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In order to solve some bigger problem I need to prove that equation

$$z^2+y^2=nx^2$$ where $ n=6 \; mod\;8$ and $x,y,z$ are all integers, doesn't have any non-zero solutions

I was given a hint how to do it when $n=6$. Say we have

$$z^2+y^2=6x^2$$ Then we can consider that $gcd(x,y,z)=1$. Then $z^2+y^2$ is divisible by 3. It can only be if both $z^2$ and $y^2$ are divisible by 3, hence $z$ and $y$ are divisible by 3. Hence $6x^2$ has to be divisible by 9, it can only be if $x$ is divisible by 3 and we have a contradiction because $gcd(x,y,z)=3.$

I can do that trick when $n=14$ and I guess with other $n$. But I can't get the general principle. If you have any ideas. please share.

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  • $\begingroup$ $n$ is divisible by some prime $q \equiv 3 \pmod 4$ to an odd power, that is how you get $\frac{n}{2} \equiv 3 \pmod 4.$ However, $-1$ is not a quadratic residue $\pmod q,$ so this $q$ can only divide $x^2 + y^2$ to an even power. In brief, $x^2 + y^2 \neq n z^2 $ mathoverflow.net/questions/208158/isotropic-ternary-forms $\endgroup$
    – Will Jagy
    Oct 16, 2021 at 23:29
  • $\begingroup$ math.stackexchange.com/questions/1767109/… $\endgroup$
    – individ
    Oct 17, 2021 at 6:41
  • $\begingroup$ If n is a square and both n and z are of the form $\space(4k+1)\space$ there "may" be as many solutions as there are distinct prime factors of the nz pr9duct. If n is not a square, there are no solutions because the nz product must be a square. $\endgroup$
    – poetasis
    Oct 20, 2021 at 9:59

1 Answer 1

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Hint:

The only values of $z^2$, $y^2$ and $x^2$ could be, mod $8$: $0,1,4$. The only values of $nx^2$ is $0$ and $6$.

More Hint:

$6$ cannot be achieved by adding, so the only possible case is where $x^2,y^2,z^2$ is all $0$ mod $8$, or $x^2,y^2$ are both $4$ mod $8$. In other words, $x^2,y^2,z^2$ are all multiples of $4$.

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  • $\begingroup$ $z^{2},y^{2}$ can both be congruent to $4 \mod 8$. So that $0=4+4=z^{2}+y^{2}=6x^{2} \mod 8$ Which we can't have since then $2|z,y$. But it is not necessary that all the squares are $0 \mod 8$. $\endgroup$
    – Derek Luna
    Oct 17, 2021 at 0:15

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