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!
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.