After sniffing that hagoromo chalk, Papa Flammy could have proven that there are infinitely many Mersenne primes, but this video is too short to contain it.
wrong! 12 is a sucessor of another funny number (one behind another one). in other languages (eg. portuguese) there's even more funny numbers: 60 = sessenta = cê senta = you sit 61 = cê senta em um = you sit in one 62 up to 69 also means "you sit in number" (of whatever you thinks about) 70 = setenta = cê tenta = you try. we also say 70 70, if it does not work, 70 again 101 (sento em um ) up to 199 means "I sit in one up to 99 whatever I want to sit on
In fact, this is true of all number one bellow a power of two (except for 2^0 - 1, which is 0): for all integer k superior or equal o 2, 2^k - 1 is written in binary as k ones. This obviously follows for Mersenne primes, which are therefore written as a prime number of ones in binary ;)
2:54 I tried factoring n³ - 1 as (n - 1)(n² + n + 1) I set n - 1 = 1 , and immediately we find n = 2 and n² + n + 1 = p, therefore n² + n + 1 = n³ - 1 Rearrange n³ - n² - n - 2 The only real root is n = 2 Substituting it back, we get n² + n + 1 = p (2)² + 2 + 1 = p p = 7 I also tried setting n - 1 = p and n² + n + 1 = 1 But the same steps lead to the negtive numbers so we leave this part out
If you don't mind checking it for me, which negative number does the second setting lead to? I'd check it myself, but I'm honestly a little lost in the steps and I know negatives aren't prime but I'm just a bit curious.
@@BadMathGavin Difference of cubes factorization: n^3 - 1 = (n - 1)(n^2 + n + 1) Inputs: n - 1 = p n^2 + n + 1 = 1 n^3 - 1 = p Evaluation: a) n = p + 1 (find n from n - 1 = p) b) n^3 - 1 = p = n - 1 (transitive property) c) p + 1 - 1 = (p + 1)^3 - 1 (substitute p + 1 for n) d) p + 1 - 1 = p^3 + 3p^2 + 3p + 1 - 1 (expand) e) p = p^3 + 3p^2 + 3p (simplify) f) 1 = p^2 + 3p + 3 (÷p) g) 0 = p^2 + p + 2 (-1) h) p = -1/2*1 ± √(1^2 - 4*1*2)/2*1 (quadratic formula) i) p = -1/2 ± i*√(7)/2 (simplify) Side note: If we take the result of b, and divide both sides of the equation by n, we find that n^2 = 1. Continuing in this manner, we find the following results: j) n = √(1) = ±1 (√) k) p + 1 = ±1 l) p = 0 p = -2 (-1) Conclusion : The only real solutions are p = -2 and p = 0.
@@RedFoxtail26 p=n³-1 we know p is prime and all primes, except 2, are odd. Thus, n³ is odd. If n³ is odd, then p is even (because of -1). Which is false, because p would have to be 2. So, n=2 and p=7.
@@alissonp.b.freitas2538 -2 and 0 are not prime. This fact alone is enough to explain why they are not valid values for p, but congrats on your alternate proof!
if k is a composite, then k = ab where a,b >1. then 2^(k) - 1= 2^(ab) -1 = (2^a)^b -1 = (2^a-1)((2^a)^(b-1)+...+1) Since a>1, 2^a>2, 2^a-1>1. Thus, 2^k-1 is composite
Or you can write (2^k)-1 in binary. Then you get a string of k ones. If k=m*n, it is easy to see that this number must be divisible by the binary number that is a string of m ones. Example: 111111=111*1001 (and also: 111111=11*10101)
@@PapaFlammy69 And I also have never seen the difference of squares/cubes etc. formula derived from the formula for the k-th geometric partial sum. Really sweet connection.
I'm amazed that a mathematician, who should be totally precise in everything, writes the letter k as if it is a random squiggle, or a bent piece of wire which has been mangled by a hydraulic press.
The best mathematicians that I've met draw and write like complete shit They real model of what they're describing is in their minds, they only write to re-state not because they need to read it afterwards
@@grmancool Thank you. I am blind and just finished my last undergrad maths exam. I model and write almost everything in my head, but still use a blackboard from muscle memory literally just to solidify the flow of ideas. I have basically no hope of reading it afterwards.
This made us think: Could 7 be the only prime that is the difference between two cubes? We put on the coffee. We started our quest with a thorough search, sifting patiently through the endless ranks of natural numbers for a counter-example. 27 - 8 = 19. Doh, 19, our arch-enemy, the most prime number below 20!
Here's something though: x^3-y^3=(x-y)(x^2+xy+y^2) (Difference of cubes formula) Thus, it's only prime if x=succ(y) (which affirms 7 as the only 1 one less than a cube, as succ(1)=2), or (x^2+xy+y^2)=1, which is possible if y
ok before I watch this I'm gonna try to prove it myself and see if he does it the same way: let the prime be p and the cube be n^3, so p + 1 = n^3 thus p = n^3 - 1 and therefore p = (n - 1)(n^2 + n + 1) since p is prime and its two factors (1 and itself) are (n - 1) and (n^2 + n + 1), we have n - 1 = 1 and n^2 + n + 1 = p therefore n = 2 and the only possible value of p is 2^2 + 2 + 1 = 7 edit: kind of skipped the whole n^(k+1) - 1 step as I'd forgotten about that general formula, however I remembered how to factorise n^3 - 1 as it's a common special case that I was taught
I guess you could use the factor theorem to check that n - 1 is a factor of n^3 - 1 so I guess even if you don't know it by heart, it's something you could work out. Then you just work out the quadratic term.
If I'm not mistaken, the same is true with 3 in the case of squares. To express an integer that comes before a square, you right n²-1. Since one is a square number, this equation is the difference of 2 squares, and can thus be written as (n+1)(n-1). For any number n>2, the above factorization will give at least 2 different factors greater than 1. For n=2, you get the exception, which is 3, while n=1 gets you 0, which is neither prime nor composite.
I just factorised n^3 - 1 as (n^2+n+1)(n-1) so that means n-1 = +1( if it was -1 we would get -9 which isn’t prime) because it is a int multiplied by another int so for that to be prime 1 of them needs to = 1 or else the answer will be divisible by itself, 1, the first int and the last int and for n-1 =1 we get n=2 where 2^3-1=7
did he say 51 was divisible by 7? Confusion beside, this was a very fun and entertaining exercise that made me feel better about going to uni this year to be a math major
7 is the only prime before a cube as n^3-1 can be factored into (n-1)(n^2+n+1). Since if nn=2 or n^2+n+1=1=>n^2+n=0=>n=0 or n=-1, which is an invalid solution. for n=2, we get 2^3-1=7
I found this same result when playing with primes in different bases: the number 100…0 (all 0’s in the middle there) in base n is n^k, where k is the number of 0’s following the initial 1 digit. If you then subtract 1, it is trivial to see that this number must be a factor of n-1 (bc the digits in base n is just n-1 repeated k times). Thus this can only be a prime if n-1=1, and thus n=2 (with obvious exception of the k=1 case).
