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Find all integers $x$ and $y$ such that $4y^2=x^7+1$.

I first subtract one from both sides to get $4y^2-1=x^7$, which can be factored into $(2y+1)(2y-1)=x^7$. From here, I can not proceed.

Any ideas?

egreg
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Hyowang Kim
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2 Answers2

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You have $(2y+1)(2y-1)=x^7$. Then $2y+1$ and $2y-1$ have no common prime factor (why?) so each prime factor of $2y+1$ must occur seven, or fourteen, or .... times. Similarly for $2y-1$.

Angina Seng
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  • Could you please expand on your explanation of "why"? I do not understand why (2y-1) and (2y+1) cannot have the same prime factor. Why cant it happen that one of the factors of (2y-1) is p^(7k-m) and one the the factors of (2y+1) is p^m, which when multiplied comes to p^(7k)? I hope my wording is clear. – LeastSquaresWonderer Sep 15 '17 at 21:07
  • $2y+1$ and $2y-1$ aren't very far apart.... @LeastSquaresWonderer – Angina Seng Sep 15 '17 at 21:08
  • Yes, I understand it intuitively, but I do not know how to derive it or prove it. Could you assist me by expanding a bit? – LeastSquaresWonderer Sep 15 '17 at 21:10
  • @LeastSquaresWonderer Thenks to Euclid algorithm $$\text{GCD}(2y+1,2y-1)=\text{GCD}(2y-1,(2y+1)-(2y-1))=\text{GCD}(2y-1,2)=1$$ So $2y+1$ and $2y-1$ are coprime. BTW I got the solution $x=-1;;y=0$ and nothing else – Raffaele Sep 15 '17 at 21:28
  • $2y+1=p^7$ and $2y-1=q^7$ implies $2=p^7-q^7$ which is impossible (why?). – Ataulfo Sep 16 '17 at 00:47
  • @Piquito $2=p^7-q^7$ is possible: $(p,q)=(1,-1)$. But the function $x\mapsto x^7$ grows quickly; unless $|x"$ is very small $(x+1)^7$ is a lot bigger than $x^7$. – Angina Seng Sep 16 '17 at 06:21
  • $p$ and $q$ are obviously supposed to be prime numbers, not $1$, in your reasoning. Cordial greetings . – Ataulfo Sep 16 '17 at 12:51
  • @Piquito $p,q$ don't have to be prime. If $ab=x^{k}$, $k\ge 1$, $a,b,x,k\in\mathbb Z$, $\gcd(a,b)=1$, then $|a|=c^k$, $|b|=d^k$ for some $c,d\in\mathbb Z$, where I used absolute value signs, which are important when $k$ is even. $c,d$ don't have to be prime. – user236182 Sep 16 '17 at 23:26
  • I think I should say here that if we had $ab=x^{2k}$, $\gcd(a,b)=1$ for some $k\ge 1$, $a,b,x,k\in\mathbb Z$, then generally you would have to say that $|a|=c^{2k}$, $|b|=d^{2k}$ for some $c,d\in\mathbb Z$, where I used absolute value signs. Notice that, e.g., when $a=-4$, $b=-9$, $x=6$, $k=1$, then $\gcd(a,b)=1$, $k\ge 1$, $a,b,x,k\in\mathbb Z$, $(-4)\cdot (-9)=6^{2\cdot 1}$, but $-4<0\le \left(c^1\right)^2$, so $-4\neq c^{2\cdot 1}$. – user236182 Sep 16 '17 at 23:34
  • My comment goes about the first answer to this problem! – Ataulfo Sep 17 '17 at 00:39
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$4y^2=x^7+1$ is clearly a case of the equation $y^2=x^p+1$ where $p$ is odd prime and has always the trivial solution $x=-1$.

$► p=3$: Euler proved that the only solutions are $x=0,-1,2$.

$►p\gt3$: Nagell proved that in order there is solution it is necessary that $$p\equiv1\pmod8\text{ and }y\equiv0\pmod p$$ Thus $4y^2=x^7+1$ has only the trivial solution $x=-1$ since $7$ is not congruent to $1$ modulo $8$.

(T. Nagell. Sur l'impossibilité de l'équation indéterminée $z^p+1=y^2$. Norsk Mat. Forenings Skrifter, 1 (1921),Nr. 4)

Ataulfo
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