a nice 'mixed' Diophantine equation

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[luckycandy]

n=0 case is pretty easy, p=2 or q=2 gives a contradiction, So both are odd. But p is odd implies p²+1 is even so q is even, so no sol here

[giorgiobarchiesi]

Great and super-interesting, as usual.
Just, at 6:20, I think Mersenne primes have the form 2^n - 1, not 2^n + 1
Not fundamental, since the fact is not used in the subsequent reasoning.

[PRIYANSH_SUTHAR]

Mathematics is amazing that sometimes showing that there are no solutions itself is the solution.

[clarejang]

Another way to prove the non-existence is:
LHS is 1 modulo p. Moreover, by Fermat's little theorem (or its proof), only q^(m(p-1)) for some m is 1 modulo p. Thus, 7 should be m(p-1). However, It is possible only for p=2 as 7 is a prime. Now, 2^(n+2) = q^7 - 1 = (q-1)(q^6+...+1), so the odd part (q^6+...+1) should be 1, but q =/= 0 thus contradiction.

[xizarrg]

Channeling a bit of Sherlock Holmes @ 5:55

[dragonite]

Did a similar thing until q = 2^n +1, but then put that into the original equation and divided by 2^n, which gives you 7 + (2^n)a = p^2, so p^2 +1 is 0 mod 8 if n>=3, which is impossible, so only n = 1, 2 must be checked. The rest is a bit heavy with computing 5^7, but it can be checked you don't get an integer prime from that because luckily 140^2 falls close enough.

[Touijarsoulaimane]

Why q-1 would not be equal to 2^m with m<n ?

[deebd]

If I'm not mistaken, there are no solutions for setting n = 0: p^2 = q^7 - 1 => p^2 must be divisible by (q - 1), since q^7 - 1 is => impossible

[張謙-nl]

10:15
"All primes are of the from 4m+1 or 4m-1"
2: What about me?

[Alan-zftt]

I am sure there must be an elegant one liner answer to this.

[Shawty-fisn]

n=0 trivial bc of quadratic nonresidues mod q

[konraddapper]

even if p and q are allowed to be gausian primes theier are no solution

[chrissquarefan]

Why is n greater than or equal to 1? 0 is also a natural number. N* is non zero naturals not N

[fitmtla]

Its funny how in Eastern Europe we consider N to always include 0 and we annotate N* when 0 is not included

[johns.]

I thought a Mersenne prime was of the form 2^n - 1

[restcure]

Been following the channel for couple of years, I just now realized: that's not an asterisk, but two opposing arrows (contradiction, duh)

[Alan-zftt]

uh-huh? @ 2:37: q odd ⟹ q^n odd and sum of even number of odd terms is even ⟹ (q^6 + ... + q + 1) is odd therefore (q^6 + ... + q) is even
Same result, different route hence the math is good

[orionspur]

Nonexistence proofs are a good exercise, but never satisfying. Are there any such problems with 1) a *nontrivial* solution that is 2) findable without extraordinary efforts?

[iMíccoli]

I got a completely incorrect proof because for some reason i thought ._.

[RexxSchneider]

At 3:10 "That only gives us three distinct cases." No, there can only be one way to factorise 2^n * p^2 into an odd and an even part, and that's if the even part is 2^n and the odd part is p^2 because 2 cannot be a factor of an odd number. So (q-1) = 2^n and (q^6+...+1) = p^2.
That saves you about three minutes of faffing about.