime p. Obviously 50|100 but 50 does not | 10.
Great problem!
nicely donee !! :) great job !
I need help
Suppose p>2 and a and b are coprime with p then
Rearrenging we get that p^3=(b^2)^2-a^2
By LTE we get that 3=v_p((b^2)^2-a^2)=v_p(b^2-a)+v_p(2). Since v_p(2) for p>2 is 0 then v_p(b^2-a)=3. In other words p^3|b^2-a and that b^2-a >= p^3
Since p^3=(b^2-a)(b^2+a)>=p^3(b^2+a) which means that b^2+a=1 but this is ridiculous.
What did I do wrong?
Lazy way: taking b^2-a=1, b^2+a=p^3, solve for b^2 = (p^3+1)/2. p=23 is the first prime that works, b=78, a=6083
Very nice problem involving lots of elementary number theory!
I was delighted to notice that for Case 1, we find that b^2 equals p(p+1)/2, which is 1+2+...+p (a.k.a. the triangle number of p). This harkens back to the balancing number video. It's easy to show that balancing numbers are the square roots of triangle numbers (though this wasn't mentioned in the video). Thus for Case 1, b has to be a balancing number! Now, it appears that the sequence of numbers whose triangle number is a perfect square (8, 49, 288, 1681, 9800, 57121, ...) contains only even numbers alternating with odd perfect squares -- i.e. no primes! Thus, Case 1 yields no solutions. For my next trick, I need to prove that last assertion...
This has led me to a surprising but easily provable observation: The difference of the squares of consecutive triangular numbers is the cube of the triangular root of the larger. In other words:
(1+...+n)^2 - (1+...+(n-1))^2 = n^3.
I suppose this is already well known to real number theorists. :-)
@@jasonwoolf It has been a year, but perhaps you want to look at Michael's sweatshirt.
@@themathhatter5290 Ha, yes, I’ve noticed it in many videos over the past year, but I missed it while watching and commenting on this one. I was an MP newbie back then.
I really enjoyed this problem, Michael, thank you!
I would like to suggest another solution, that doesn't need the use of the gcd's tricks.
Same procedure up to minute 06:55, so we know 2b²=p³ + 1.
Since p is odd, let's write p=2n-1. Then we have:
2b² = p³ + 1 = (8n³ - 12n² + 6n - 1) + 1 = 8n³ - 12n² + 6n
b² = 4n³ - 6n² + 3n = n × (4n² - 6n + 3)
Let's factorize n in prime numbers:
If c is a prime dividing n, then c also divides 4n² - 6n, so c divides (4n² - 6n + 3) ONLY IF c=3.
Any prime d≠3 dividing n IS NOT a factor of (4n² - 6n + 3), so d must have an even exponent in the factorization of n, else n×(4n² - 6n + 3) CANNOT be a perfect square.
In other words, the only prime number than can appear with an odd exponent in the factorization of n is the number 3.
So we must search for n such that 2n-1 is prime and:
1) n is a perfect square, OR
2) n is 3 times a perfect square
Then n is in the set {3, 4, 9, 12, 16, ...}
The first value for n in this set that works is n=12:
n = 3×2²
(4n² - 6n + 3) = 4×144 - 6×12 + 3 = 507 = 3×169 = 3×13²
b² = (3×2²) × (3×13²) = (2 × 3 × 13)²
That yields p = 2×12 - 1 = 23
We've got the smallest solution, but is it the only one? I found no other below 10^15 (Python is fast!). I suspect it's the only solution.
best video ever! i could actually see myself solving this myself
Wow this is something beautiful and incredible thank you for this video
This one required some patience, but nonetheless nice problem
p=23, OK, but what about a and b?
I feel no shame in admitting I wasn't able to solve this on my own.
Ok maybe a little.
But guess what I could solve it
(After seeing the solution)
Is the symbol Michael uses for "contradiction" a standard one? I've never seen it before. 🤔
No it isn't. The "standard" symbol is an upsidedown "T".
@@SuperMerlin100 I've seen two implication arrows pointing against each other used for "contradiction". Are there more in use that I've missed?🤔☺️
michael does it like this -->
@@djvalentedochp Really? Looks like a + with a strike-through to me. 🤔
That doesn’t add up (plus sign with strike through it, joke)
p=23 is the only prime that works for p < 1.000.000.
Yeh I noticed that too. Wonder how to prove its a unique answer.
Interesting.
So 6083^2 + 23^3 = 78^4. Yay science!
