2 is the only prime number followed by another prime.

8 is the only cube that comes after a prime.

3+4=7

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.

70 is the only number after the funny number. Idk why it is relevant but eh

Another fun fact about Mersenne primes - written in binary, they all have form of 11, 111, 11111, 1111111, 1111111111111 and so on. No zeros.

Can we start a petition to make 51 a prime number, please?

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.

Did you get the package of wet hamster fur I sent you?

I never considered Mersenne Primes from this angle, really interesting!

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!

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

The greeting is getting so shortened that it's become just a single sound

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

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

That's a crazy way to introduce the concept of Mersenne primes! There's 51 of them we know of so far!

Me: *factors n^3-1 and confirms claim* Flammable Boi: *17 minute video*

5:54 Papa being like -1 is not. A Good. PRIME. N U M B E R.

12:43 Flammy: " 51 | 7 " Guess 7 ≈ 17.

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.

His "m" just transformed to a single line 😂

Video lasts 17 minutes as a reminder to us all why 51 cannot be prime (no I'm not crying, you are)

Well in my defence 7 is a cannibal

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

"-1 does not make a good prime number" factorisation in the integers : 😑

Papa Flammy, why must you write like a doctor? 😢😢😢😢

I would say even more: 7 is the only prime followed by a cube and then followed by a square!

0:12 Brilliant: Am I a joke to you?

My brain melted and I enjoyed every minute of it

Flammy oto-san back with a proof

It's always nice seeing you in new videos. Awesome!

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

16:38

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

Been learning German to speak to Papa in his purest form, so far I can say "Hallo , Donde Esta la Biblioteca?"

13:15 "Fit into the margin " hahaha

I enjoyed delving deeper after the initial theorem, thanks!

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 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"

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.

Less than a minute ago, never have I ever clicked so fast on a papa flammy video

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.

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.

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).

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.

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)

0:08 when the vocal chords are still booting up

Wow holy sh*t I didn't know that was a thing, that's OP

Still wondering why the board is moving by itself 😳

“51 best prime number” Half a second later… “No iTs nOt PrImE ItS dIvIsiBlE bY 7”

What is the reasoning behind the 2^ab - 1 = composite number. Is this a well known statement or just something which is glossed over?

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.

Smaller numbers are more probable to have a unique trait. Just an axiom

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.

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.

Oh yes, the Mersenne primes.

I dont know why but the fact that papa said zero is not a good prime got me chuckling

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)

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

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.)

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.

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 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?

The opening was amazing NiCe

reading on elementary number theory rn. This is cool!

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.

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.

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

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.

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)

if 1 had double personality (as itself and the divisor), will it then be admitted into the hall of primes?

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?

Loving the linear algebra diagram on the background, used that on my final this semester!

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.

That was cool. Number theory is a neat area of math. Thanks!

What about Mersenne-Mersenne-Primes: primes p for which 2^(2^p-1)-1 is prime?

You should make Supremum merch too

before I watch X is a whole number X^3-1 (x-1)(x^2+x+1) This polynomial has 2 factors, so it can only be prime if one of them is 1 and the other is prime x-1=1 x=2 x^2+x+1=1 x^2+x=0 x+1=0 x=-1 This solution doesn’t work because the other factor ends up as -2, and negative versions of primes aren’t considered primes

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?

That's sick dude, I never thought about it.

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.

23 is a prime followed by 27 which is a cube. Checkmate.

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?

7 is the only prime which devoured a square number

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

Ha! Fermat joke at 13.26.

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 .//

7 is also the only predatory number, hunting down and eating packs of square 3s

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.

Very interesting fact. If p

This also means 3 is the only prime followed by a perfect square

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

finally a video I understand

You could follow up with Brazillian primes!

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.

Given there's only 1 even prime (2), this is a fairly trivial observation.

finally a video i understand!

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.

Hey Flammy, can you recommend books to learn pure mathematics?

Seems surprising until you remember how x³-1 factorises haha