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As we all know that $x^2+y^2=z^2$ has infinite integer solutions. Can we use this fact to give an intuitive explanation for $ x^2+y^2=7 z^2 $ doesn't have any integer solution? $x, y, z $ are positive integers and $(x,y)=1$

Wieshawn
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4 Answers4

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Imagine you have a solution $(x,y,z)$ with $x$ and $y$ relatively prime. Then if you reduce modulo $7$, you find $x^2 + y^2 = 0$ in $\mathbf{F}_7$. If $x$ or $y$ is $0$ so is the other, meaning that $7$ divide $x$ and $y$ which is impossible as they are relatively prime. Then you have $(x/y)^2 = -1$ in $\mathbf{F}_7$ which is impossible as $-1$ is not a square in $\mathbf{F}_7$, as $\mathbf{F}_7$'s squares are $0, 1, 2$ and $4$. Remark : I noted by abuse $x$ and $y$ the images of $x$ and $y$ in $\mathbf{F}_7$, hope it doesn't cause confusion.

Olórin
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  • That is neither a answer to the question ("Does this fact ...") nor does this give us a complete answer to the more usefull question: what are the integer solutions of the equation. – miracle173 Jan 12 '15 at 09:31
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    Read the question again, and then read my answer again, and convince yourself that my answer is an answer to the question. The intuitive reason is that $-1$ is not a square in $\mathbf{F}_7$. If you find that your question (knowing integer solutions of the equation) is more useful, either ask it, or convince yourself that its answer results trivially from the answer to the initial question. – Olórin Jan 12 '15 at 09:34
  • The real point is not $x$ and $y$ being relatively prime or not, but being both divisible by $7$ or not. If $7$ doesn't divide $x$ and $y$ gcd, then $-1$ would be a square in $\mathbf{F}_7$. If $7$ does divide $x$ and $y$, take out from then the biggest power of $7$ dividing them and see what happens to the equation... something like $7^{2d} ({x'}^2 + {y'}^2) = 7 z^2$ with $d\in\mathbf{N}^{*}$. What could you say then ? – Olórin Jan 12 '15 at 09:46
  • I still I can't see where you use the fact that $x^2+y^2=z^2$ has infinite integer solutions. – miracle173 Jan 12 '15 at 10:01
  • Try harder to see. According to values of $d$. – Olórin Jan 12 '15 at 10:05
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Hint: $x^2+y^2 = 0 \pmod 7 \to x = y = 0 \pmod 7$. Then use Descending method.

DeepSea
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Reduce mod $4$.

$x^2 + y^2 \equiv 0,1,2 \bmod 4$

whereas

$7z^2 \equiv 3z^2 \equiv 0,3 \bmod 4$.

So if there were a solution both sides would have to be $0 \bmod 4$, hence all three of $x,y,z$ would be even, contradicting the coprimality assumption.

fretty
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For your problem, we use the result: if $p$ is a prime of the form $4k+3$ and $p|x^2+y^2$, then $p|x,p|y$ (use Fermat theorem to prove this).

More generalizing result: the equation $x^2+y^2=nz^2$ has root different from $(0,0,0)$ ($x,y$ need not to be relatively prime) if and only if each of the prime factor of the form $4k+3$ of $n$ has an even exponent.

Tien Kha Pham
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