Thank you!
ok so (I'm omitting all the exponentiation symbols : p3 means p^3 etc.. but 2p still means 2*p.)
p3 = b4 - a2 = ( b2-a)(b2+a)
So either:
1) b2-a=1 and b2+a=p3
or 2) b2-a = p and b2+a = p2
Since b2+a == b2-a + 2a -->
For 1) we get that 1+2a = p3 so a = ( p3-1)/2 and b2 = (p3+1)/2
For 2) we get p + 2a = p2 --> a = (p2 - p) /2 = p (p-1) / 2 and b2 = p (p+1) / 2
Now, by successively trying for p=2,3,... ; I do get that the smallest solution for p is p=23, a=6083, b=78
I can get to that solution with just pen and paper but it is tedious. I used a spreadsheet and got the solution in minutes. Using a simple calculator would also get me the solution in just a few more minutes.
I don't really know how to reduce the amount of manual calcul to do to get to the solution.
Looks like my approach was basically the same as Michael's:
a²+p³=b⁴
p³=(b²+a)(b²-a)
Case 1:
b²+a=p²
b²-a=p
2b²=p(p+1) p would make RHS ≥ (p+1)p, so n+1=p, but then RHS = p(p-2)).
Case B: gcd = 3:
p+1=6m²
p=6m²-1
Try values of m:
p=5: no
p=23: yes, 23²-23+1=507=3.13²
p=23, b=78, a=6083
16:42 My comment was shadowbanned by TH-cam so let’s try this again...
Geometry homework today! Good luck and have a good day...
Squares are constructed externally on the sides of an arbitrary quadrilateral. Show that the line segments joining the centers of opposite squares lie on perpendicular lines and are of equal length.
Oh I see, it didn’t like the link. So the solution is at this adress : qbyte.org/puzzles/p062s.html
Thank u
Ahhh midpoint theorem. Brings me back to middle school days!
@@tonyhaddad1394 You're welcome
@@blazedinfernape886 Also known as van Aubel's Theorem: mathworld.wolfram.com/vanAubelsTheorem.html
We proved that IF the solution exists, thus it has to be 23. But we could have that the solution doesn't exist, right?
we are already given the fact that the solution exists in the problem
He always knows a good place to stop
any idea about any other possible solutions? Exhaustive search for x up to 10^8 yielded none.
Sounds like you chopped off some explanation at the beginning there!
4:58 mp michael penn
I ran a Python script to check all possible combinations of a, b, and p up to a certain b-value, and I only found one combination: p=23, a=6083, and b=78 when searching up to a b-value of 3000. The numbers are getting quite large and the script runs so slow at that point, even with all 16 threads running on my Ryzen 5800X.
that was amazing
How do you prove the GCD trick?
Prof Michael
Can you check this problem out
Define a positive integer $n$ to be a factorial tail if there is some positive integer $m$ such that the decimal representation of $m!$ ends with exactly $n$ zeroes. How many positive integers less than $1992$ are not factorial tails?
It's such a beautiful question :) also involves floor functions
🤔🤔 I am able to see the legendary legendre formula here
When I came to case 2 I just started to calc sqrt((p^3+1)/2) for small primes which gave me 23 as the smallest solution. I had more test cases but got the solution in less time.
btw: I as far as I can see 23 is also the ONLY prime number which solves this problem.
No solution besides 23 bellow 100 000 000 using python
Wait a minute at 2:25 why can't he switch the cases he forgot..you can have p divide the bigger term and p squared divide the b squared minus a..I dont see why not. Same with p cubed and 1 factors..
Essentially because p^2 > p and b^2+a > b^2-a, they need to match up if that makes sense. The two bigger terms have to be equal, and the two smaller terms have to be equal.
Lol why I think you should put your silver play button above the blackboard
@Jon Jukoba yeah lol
How is mathematics not entertaining?
I like to write my thoughts before seeing too many hints:
p^3=(b^2 - a)(b^2 +a)
so b^2-a must be 1 and b^2+a must be minimized. I imagine just picking the smallest square other than 1 to be equal to b^2 would work. b=2, a=3 so p=7
Edit: I see I just forgot the cubed ... so b^2-a does not have to be 1.
Sorry to everyone who has to read my bad ideas
what is the inspiration to look at the GCD of the 2 numbers?
If you ask why there should be small number of cases of outcome of GCD, then I have no clue.
But if you accept that you just try to factorize X=p^2-p+1 and Y=(p+1)/2 to further restrict number of possibilities, then from XY being perfect square you know that X=x^2*GCD(X,Y) and Y=y^2*GCD(X,Y) for x and y coprime natural numbers (i.e., x in N, y in N, GCD(x,y)=1). (You can imagine full factorization of X and Y and see how are the primes distributed.)