9:50 : You could have proved this! You don't need binary expansions or anything; you just need the factorization rule that we've already covered. Specifically, if k is composite (so k = ab where a, b > 1), then 2^k - 1 = 2^(ab) - 1 = (2^a)^b - 1 = (2^a - 1)(1 + … + (2^a)^(b-1)). And neither factor here is 1 (because 2^a is neither 2 nor 0), so 2^k - 1 is composite whenever k is.
Here's a cool fact about cubes I just found, but can't find anywhere else worth posting it. For any base n where n-1 is a perfect square, the √(n-1)ᵗʰ doubleback number is (√(n-1))³ . For example in decimal, where 10-1=9=3², the doubleback sequence goes 09 18 27 36 45 54 63 72 81 90, the 3rd of which is 27 or 3³.
7:40 : Another possibility here is that k = 1, then the second factor can be 1 without having n = 0. in fact, when k = 1, the second factor is always just 1, for any n. (Of course, as with the case when the first factor is 1, not every possible n will work when k is one, only the successors of prime numbers.)
This can be solved with easy algebra. Supose a cube (x+1)^3 where x > 1. Writing it out gives [ x^3 + 3x^2 + 3x + 1 ] Subtracting 1 gives [ x^3 + 3x^2 + 3x ] which is divisible by x. [ x * ( x^2 + 3x + 3 )] and since x isn't 1 and the division product isn't 1 it contradicts any number above the cube of 2 to be prime. Similarly [ (x+1)^n - 1 ] will always be divisible by x, since [ (x+1)^n ] will always be some coagulation of a*x^b powers +1. (I can see logically how this is true but I don't know how to prove it for real, I'm not a mathematician or mathwiz, I'm just happy I managed one on my own :'). Love to see if people have comments or if i made mistakes)
I remember I once was able to sneak Mersenne primes into a proof (can't remember the statement) and when my teacher asked why I basically said "beacuse I can"
At 10:34 the contrapositive is really: if 2^k-1 is prime, then k is not equal to ab for a,b natural numbers (that is, k is prime) Remember that (A implies B) iff (not B implies not A)
IMO it's more fun to look at cyclotomic polynomials than just using 2 as the base. n^k - 1 is always divisible by n - 1, which is redundant for n = 2, and if k is composite, there are more factors. Every cyclotomic polynomial should be able to generate primes (possibly infinitely many). The 24th one, n^8 - n^4 + 1, finds quite a few primes as the first two primes that could ever divide it are 73 and 97. Just as fun, cyclotomic polynomials are great for predicting lengths of recurring periods of various fractions in base n. There are a couple of subtleties to watch out for.
@@danielwimmer4698 There are (probably) infinitely many prime-rich polynomials. The reason I was discussing cyclotomic polynomials was their connection to recurring fractions. If you just want to find primes, there are a lot of superior options.
Is there a counterexample to the statement? " If p is a Mersenne prime, then (2^p)-1 is prime " This would imply that there are infinitely many Mersenne primes
I really enjoyed seeing the derivation of a³-b³ from the geometric summation; i have always used Binomial expansion for this identity. It was really cool to see the intuition of Mersenne primes as well?
For those like me who thought papa was just being lazy at 9:42 and can't find the answer, here's how I did it. Consider the following sum, which is evidently an natural number for p, q, two natural numbers: \sum_{k=0}^{q-1} 2^{pk} = (2^pq - 1)/(2^p - 1) Hence we've proved 2^pq -1 is not prime, for any p, q, different than 1.
Let n be positive integer, so n^3 is a cube, so n^3-1 is a number before a cube. Which is also equal to (n-1)(n^2+n+1) which is divisible by at least two primes for n greater than 2. For n=2 we have the case of 7 and 8. For n=1, we have 0 and 1. Thus 7 is the only prime before a cube.
2^11-1=23*89. So 2^k-1=p, if p is prime, then k must be prime. Not the other way around. I think that is what he was trying to clarify in the onscreen text, but I had to find the first counter example.
n^3-1=p for some positive integer p. (n-1)(n^2+n+1)=p. One of these factors must be one and the factor on the right is always at least 3 and n-1=1 when n=2. (2)^2+2+1=7 QED.
You can also let n=-1, yielding p=-2, which I claim is prime; it's not a unit, and, if it divides a product, then it divides at least one of the factors.
I saw this, and had to figure it out. It was satisfyingly simple. p + 1 = n³ => p = n³ - 1 = (n - 1)(n² + n + 1) `p` cannot equal a product of two integers, unless one is one and the other is `p` n = 2 & p = 7 | n = p + 1 & n² + n + 1 = 1 The first case is what was promised, and the second requires -1 to be prime. Edit: also, in general for p + 1 = n^k, p = (n - 1)(sum{i=0; n = 2 | n = p + 1 => p = 2^k - 1 | p + 1 = (p + 1)^k So `p` is either a Mersenne prime, or `k` is 1
Let p be a prime and n an integer So we prove that: p+1=n^3 only for (n=2;p=7) p=n^3-1 p=(n-1)(n^2+n+1) Since p is a prime & n-1 < n^2+n+1 We get the system: n-1=1 ; n^2+n+1=p (by definition) So by replacing we conclude that: The only sol'n are (n=2;p=7)
You can prove it in a much simpler way. All primes except 2 are odd numbers. 2 is followed by 3, which isn't a k-th power of a natural number since it is a prime. Therefore, consider the odd primes. Odd primes are always followed by an even number, let's call it q. Since q is even, it must be divisible by 2. We assume q is of the form n^k, therefore all its factors must be the same and it follows that q = 2^k. The number preceding it is p = 2^k - 1, which is possibly prime. This means that for any given k, there is at most 1 prime that is followed by a k-th power of a natural number and this natural number must be 2.
I know it's a little random and off-topic but at 4:35 is it proper to have the + on both sides of the ellipsis? It's something that I get confused about and often end up leaving the 2nd one out. I know it probably doesn't *really* matter and could just be up to personal preference but I just thought I'd ask anyway.
