Find all integers $x, y, u, v$ for which holds: $x^2 + y^2 = 3(u^2 + v^2)$. My approach was to say that $u^2+v^2$ is the divisor of $x^2 + y^2$ and $u^2+v^2$ is the divisor of $x^2$ and $y^2$, but I somehow need to use the information about number $3$ that is given. Can someone help me? Thanks in advance!

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    $\begingroup$ $$x^2+y^2\not\equiv-1\pmod4$$ $\endgroup$ – lab bhattacharjee Jan 18 at 16:32
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    $\begingroup$ Reduce so there are no common factors. Consider modulo $8$. $\endgroup$ – Mark Bennet Jan 18 at 16:35
  • $\begingroup$ Wait, I'm confused again. Why is the right side $-1 (mod 4)$? I understand the fact that left side is congruent to 0, or 1 $(mod 4)$ but I don't know why the right side is $-1$. $\endgroup$ – Wolf M. Jan 18 at 16:36

There are no solutions besides the trivial $x=y=u=v=0$. To show this, suppose for a contradiction that non-trivial solutions exist, and choose one with $|x|+|y|+|u|+|v|$ as small as possible. From $3\mid x^2+y^2$ it follows that both $x$ and $y$ are divisible by $3$; say, $x=3x_1$ and $y=3y_1$. Substituting, we get $3(x_1^2+y_1^2)=u^2+v^2$, showing that $(u,v,x_1,y_1)$ is also a solution. Moreover, we have $|u|+|v|+|x_1|+|y_1|<|x|+|y|+|u|+|v|$, contradicting the choice of $(x,y,u,v)$ as the "smallest" non-trivial solution.

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    $\begingroup$ This is a good solution. However, I believe the $x$ and $y$ in "... showing that $\left(u, v, x, y\right)$ is also a solution" should be $x_1$ and $y_1$ instead. Also, there is no $x_2$ so, in the absolute value inequality, I believe $y_1$ is what you want to use. $\endgroup$ – John Omielan Jan 18 at 22:42
  • $\begingroup$ @JohnOmielan: Thanks for the remarks, I will fix the things immediately. $\endgroup$ – W-t-P Jan 19 at 7:41

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