But you can as well stop at 2b^2 = p^3+1, rewrite it as b = sqrt( (p^3+1)/2 ) and start testing primes. p=23 is ninth prime, not being too far if you have no other clue how to restrict number of primes to test.
i'm convinced the 23 enigma is real
After I got 2*b^2=p^3+1, I just checked b and p mod 3,5,8. And had the following conditions:
p must be 1,5,7 mod(8)
p cannot be 1 mod(3)
p cannot be 2 mod(5)
the first prime that verifies these 3 conditions is p=23 which also verifies the equation.
How do you get the "p cannot" parts?
Perfect squares mod 3 are 0 and 1
So 2b^2 is 0,2 mod 3
=> p = 2b^2 - 1 is -1,1 or 1,2 mod 3
Perfect squares mod 5 are 0,1,4
So 2b^2 can be 0, 2, 3 mod 5
And thus p can be 1, 2, 4 mod 5
@@reeeeeplease1178 for mod 3, it was a typo, p cannot be 0 mod (3), so basically eliminates 3.
For mod 5, if p is 2 mod 5, then p^3+1 becomes 4 mod 5, which 2b^2 cannot be.
Still, with only mod 8 and mod 5 conditions, we can easily eliminate primes up to p=23.
@@riadsouissi ohhhh i missed the p^3
Good
Would it not be faster to write a script that tests different primes till it finds a solution?
def getPrimes(max=2000):
yield 2
r=3;
while r
We could also factorize p as p^(1 / 4 ) x p(3 /4) , can't we ?
But then we wouldnt have integers
Great!
Like it. Especially because the answer isn't 0,1 or 2.
9:48 isnt gcd of p+1 and 3 = 1?
No
8:30 I think this is the proof of the property used but is it correct?
Let gcd(a,b)=k and let gcd(a,b+an)=m where n is an integer
By definition of the gcd, we know k is the greatest natural number such that k|a and k|b simultaneously. Similarly, m is the greatest natural number to divide both a and b+an at the same time.
But if m|a, then m|b+an implies m|b as a*n is a multiple of m -and therefore m is the greatest† number to divide both a and b and so is their gcd and therefore m=k- QED
But I think just because m|a and m|b doesn't imply it is the _greatest_ number to do so and it might not be the gcd of a and b necessarily. Can someone please help me?
*EDIT* : The crossed-out part is the non-sequitur part of the original attempt at a proof
†We know both m and k divide a and b and k is the greatest possible number that can do so. So m
Search for gcd Euclid lemma
From what you did you can conclude that k|m, so now show that m|k and you are done.
An easy way to convince yourself of the symmetry of this argument (i.e. that it gives both m|k and k|m), may be to substitute c=b+na and run through the exact same logic
@@ichtusvis I see, we get b=c-an and so the roles of m and k are reversed here and since we saw one divides the other in the original comment and we get the opposite here and so m|k, k|m and therefore m=k
Thanks a lot for commenting :)
@@wasselkun5015 Your comment with the comment below yours helped a lot. Thanks a lot for commenting :)
Honestly one can just start putting in primes after we get the relatively simple criterion that (p^3+1)/2 has to be a square, since we're just looking for the smallest solution... not as elegant, but way faster
please someone explain this 14:17
Aren't there 2 more cases; b^2-a being p^2, b^2+a being p and b^2-a being p^3, b^2+a being 1?
No, because -a < a since a > 0, and b^2-a < b^2+a so p^x < p^y and x < y, where (x, y) is (1, 2) or (0, 3)
Hi everyone, I'm not sure if this is the place, but I've wanted to ask for advice. I like maths, and I really do find them enjoyable, through videos or pop-sci books, and I think I'd be pretty good at it; however, due to personal issues I can't study that (or related topics) in college right now. What are some books, videos, lectures to help me get started learning mathematics?
I'm not sure if I'm going to major in it, but I really do want to learn more of it!
P.S. Does anyone know about career options?
I think you should start with some basic algebra and elementary geometry, that got me hooked on math. And don't forget to practice arithmetic because it'll be useful especially when you're studying algebra. Give it a try, math is fun.
@@jomama3465 I finished highschool, and I think my knowledge is about the level of AP Calc AB + algebra 1 (on khan academy), if that helps somehow
@@yahav897 what are you doing now mate for other people and you the answer maybe to start with linear algebra (good books on linear algebra are linear algebra by insel friedberg and algebra by serge lang or algebra by artin) or you might try differential equations (math sorcerer has video lectures on it I am too lazy to provide a link sed) or you might wanna go for real analysis ( best books are mathematical principles or principles in mathematics I don't remember although it is very famous by the author Walter Rudin or understanding analysis by Stephen Abbott is also a good one ) or try number theory (best book is by david Burton)
MP = Michael Penn
Good!