Okay, here I did this monster of a comment for fun. I know it gets out of context sorry if it bothers anyone. Theorem: For a, b are Z+>1 and k=ab, (2^k)-1 has factor(s) (2^a)-1 and (2^b)-1 and so has >1 factor >1. Proof: Without loss of generality consider only a, (2^ab)-1=((2^a)^b)-1. (2^k)-1=Sum(i=0 to k-1, 2^i) substituting we have (2^ab)-1=Sum(i=0 to ab-1, 2^i)=((2^a)^b)-1=Sum(i=0 to b-1, ((2^a)-1)((2^a)^i))=((2^a)-1)(Sum(i=0 to b-1, (2^a)^i) thus showing that (2^a)-1 is a factor of (2^ab)-1 and of course the other factor(s) here are also >1. It's been awhile since I proved this myself after first learning about the exponent for Mersenne primes being prime, so I've forgotten how I did it, but I'm pretty sure the method was different. That being said in the spirit of full disclosure I did glance over a few of the other comments in an attempt to refresh my memory (after searching in vain for my original proof). ■ 2 proofs of a generalization of the sum used above: Theorem: Sum(k=0 to n, a^k)=((a^(n+1))-1)/a-1, a>1. Proof 1: Consider that a^(k+1)-a^k=(a-1)(a^k) and so (a-1)(a^k)+a^k=a^(k+1), then Sum(k=0 to n, a^k)+Sum(k=0 to n, (a-1)(a^k))=Sum(k=1 to n+1, a^k), rearrange the equation, Sum(k=1 to n+1, a^k)-Sum(k=0 to n, a^k)=Sum(k=0 to n, (a-1)(a^k)) and the difference of the left side is (a^(n+1))-1, then after factoring (a^(n+1))-1=(a-1)(Sum(k=0 to n, a^k), divide, Sum(k=0 to n, a^k)=((a^(n+1))-1)/a-1. ■ Proof 2: Basis for induction: Sum(k=0 to 0, a^k)=(x-1)/(x-1) Induction hypothesis: If Sum(k=0 to n, a^k)=(a^(n+1)-1)/a-1, then Sum(k=0 to n+1, a^k)=(a^(n+2)-1)/a-1. Induction Step: (a^(n+1)-1)/(a-1)+a^(n+1)=(a^(n+1)-1)/(a-1)+(a-1)(a^(n+1)/a-1=(a^(n+1)-1+a^(n+2)-a^(n+1))/a-1=(a^(n+2)-1)/a-1. ■ Connection to geometric progressions: a(Sum(k=0 to n, a^k)+1=Sum(k=0 to n+1, a^k). Proof: a(Sum(k=0 to n, a^k)=Sum(k=1 to n+1, a^k) and a^0=1. Note: this only holds for initial values of that form. For a general solution we need to expand deeper (this comment is already much too long). ■ Edit: Removed some minor variable mishaps I noticed after the heart and I shouldn't have missed the first time through. Specifically, I habitually replaced k with x in the first proof, I replaced the final term n with the index variable in the statement of the second theorem, and I wrote n instead of n+1 for the exponent of a in the numerator of the closed form for the 1st proof of the 2nd theorem. That being said it wasn't that bad for like 3 in the morning or whatever it was. Also, I got Covid early unfortunately. But unlike many others I'm young and have a strong immune system so it wasn't bad at all for me. R.I.P victims of Covid-19.
I'm kinda confused about you saying the pattern doesn't hold for k = 4, that's not what we were doing in the first place, right? Like the only thing that's saying is that for any power k, the only prime below any number to the power of k if one exists is 2^k - 1, it's not saying that it has to be a prime, or did I just misunderstand the whole thing?
Assume that there exists a set of numbers that do not have unique traits. This would imply there is a smallest number without a unique trait. But *the fact this number is the smallest number without a unique trait* is a unique trait in itself, of which is belonging to that number. This is a contradiction, as we said this number does not have any unique traits. Hence, all numbers must have unique traits. So the probability that any number has a unique trait is 1, so your axiom is wrong.
@@CAG2 @CAG2 Ah I see we made the same joke 🤭 However I have a few interesting comments to make about your answer. First, what is the probability you're using on the set of naturel integers ? This is important to mention as there is no canonic or general one, and even finitely additive probabilities on N turn out to be very unpleasant. Then something is not right about your assumption. I believe your intention was to say "assume that the numbers which do not have a unique trait form a nonempty set", I agree that from this assumption follows a contradiction, and this it must be false. But what is the negation of "the numbers which do not have a unique trait form a nonempty set" ? At first sight it would appear that the negation is "the numbers which do not have a unique trait form the empty set" or in other words, there are no such numbers. Actually this is not true, in your assumption you say two important things instead of one : 1/ you say that the numbers which do not have a unique trait do not form the empty set 2/ you say that the numbers which do not have a unique trait form a set This second point is not trivial at all ! In mathematics, not anything that contains elements and can be named is a set. So it could possibly be the case that those numbers just do not form a set, in which case you can't prove that there is a smallest such number. I find that very interesting that there is a mathematical theory, called IST, whose goal is to interpret what it means for a mathematical object to have "a unique trait" (such objects are called standard in the theory), and one of the first results of that theory is that not every integer is standard ! However you still can't infere that there is a smallest non-stantard integer, simply because standard integers do not form a set (and you can't consider the measure of the standard elements because only sets can be measured)
@@swenji9113 This sounded fascinating so I took a moment to look up Internal Set Theory and I have a few questions. First, how is the existence of non-standard numbers justified (it's not obvious that any exist at all)? Second, how do non-standard numbers actually relate to a concept of having unique traits? I only glanced at the idea but it seemed to me that non-standard numbers are defined around the concept of accessibility rather than true uniqueness (is it that unique traits, by making numbers more distinguishable, also make them more accessible/possible to isolate)? And finally third, how does this theory relate to ZFC? Wikipedia says that it is a "syntactic enrichment" and that it adds additional axioms that are consistent to ZFC but this does still mean that it is distinct from ZFC alone? Also even though CAG2 mentioned probability I don't think their argument is actually probabilistic at all. I think they were just trying to find a way to articulate the concept that every positive integer must have some unique trait even if it is only that it is the smallest element with a unique trait. In fact, it seems that every integer should have several unique traits by virtue of their distinguishability and that non-standard numbers (albeit unique) are not accessible with some finite amount of resources? Also, having seen this before, I think Kaushik Raj meant that smaller numbers are less likely to have many traits that are not unique rather than more likely to many traits that are unique. My intuition tells me that larger numbers should almost always have more unique traits (and many more non-unique traits overall). But that small numbers will succeed at having comparatively few non-unique traits.
@@danielwimmer4698 [2/3] Now your other two questions are linked to the 3 axiom schemes which rule the behaviour of s in IST, and to their interpretation (at least my interpretation of what it means to be standard). When I said that standard numbers were numbers which are defined by a unique trait, I cheated a little by not defining precisely what would be a unique trait. I'll try with the example of Peano arithmetic. We denote the successor map by f, since s is already taken. The question is : which positive integers have a unique trait that we can express in Peano arithmetic ? Let's start with 0. It has the unique trait that it is the only integer which is not the successor of any integer. So 0 indeed would be called "standard". Then what about 1 ? Well since 1 is the successor of 0, it has the unique trait that it is the successor of an integer which is the successor of no integer. So 1 does also have a unique trait. In fact, once you know an integer n which does have a unique trait, any integer that could be expressed uniquely from n does have the unique trait of being the only integer expressible in this way from the only integer staisfying the unique trait of n. From this observation arises the notion of standard number. One of the 3 axiom schemes in IST, called the scheme of transfer, states that for any property written only with € and = and talking about standard objects only, there exists an object satisfying this property if and only if there exists a standard object satisfying this property ! That's where the unique traits meet the standard notion. Suppose that an object x has a unique trait (by a unique trait I mean that x is the only object satisfying a certain property expressed only with € and = and with other standard elements). We know there exists an object satisfying this unique trait, namely x, thus we can apply the axiom above and we get that there exists a standard object satisfying the unique trait of x. Since only x satisfies this trait, x must be a standard object. Conversely, if x is standard, then x has a very simple unique trait : that of being equal to x. Of course since we have to define properly what it is to have a unique trait, I took the liberty of defining it so that it would be equivalent to being standard. But historically the notion of standard was defined in order to get a proper definition of was it is to be uniquely determined by its own properties, so you will have to accept that this is what I call "having a unique trait" if you want to link IST to unique traits
Sir I have observed that : digital root/ digital sum (sum of the digits ) of every prime except 3 is never 3 or 6 or 9 .That is digital root of every prime except 3 is either 1 or 2 or 4 or 8 or 7 ( 1 + 6 ) or 5 ( 3 + 2 ) .Is it a mere consequence or some mathematics behind it ? Plz explain. With Ragards . Dr Rahul Rohtak .//
There is an error on the board introduced at 11:17 It should read: 2^(k-1) is prime implies k is prime. The converse is false: k is prime does not imply 2^(k-1) is prime.
