# Find all integers $x, y, u, v$ for which holds: $x^2 + y^2 = 3(u^2 + v^2)$.

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!

• $$x^2+y^2\not\equiv-1\pmod4$$ – lab bhattacharjee Jan 18 at 16:32
• Reduce so there are no common factors. Consider modulo $8$. – Mark Bennet Jan 18 at 16:35
• 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 $-1$. – 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.
• 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. – John Omielan Jan 18 at 22:42