Jesus, I did this thinking before the video, and thought I had to prove each prime. Glad to know I’m not the only one totally lost on that front. Also, this is my solution using the last digits of numbers as my contradictions.
The first step is the factoring then proving that bb-a = 1 not p
Since that would say p(p+1)=2bb which means p2M=2bb or pM=bb so p=b since m doesnt have p as factor since ita smaller than p, but p=b means p+1 = 2p
Ppp= bb + a And bb - a = 1
So ( ppp+1 )/2 = bb, a square
A square only ends in 014569
But primes only end in 1379
Then, this function shows that the only primes that can work are those ending in 13 or 9, and bb ends in 014
Ppp -1 )/2 = a
But bb - a = 1
Therefore a can only end in 903
The only primes that do this end in 0 or 3
This was tricky: if a ends in 0
Then p ends in 1
But then appp = ???10 = 2aa +a = ??00 + a
Therefore a ends in 10
But if a ends in 10, bb ends in 11
Since bb-a = exactly 1
But no number squared ends in 11
Proving that is confusing but generally the first two digits of square are created by {aa+20ab} where a and b are the digit of the ones and the tens places. There is no combination of the digits a and b that make that end in 11, since a must be 1 or 9, but that means 1+20b ends in 11, so 2b ends in 1, or 81+180b ends in 11 which brute force shows simply doesnt.
Therefore only primes ending in 3 will ever work.
There is no way to show what the next prime beyond 23 would be, but I tried: p is z10 +3
And putting that into ( ppp+1) /2 equals a square, perhaps you could show which values of z are possible and more importantly why.
Actually, it might be possible to prove that similar formulas have only one solution, and derive from that that this formula also has one solution. Cuz comments say its 1 in an eternity to find the next solution, implying its just 23.
Why would anyone think to break up p into p squared times p ..i dont see anyone thinking of that..
It's a common trick to think about prime factorisation of a power of prime (p^n) like this.
@@ahzong3544 well tricks are dirty and sneaky and shouldn't count..
@@leif1075 there is no dirty way to solve a puzzle, only a solution
@@leif1075 If you mean that tricks shouldn't be counted as a solution, then this statement does not seem to be true.
To be good at solving math olympiad problems, it is necessary to master certain commonly used tricks. These tricks can then help you do well in math olympiad.
I won't say tricks are dirty, since it doesn't violate the rules. They are something you can have in your mind to help you when solving math olympiad problems.
I hate this problem. Relentess hopeless handwaving that makes no sense. Just like the worst problems from school.
Which semester is this?
Beautiful ❤️
Micheal you are my drug 🤩🤩🤩🤩🤩🤩🤩🤩🤩🤩🤩🤩🤩🤩
Can anyone help mewith the GCD trick....can anyone help..plsssss
Let's say d=gcd((p+1)/2,p^2-p+1) and d'=gcd(p+1,p^2-p+1) for p an odd prime. Then d/(p+1)/2=>d/(p+1)=>d/d'. Because p^2-p+1 is odd, d' has to be odd, thus gcd(d',2)=1. So d'/(p+1)=>d'/2(p+1)/2 and gcd(d',2)=1, that means d'/(p+1)/2=>d'/d. So d=d' because they divide each other.
For the second trick we have that gcd(a,b)=gcd(a,b-ka), because d/a and d/b=>d/b-ka=>d/d' and d'/a=>d'/ka, so d'/ka and d'/b-ka=>d'/ka+b-ka=>d'/b=>d'/d, so again d=d'.
Are there infinitely many such primes?
Is there any other such prime? If my computer program is correct, then no number 3
that was not a good places to stop. You have tested the values in the equation.
So don t be late 🙏
33333rd view, lol
Bruh why am I wasting my early teenager days with calculus.
*with "arithmetic"
@@bourhinorc1421 *cries hysterically*
Then stop wasting
Integration and differentiation is interesting to a point, but once you realise that computers can solve these questions, integration and differentiation becomes less interesting as a mathematical pursuit. Real & complex analysis are of course of great interest.
First comment! Nice vid!
Bruh why am I wasting my early teenager days with calculus.
Are you?
@@xCorvus7x YES, I AM!
@@AnatoArchives
Is this a JoJo reference?