Hey papa flammy, nice video. BTW I am stuck with an integral Integral from 0 to a cos(pi n x/a)/(x²+a²)dx where n is an integer. Please help me, make a video on it, as long time no integral video.
This insight suddenly dawned on me: 1. For every Mersenne prime M = 2^p - 1, we can state: M is the only prime followed by a p:th power. 2. For any integer k > 1, where 2^k - 1 is not a prime, we can state: There is no prime that is followed by a k:th power. For a formal proof of these statements, just take all the comments for this video and remove the goofy ones. I just love how you can gain deep mathematical insights by just goofing around.
mathematicians explaining why 7 is the only prime followed by a cube: th-cam.com/video/mdquYEw36TU/w-d-xo.html seriously, i love how fast you're able to speak during these lessons
@@rishikaushik8307 You're right, what's written on the board implies your statement. There is a mistake on the board there, the second line should be «2^k-1 is prime => k is prime». This provides a necessary condition for a number to be a Mersenne prime yet it is not a sufficient one
At 8:50 , you can also show that for all even k, 2ᵏ = 1 (mod3) which implies 2ᵏ-1 = 0 (mod3). This will show that only odd k values will produce mersenne primes.
I saw this 7 being one from 8. I then wondered about 27 and the closest prime is 23, but that is four from the cube. I am noticing a pattern. I then test 4 cube. 55 is 9 from 64 but not prime. 5 cube I find 109 from 125. I am seeing a parabola that seems to agree with the 7 to 2 cube affect.
After sniffing that hagoromo chalk, Papa Flammy could have proven that there are infinitely many Mersenne primes, but this video is too short to contain it.
exactly!
He could even have proven that there are infinitely many perfect numbers.
@@RecursiveTriforce the smell is actually just the value of the lowest odd perfect number, concentrated and disguised as chalk
3+4=7
:^)
@@PapaFlammy69 can you Please teach calculus from basic to complex or more it will help
57 is the only prime between 42 and 69
and it ate nine if I remember correctly
7 is also the closest prime to the square root of 69
8 is the only cube that comes after a prime.
Beautiful. You should make a video about this!
I just proved it!
8 is the only cube that comes before a cube
@@isaacgates5859 9 is a square
@@attenonmj3708 yeah I'm also a dumbass who just came back from working a double, I'm going to bed now
2 is the only prime number followed by another prime.
Ez
Omg mind blown
3 is the only prime number preceded by a prime number.
2 is only number which satisfies 2x=4, x^2=4, x^^2=4, x^x=4
2 is the only number satisfying x+x=x*x=x^x=...
70 is the only number after the funny number. Idk why it is relevant but eh
And is the first weird number.
*sad 421 noises*
wrong!
12 is a sucessor of another funny number (one behind another one).
in other languages (eg. portuguese) there's even more funny numbers:
60 = sessenta = cê senta = you sit
61 = cê senta em um = you sit in one
62 up to 69 also means "you sit in number" (of whatever you thinks about)
70 = setenta = cê tenta = you try.
we also say 70 70, if it does not work, 70 again
101 (sento em um ) up to 199 means "I sit in one up to 99 whatever I want to sit on
421? 69,421? 22?
@@valeriewithsalt 421 comes after another funny number, the funny number being 420. As in "420 blaze it".
Another fun fact about Mersenne primes - written in binary, they all have form of 11, 111, 11111, 1111111, 1111111111111 and so on. No zeros.
In fact, this is true of all number one bellow a power of two (except for 2^0 - 1, which is 0): for all integer k superior or equal o 2, 2^k - 1 is written in binary as k ones. This obviously follows for Mersenne primes, which are therefore written as a prime number of ones in binary ;)
@@不眠社畜 and? still a cool little fun fact.
That's why Mersenne Primes are so valuable. They're instrumental to modern encryption software.
That is fun indeed!
@@jonathanshapiro6593 and it makes it less interesting.
2:54
I tried factoring n³ - 1 as
(n - 1)(n² + n + 1)
I set n - 1 = 1 , and immediately we find n = 2
and n² + n + 1 = p, therefore
n² + n + 1 = n³ - 1
Rearrange
n³ - n² - n - 2
The only real root is n = 2
Substituting it back, we get
n² + n + 1 = p
(2)² + 2 + 1 = p
p = 7
I also tried setting
n - 1 = p and
n² + n + 1 = 1
But the same steps lead to the negtive numbers so we leave this part out
After you find n=2, it definitely hold true because (n-1)(n²+n+1)=n³-1 and n-1=1 so n²+n+1=n³-1, you don't need to test it again
If you don't mind checking it for me, which negative number does the second setting lead to? I'd check it myself, but I'm honestly a little lost in the steps and I know negatives aren't prime but I'm just a bit curious.
@@BadMathGavin
Difference of cubes factorization:
n^3 - 1 = (n - 1)(n^2 + n + 1)
Inputs:
n - 1 = p n^2 + n + 1 = 1 n^3 - 1 = p
Evaluation:
a) n = p + 1 (find n from n - 1 = p)
b) n^3 - 1 = p = n - 1 (transitive property)
c) p + 1 - 1 = (p + 1)^3 - 1 (substitute p + 1 for n)
d) p + 1 - 1 = p^3 + 3p^2 + 3p + 1 - 1 (expand)
e) p = p^3 + 3p^2 + 3p (simplify)
f) 1 = p^2 + 3p + 3 (÷p)
g) 0 = p^2 + p + 2 (-1)
h) p = -1/2*1 ± √(1^2 - 4*1*2)/2*1 (quadratic formula)
i) p = -1/2 ± i*√(7)/2 (simplify)
Side note:
If we take the result of b, and divide both sides of the equation by n, we find that n^2 = 1.
Continuing in this manner, we find the following results:
j) n = √(1) = ±1 (√)
k) p + 1 = ±1
l) p = 0 p = -2 (-1)
Conclusion :
The only real solutions are p = -2 and p = 0.
@@RedFoxtail26 p=n³-1
we know p is prime and all primes, except 2, are odd.
Thus, n³ is odd. If n³ is odd, then p is even (because of -1). Which is false, because p would have to be 2. So, n=2 and p=7.
@@alissonp.b.freitas2538 -2 and 0 are not prime. This fact alone is enough to explain why they are not valid values for p, but congrats on your alternate proof!
if k is a composite, then k = ab where a,b >1.
then 2^(k) - 1= 2^(ab) -1 = (2^a)^b -1 = (2^a-1)((2^a)^(b-1)+...+1)
Since a>1, 2^a>2, 2^a-1>1. Thus, 2^k-1 is composite
How do you get to (2^a-1)((2^a)^(b-1)+...+1)?
