Solve an equation in positive integers Does $$x^2+y^2=3(z^2+ u^2)$$ have solutions in positive integers? 
I was assigned this problem, but I am struggling to find a solution. I guess that a proof by contradiction is required.
 A: Hints:
$$x^2+y^2=0\pmod 3\iff\;\;\begin{cases}x=0\pmod 3\\y=0\pmod 3\end{cases}$$
But then 
$$9\mid(x^2+y^2)\;\implies 3\mid(z^2+u^2)$$
Well, check those powers of three in both sides and get a contradiction...
A: Suppose, for the sake of contradiction, that the equation 
$$x^2+y^2=3(z^2+u^2)$$
has solutions in positive integers. Then among them must be one, which we will denote by $(x_1,y_1, z_1, u_1)$, with the smallest possible value of $x$. From the equation
$$x_1^2+y_1^2=3(z_1^2+u_1^2)$$
it follows that $x_1^2+y_1^2$ is divisible by $3$. Hence, both $x_1$ and $y_1$ are divisible by $3$ (indeed, if an integer is not divisible by 3, then its square is of the form $3k+1$ for some integer $k$). Therefore,$x_1=3x_2$ and $y_1=3y_2$ for some positive integers $x_2$ and $y_2$. It follows that $$9x_2^2+9y_2^2=3(z_1^2+u_1^2),$$ that is $$3x_2^2+3y_2^2=z_1^2+u_1^2.$$ This implies that $u_1$ and $z_1$ are divisible by $3$. Therefore, $z_1=3z_2$ and $u_1=3u_2$ for some positive integers $z_2$ and $u_2$. It follows that $$x_2^2+y_2^2=3(z_2^2+u_2^2).$$ This means that $(x_2,y_2, z_2, u_2)$ is a solution of the first equation. However, $x_2= \frac{1}{3}x_1 < x_1$, which contradicts the assumption that $x_1$ was the smallest possible value for all solutions of the equations. Hence, the equation has no solution in positive integers.
References:
This problem is discussed in J. Cofman, What to Solve? Problems and Suggestions for Young Mathematicians, Oxford University Press, 1990.