@@mariocervantes6163 using the x^n formula in the video
@@mariocervantes6163 for any k, k^b-1 = (k-1)(k^(b-1)+...+1). Now sub 2^a into k, u get (2^a)^b -1 = (2^a-1)((2^a)^(b-1)+...+1)
Or you can write (2^k)-1 in binary. Then you get a string of k ones. If k=m*n, it is easy to see that this number must be divisible by the binary number that is a string of m ones. Example: 111111=111*1001 (and also: 111111=11*10101)
beatiful
I never considered Mersenne Primes from this angle, really interesting!
=)
@@PapaFlammy69 And I also have never seen the difference of squares/cubes etc. formula derived from the formula for the k-th geometric partial sum. Really sweet connection.
glad you liked the insights! :3
@@SeeTv. same, i've only seen it using synthetic divison
Did you get the package of wet hamster fur I sent you?
What the fuck
bruh
Papa Flammy is officially in Math Jail. The package was confiscated immediately.
Yes
Can we start a petition to make 51 a prime number, please?
yes, pls
17 begone
It is a prime number in the octal base
I like the octal base because then I can write 3^3=33
@@luggepytt oh thats a brain moment
I'm amazed that a mathematician, who should be totally precise in everything, writes the letter k as if it is a random squiggle, or a bent piece of wire which has been mangled by a hydraulic press.
The best mathematicians that I've met draw and write like complete shit
They real model of what they're describing is in their minds, they only write to re-state not because they need to read it afterwards
@@grmancool Thank you. I am blind and just finished my last undergrad maths exam. I model and write almost everything in my head, but still use a blackboard from muscle memory literally just to solidify the flow of ideas. I have basically no hope of reading it afterwards.
The greeting is getting so shortened that it's become just a single sound
xD
And I like it the best
This made us think:
Could 7 be the only prime that is the difference between two cubes?
We put on the coffee.
We started our quest with a thorough search, sifting patiently through the endless ranks of natural numbers for a counter-example.
27 - 8 = 19.
Doh, 19, our arch-enemy, the most prime number below 20!
thus ends the Gauteamus difference of cubes saga
I don't think 19 is the greatest prime less than 20!
Certainly the greatest prime less than 20.
@@connorhorman ha! Classic :-)
Here's something though:
x^3-y^3=(x-y)(x^2+xy+y^2) (Difference of cubes formula)
Thus, it's only prime if x=succ(y) (which affirms 7 as the only 1 one less than a cube, as succ(1)=2), or (x^2+xy+y^2)=1, which is possible if y
Interestingly, this implies that 19 is also the only prime number thats the difference of cubes of 2 primes.
ok before I watch this I'm gonna try to prove it myself and see if he does it the same way:
let the prime be p and the cube be n^3, so p + 1 = n^3
thus p = n^3 - 1 and therefore p = (n - 1)(n^2 + n + 1)
since p is prime and its two factors (1 and itself) are (n - 1) and (n^2 + n + 1), we have n - 1 = 1 and n^2 + n + 1 = p
therefore n = 2 and the only possible value of p is 2^2 + 2 + 1 = 7
edit: kind of skipped the whole n^(k+1) - 1 step as I'd forgotten about that general formula, however I remembered how to factorise n^3 - 1 as it's a common special case that I was taught
you forgot the case n-1=p and n^2 + n + 1=1 which is clearly false but still deserves a check
I did it exactly the same way too.
I guess you could use the factor theorem to check that n - 1 is a factor of n^3 - 1 so I guess even if you don't know it by heart, it's something you could work out. Then you just work out the quadratic term.
If I'm not mistaken, the same is true with 3 in the case of squares. To express an integer that comes before a square, you right n²-1. Since one is a square number, this equation is the difference of 2 squares, and can thus be written as (n+1)(n-1). For any number n>2, the above factorization will give at least 2 different factors greater than 1. For n=2, you get the exception, which is 3, while n=1 gets you 0, which is neither prime nor composite.
It's always nice seeing you in new videos. Awesome!
That's a crazy way to introduce the concept of Mersenne primes!
There's 51 of them we know of so far!
yas :)
Me: *factors n^3-1 and confirms claim*
Flammable Boi: *17 minute video*
ez
His "m" just transformed to a single line 😂
I just factorised n^3 - 1 as (n^2+n+1)(n-1) so that means n-1 = +1( if it was -1 we would get -9 which isn’t prime) because it is a int multiplied by another int so for that to be prime 1 of them needs to = 1 or else the answer will be divisible by itself, 1, the first int and the last int and for n-1 =1 we get n=2 where 2^3-1=7
5:54 Papa being like
-1 is not. A Good. PRIME. N U M B E R.
xD
nae gud!
Bad primes, bad primes, whatcha gonna do? Whatcha gonna do when they factor you?
did he say 51 was divisible by 7? Confusion beside, this was a very fun and entertaining exercise that made me feel better about going to uni this year to be a math major
Meant to say 17 lol
@@PapaFlammy69 even math teachers are prone to 51/17
Well well how the turntables
@@PapaFlammy69 thank you for the clarification sir. Keep up the good work.
I'm a grad student in number theory and I didn't know this was true. Thanks! I'm really glad there are good math channels like yours these days.
7 is the only prime before a cube as n^3-1 can be factored into (n-1)(n^2+n+1). Since if nn=2 or n^2+n+1=1=>n^2+n=0=>n=0 or n=-1, which is an invalid solution. for n=2, we get 2^3-1=7
"-1 does not make a good prime number"
factorisation in the integers : 😑
0:12 Brilliant: Am I a joke to you?
Less than a minute ago, never have I ever clicked so fast on a papa flammy video
@@PapaFlammy69 you forgot 2.99999999999999...... before "
I found this same result when playing with primes in different bases: the number 100…0 (all 0’s in the middle there) in base n is n^k, where k is the number of 0’s following the initial 1 digit. If you then subtract 1, it is trivial to see that this number must be a factor of n-1 (bc the digits in base n is just n-1 repeated k times). Thus this can only be a prime if n-1=1, and thus n=2 (with obvious exception of the k=1 case).
What is the reasoning behind the 2^ab - 1 = composite number. Is this a well known statement or just something which is glossed over?
Well in my defence
7 is a cannibal
Who's to say 9 wasn't an impostor
I enjoyed delving deeper after the initial theorem, thanks!
9:50 : You could have proved this! You don't need binary expansions or anything; you just need the factorization rule that we've already covered. Specifically, if k is composite (so k = ab where a, b > 1), then 2^k - 1 = 2^(ab) - 1 = (2^a)^b - 1 = (2^a - 1)(1 + … + (2^a)^(b-1)). And neither factor here is 1 (because 2^a is neither 2 nor 0), so 2^k - 1 is composite whenever k is.
Here's a cool fact about cubes I just found, but can't find anywhere else worth posting it.
For any base n where n-1 is a perfect square, the √(n-1)ᵗʰ doubleback number is (√(n-1))³ . For example in decimal, where 10-1=9=3², the doubleback sequence goes 09 18 27 36 45 54 63 72 81 90, the 3rd of which is 27 or 3³.
Video lasts 17 minutes as a reminder to us all why 51 cannot be prime (no I'm not crying, you are)
7:40 : Another possibility here is that k = 1, then the second factor can be 1 without having n = 0. in fact, when k = 1, the second factor is always just 1, for any n. (Of course, as with the case when the first factor is 1, not every possible n will work when k is one, only the successors of prime numbers.)
My brain melted and I enjoyed every minute of it
This can be solved with easy algebra.
Supose a cube (x+1)^3 where x > 1.
Writing it out gives [ x^3 + 3x^2 + 3x + 1 ]
Subtracting 1 gives [ x^3 + 3x^2 + 3x ] which is divisible by x. [ x * ( x^2 + 3x + 3 )] and since x isn't 1 and the division product isn't 1 it contradicts any number above the cube of 2 to be prime.
Similarly [ (x+1)^n - 1 ] will always be divisible by x, since [ (x+1)^n ] will always be some coagulation of a*x^b powers +1.
(I can see logically how this is true but I don't know how to prove it for real, I'm not a mathematician or mathwiz, I'm just happy I managed one on my own :'). Love to see if people have comments or if i made mistakes)
Been learning German to speak to Papa in his purest form, so far I can say "Hallo , Donde Esta la Biblioteca?"
xD
I remember I once was able to sneak Mersenne primes into a proof (can't remember the statement) and when my teacher asked why I basically said "beacuse I can"
At 10:34 the contrapositive is really: if 2^k-1 is prime, then k is not equal to ab for a,b natural numbers (that is, k is prime)
Remember that (A implies B) iff (not B implies not A)
IMO it's more fun to look at cyclotomic polynomials than just using 2 as the base. n^k - 1 is always divisible by n - 1, which is redundant for n = 2, and if k is composite, there are more factors. Every cyclotomic polynomial should be able to generate primes (possibly infinitely many). The 24th one, n^8 - n^4 + 1, finds quite a few primes as the first two primes that could ever divide it are 73 and 97.
Just as fun, cyclotomic polynomials are great for predicting lengths of recurring periods of various fractions in base n. There are a couple of subtleties to watch out for.
This is interesting. What about a famous polynomial like n^2+n+41?
@@danielwimmer4698 There are (probably) infinitely many prime-rich polynomials. The reason I was discussing cyclotomic polynomials was their connection to recurring fractions. If you just want to find primes, there are a lot of superior options.
@@coopergates9680 Well, I kinda just want to find primes :)
@@danielwimmer4698 How about 41n(n+1) - 33023? Do you know how to find more polynomials?
@@coopergates9680 Sadly, I'm not very familiar with this. It's not exactly my usual approach to prime numbers but maybe it should be.
Wow holy sh*t I didn't know that was a thing, that's OP
I would say even more: 7 is the only prime followed by a cube and then followed by a square!
-2
nice!
And also 6 the number before has 3 and 2 as the non zero factors both which are the following squares and cubes respectively.
Non one is what i meant
@@abhiroopreddy8673 and before that theres 5 which is 2+3
“51 best prime number”
Half a second later…
“No iTs nOt PrImE ItS dIvIsiBlE bY 7”
I sent that ASS petition to my friends and all agreed. For some reason I wasn't notified about your video or poll... I had to manually search for you.
Flammy oto-san back with a proof
Is there a counterexample to the statement?
" If p is a Mersenne prime, then (2^p)-1 is prime "
This would imply that there are infinitely many Mersenne primes
M=2^13 - 1 = 8191 , is a Mersenne prime.
2^M - 1 = 5.45374067*10^2465 is divisible by 338193759479, therefore, is not prime.
Rip
@@josevidal354 :(
@@josevidal354 Ka-pow! BOOOM!!
@@josevidal354 The sound of fond hopes dying. The sound of tears softly splashing on the keyboard.
I really enjoyed seeing the derivation of a³-b³ from the geometric summation; i have always used Binomial expansion for this identity. It was really cool to see the intuition of Mersenne primes as well?
For those like me who thought papa was just being lazy at 9:42 and can't find the answer, here's how I did it.
Consider the following sum, which is evidently an natural number for p, q, two natural numbers:
\sum_{k=0}^{q-1} 2^{pk}
= (2^pq - 1)/(2^p - 1)
Hence we've proved 2^pq -1 is not prime, for any p, q, different than 1.
Let n be positive integer, so n^3 is a cube, so n^3-1 is a number before a cube. Which is also equal to (n-1)(n^2+n+1) which is divisible by at least two primes for n greater than 2. For n=2 we have the case of 7 and 8. For n=1, we have 0 and 1. Thus 7 is the only prime before a cube.
2^11-1=23*89. So 2^k-1=p, if p is prime, then k must be prime. Not the other way around. I think that is what he was trying to clarify in the onscreen text, but I had to find the first counter example.
n^3-1=p for some positive integer p. (n-1)(n^2+n+1)=p. One of these factors must be one and the factor on the right is always at least 3 and n-1=1 when n=2. (2)^2+2+1=7 QED.
5:50 if u use -1 in the second term, then
p = (-1 - 1) (-1 + -1 + -1^2)
p = -2(-1 - 1 + 1)
P = -2(1)
p = -2
-2 + 1 = -1 = -1^3
12:43
Flammy: " 51 | 7 "
Guess 7 ≈ 17.
Alternatively, 51 ~= 3.
@@spaghettiking653 51 ≈ 3?
@@CarmenLC LMAO, didn't realise wtf I actually wrote. I meant 7≈3 lol
You can also let n=-1, yielding p=-2, which I claim is prime; it's not a unit, and, if it divides a product, then it divides at least one of the factors.
I saw this, and had to figure it out. It was satisfyingly simple.
p + 1 = n³ => p = n³ - 1 = (n - 1)(n² + n + 1)
`p` cannot equal a product of two integers, unless one is one and the other is `p`
n = 2 & p = 7 | n = p + 1 & n² + n + 1 = 1
The first case is what was promised, and the second requires -1 to be prime.
Edit: also, in general for p + 1 = n^k, p = (n - 1)(sum{i=0; n = 2 | n = p + 1 => p = 2^k - 1 | p + 1 = (p + 1)^k
So `p` is either a Mersenne prime, or `k` is 1
Let p be a prime and n an integer
So we prove that: p+1=n^3 only for (n=2;p=7)
p=n^3-1
p=(n-1)(n^2+n+1)
Since p is a prime & n-1 < n^2+n+1
We get the system:
n-1=1 ; n^2+n+1=p (by definition)
So by replacing we conclude that:
The only sol'n are (n=2;p=7)
I didn’t understand why he needed a long method like that to find expansion of (n^3-1). Doesn’t it have a direct expansion formula?
We don't take things for granted here
You can prove it in a much simpler way.
All primes except 2 are odd numbers. 2 is followed by 3, which isn't a k-th power of a natural number since it is a prime. Therefore, consider the odd primes. Odd primes are always followed by an even number, let's call it q. Since q is even, it must be divisible by 2. We assume q is of the form n^k, therefore all its factors must be the same and it follows that q = 2^k. The number preceding it is p = 2^k - 1, which is possibly prime. This means that for any given k, there is at most 1 prime that is followed by a k-th power of a natural number and this natural number must be 2.
I know it's a little random and off-topic but at 4:35 is it proper to have the + on both sides of the ellipsis? It's something that I get confused about and often end up leaving the 2nd one out. I know it probably doesn't *really* matter and could just be up to personal preference but I just thought I'd ask anyway.
Okay, here I did this monster of a comment for fun. I know it gets out of context sorry if it bothers anyone.
Theorem: For a, b are Z+>1 and k=ab, (2^k)-1 has factor(s) (2^a)-1 and (2^b)-1 and so has >1 factor >1.
Proof: Without loss of generality consider only a, (2^ab)-1=((2^a)^b)-1. (2^k)-1=Sum(i=0 to k-1, 2^i) substituting we have (2^ab)-1=Sum(i=0 to ab-1, 2^i)=((2^a)^b)-1=Sum(i=0 to b-1, ((2^a)-1)((2^a)^i))=((2^a)-1)(Sum(i=0 to b-1, (2^a)^i) thus showing that (2^a)-1 is a factor of (2^ab)-1 and of course the other factor(s) here are also >1.
It's been awhile since I proved this myself after first learning about the exponent for Mersenne primes being prime, so I've forgotten how I did it, but I'm pretty sure the method was different. That being said in the spirit of full disclosure I did glance over a few of the other comments in an attempt to refresh my memory (after searching in vain for my original proof). ■
2 proofs of a generalization of the sum used above:
Theorem: Sum(k=0 to n, a^k)=((a^(n+1))-1)/a-1, a>1.
Proof 1: Consider that a^(k+1)-a^k=(a-1)(a^k) and so (a-1)(a^k)+a^k=a^(k+1), then Sum(k=0 to n, a^k)+Sum(k=0 to n, (a-1)(a^k))=Sum(k=1 to n+1, a^k), rearrange the equation, Sum(k=1 to n+1, a^k)-Sum(k=0 to n, a^k)=Sum(k=0 to n, (a-1)(a^k)) and the difference of the left side is (a^(n+1))-1, then after factoring (a^(n+1))-1=(a-1)(Sum(k=0 to n, a^k), divide, Sum(k=0 to n, a^k)=((a^(n+1))-1)/a-1. ■
Proof 2:
Basis for induction:
Sum(k=0 to 0, a^k)=(x-1)/(x-1)
Induction hypothesis:
If Sum(k=0 to n, a^k)=(a^(n+1)-1)/a-1, then Sum(k=0 to n+1, a^k)=(a^(n+2)-1)/a-1.
Induction Step:
(a^(n+1)-1)/(a-1)+a^(n+1)=(a^(n+1)-1)/(a-1)+(a-1)(a^(n+1)/a-1=(a^(n+1)-1+a^(n+2)-a^(n+1))/a-1=(a^(n+2)-1)/a-1. ■
Connection to geometric progressions:
a(Sum(k=0 to n, a^k)+1=Sum(k=0 to n+1, a^k).
Proof: a(Sum(k=0 to n, a^k)=Sum(k=1 to n+1, a^k) and a^0=1.
Note: this only holds for initial values of that form. For a general solution we need to expand deeper (this comment is already much too long). ■
Edit: Removed some minor variable mishaps I noticed after the heart and I shouldn't have missed the first time through. Specifically, I habitually replaced k with x in the first proof, I replaced the final term n with the index variable in the statement of the second theorem, and I wrote n instead of n+1 for the exponent of a in the numerator of the closed form for the 1st proof of the 2nd theorem. That being said it wasn't that bad for like 3 in the morning or whatever it was.
Also, I got Covid early unfortunately. But unlike many others I'm young and have a strong immune system so it wasn't bad at all for me. R.I.P victims of Covid-19.
Ha! Fermat joke at 13.26.
I'm kinda confused about you saying the pattern doesn't hold for k = 4, that's not what we were doing in the first place, right? Like the only thing that's saying is that for any power k, the only prime below any number to the power of k if one exists is 2^k - 1, it's not saying that it has to be a prime, or did I just misunderstand the whole thing?
Smaller numbers are more probable to have a unique trait.
Just an axiom
Interesting theory, could you elaborate ? I wonder, what is the smallest number not to have a unique trait ?
Assume that there exists a set of numbers that do not have unique traits.
This would imply there is a smallest number without a unique trait.
But *the fact this number is the smallest number without a unique trait* is a unique trait in itself, of which is belonging to that number. This is a contradiction, as we said this number does not have any unique traits.
Hence, all numbers must have unique traits. So the probability that any number has a unique trait is 1, so your axiom is wrong.
@@CAG2 @CAG2 Ah I see we made the same joke 🤭
However I have a few interesting comments to make about your answer.
First, what is the probability you're using on the set of naturel integers ? This is important to mention as there is no canonic or general one, and even finitely additive probabilities on N turn out to be very unpleasant.
Then something is not right about your assumption. I believe your intention was to say "assume that the numbers which do not have a unique trait form a nonempty set", I agree that from this assumption follows a contradiction, and this it must be false. But what is the negation of "the numbers which do not have a unique trait form a nonempty set" ? At first sight it would appear that the negation is "the numbers which do not have a unique trait form the empty set" or in other words, there are no such numbers. Actually this is not true, in your assumption you say two important things instead of one :
1/ you say that the numbers which do not have a unique trait do not form the empty set
2/ you say that the numbers which do not have a unique trait form a set
This second point is not trivial at all ! In mathematics, not anything that contains elements and can be named is a set. So it could possibly be the case that those numbers just do not form a set, in which case you can't prove that there is a smallest such number.
I find that very interesting that there is a mathematical theory, called IST, whose goal is to interpret what it means for a mathematical object to have "a unique trait" (such objects are called standard in the theory), and one of the first results of that theory is that not every integer is standard ! However you still can't infere that there is a smallest non-stantard integer, simply because standard integers do not form a set (and you can't consider the measure of the standard elements because only sets can be measured)
@@swenji9113 This sounded fascinating so I took a moment to look up Internal Set Theory and I have a few questions. First, how is the existence of non-standard numbers justified (it's not obvious that any exist at all)? Second, how do non-standard numbers actually relate to a concept of having unique traits? I only glanced at the idea but it seemed to me that non-standard numbers are defined around the concept of accessibility rather than true uniqueness (is it that unique traits, by making numbers more distinguishable, also make them more accessible/possible to isolate)? And finally third, how does this theory relate to ZFC? Wikipedia says that it is a "syntactic enrichment" and that it adds additional axioms that are consistent to ZFC but this does still mean that it is distinct from ZFC alone? Also even though CAG2 mentioned probability I don't think their argument is actually probabilistic at all. I think they were just trying to find a way to articulate the concept that every positive integer must have some unique trait even if it is only that it is the smallest element with a unique trait. In fact, it seems that every integer should have several unique traits by virtue of their distinguishability and that non-standard numbers (albeit unique) are not accessible with some finite amount of resources?
Also, having seen this before, I think Kaushik Raj meant that smaller numbers are less likely to have many traits that are not unique rather than more likely to many traits that are unique. My intuition tells me that larger numbers should almost always have more unique traits (and many more non-unique traits overall). But that small numbers will succeed at having comparatively few non-unique traits.
@@danielwimmer4698 [2/3] Now your other two questions are linked to the 3 axiom schemes which rule the behaviour of s in IST, and to their interpretation (at least my interpretation of what it means to be standard).
When I said that standard numbers were numbers which are defined by a unique trait, I cheated a little by not defining precisely what would be a unique trait. I'll try with the example of Peano arithmetic. We denote the successor map by f, since s is already taken.
The question is : which positive integers have a unique trait that we can express in Peano arithmetic ?
Let's start with 0. It has the unique trait that it is the only integer which is not the successor of any integer. So 0 indeed would be called "standard". Then what about 1 ? Well since 1 is the successor of 0, it has the unique trait that it is the successor of an integer which is the successor of no integer. So 1 does also have a unique trait. In fact, once you know an integer n which does have a unique trait, any integer that could be expressed uniquely from n does have the unique trait of being the only integer expressible in this way from the only integer staisfying the unique trait of n. From this observation arises the notion of standard number.
One of the 3 axiom schemes in IST, called the scheme of transfer, states that for any property written only with € and = and talking about standard objects only, there exists an object satisfying this property if and only if there exists a standard object satisfying this property !
That's where the unique traits meet the standard notion.
Suppose that an object x has a unique trait (by a unique trait I mean that x is the only object satisfying a certain property expressed only with € and = and with other standard elements). We know there exists an object satisfying this unique trait, namely x, thus we can apply the axiom above and we get that there exists a standard object satisfying the unique trait of x. Since only x satisfies this trait, x must be a standard object.
Conversely, if x is standard, then x has a very simple unique trait : that of being equal to x.
Of course since we have to define properly what it is to have a unique trait, I took the liberty of defining it so that it would be equivalent to being standard. But historically the notion of standard was defined in order to get a proper definition of was it is to be uniquely determined by its own properties, so you will have to accept that this is what I call "having a unique trait" if you want to link IST to unique traits
Sir I have observed that : digital root/ digital sum (sum of the digits ) of every prime except 3 is never 3 or 6 or 9 .That is digital root of every prime except 3 is either 1 or 2 or 4 or 8 or 7 ( 1 + 6 ) or 5 ( 3 + 2 ) .Is it a mere consequence or some mathematics behind it ? Plz explain. With Ragards . Dr Rahul Rohtak .//
" They pop up there out of nothing..." Virtual Mersenne Primes, perhaps? 😁
There is an error on the board introduced at 11:17
It should read:
2^(k-1) is prime implies k is prime.
The converse is false:
k is prime does not imply 2^(k-1) is prime.
Papa Flammy, why must you write like a doctor? 😢😢😢😢
more like Doctorate in Mathematics
It is a side effect of sniffing the hagoromo chalk, sometimes it gives you the wrong doctor power.
Ehhh that was way more readable than a doctor's note imo
What about Mersenne-Mersenne-Primes: primes p for which 2^(2^p-1)-1 is prime?
16:38
when p = (n - 1) (1 + n + n*n), wouldn't n = -1 also make (1 + n + n*n) = 1? or is n a natural number only?
What about n = -1 ? For all odd k, (1 + ... + n^k-1) = 1, and -2 is one less than -1, which is a number that is equal to (-1)^k for all odd k
Oh yes, the Mersenne primes.
if 1 had double personality (as itself and the divisor), will it then be admitted into the hall of primes?
ye
I dont know why but the fact that papa said zero is not a good prime got me chuckling
13:15 "Fit into the margin " hahaha
That was cool. Number theory is a neat area of math. Thanks!
I don't get how it narrows it down that it has a factor of 1. Literally all numbers have a factor of one, so how does it checking if it does help?
Hey papa flammy, nice video.
BTW I am stuck with an integral
Integral from 0 to a cos(pi n x/a)/(x²+a²)dx where n is an integer.
Please help me, make a video on it, as long time no integral video.
Nice integral. Idk how to do it, but Wolfram does
wolframalpha.com/input/?i=cos%28pi*n*x%2Fa%29%2F%28x%5E2%2Ba%5E2%29+dx
@@AlexCoglin I think this can be solved using differentiating under the integral sign and creating differential equations.
That's sick dude, I never thought about it.
You should make Supremum merch too
I am abit afraid of legal issues there lol
The opening was amazing
NiCe
14:00 another crucial application of massive primes worth mentioning: public key encryption.
:D
This insight suddenly dawned on me:
1. For every Mersenne prime M = 2^p - 1, we can state: M is the only prime followed by a p:th power.
2. For any integer k > 1, where 2^k - 1 is not a prime, we can state: There is no prime that is followed by a k:th power.
For a formal proof of these statements, just take all the comments for this video and remove the goofy ones.
I just love how you can gain deep mathematical insights by just goofing around.
0:08 when the vocal chords are still booting up
mathematicians explaining why 7 is the only prime followed by a cube:
th-cam.com/video/mdquYEw36TU/w-d-xo.html
seriously, i love how fast you're able to speak during these lessons
if k is prime => 2^k -1 is prime, cant we replace k with 2^k-1? i mean recursively call 2^k -1 with last output as input?
If it was true then yes it would work. However, this is far from being true, only the converse is
@@swenji9113 thanks for the reply, canyou explain how what I'm inferring is different from 10:25?
@@rishikaushik8307 You're right, what's written on the board implies your statement. There is a mistake on the board there, the second line should be «2^k-1 is prime => k is prime». This provides a necessary condition for a number to be a Mersenne prime yet it is not a sufficient one
@@swenji9113 oh, it makes sense now, i should have spotted it (long time since I've touched implications :P)
If p + 1 = n^3 then p + 1 = 2r -> (2^3)r^3 = 8r^3 = p + 1 -> p+1/r^3 = 8
So factors of 8 do not follow primes other than 7 for some reason
At 8:50 , you can also show that for all even k, 2ᵏ = 1 (mod3) which implies 2ᵏ-1 = 0 (mod3).
This will show that only odd k values will produce mersenne primes.
Except for k = 2 lol kek. 3 be a prime number.
Yes, this actually is a particular case of the statement later on on the video that 2^a-1 | 2^ab-1
11:41 when you are a mathematician, even your letters(m in this case) converge :P
I saw this 7 being one from 8. I then wondered about 27 and the closest prime is 23, but that is four from the cube. I am noticing a pattern. I then test 4 cube. 55 is 9 from 64 but not prime. 5 cube I find 109 from 125. I am seeing a parabola that seems to agree with the 7 to 2 cube affect.
Wait, do youtube comments actually have margins? How much have I missed . . .
r i p
Waaaahhh, I love your vids :'3
that one was super interesting
_gonna look for a proof-_
( _will not find any don't worry_ )
How can we prove that (n-1)(1+n+n^2) is the only way to factorize n^3-1 tho?
8.. is the only nonzero perfect power that is one less than another perfect power
that fermat reference is why i watch u XD
:D
This also means 3 is the only prime followed by a perfect square
finally a video I understand
Have seen you after 4 months. Now you have more subscribers then Andrew 😎 .