Find all integer solutions of the equation:

$$
x^{2}+y^{2}=3\left(u^{2}+v^{2}\right)
$$
First, we show an initial result: if 3 divides $x^{2}+y^{2}$, then 3 divides $x$ and 3 divides $y$. Indeed, squares modulo 3 are always 0 or 1. Thus, the sum of two squares is zero modulo 3 if and only if both squares are congruent to zero modulo 3, hence the result. Let $x, y, u, v$ be non-zero integers. Then 3 divides $x^{2}+y^{2}$, so $x$ and $y$ are divisible by 3 by the previous result. But then $3\left(u^{2}+v^{2}\right)$ is divisible by $3^{2}$, so $u$ and $v$ are divisible by 3 (by the previous result).

Thus $x, y, u, v$ are divisible by 3, and $x^{\prime}=\frac{x}{3}, y^{\prime}=\frac{y}{3}, u^{\prime}=\frac{u}{3}, v^{\prime}=\frac{v}{3}$ satisfy the same equation. Thus these four integers are divisible by three by the same reasoning, and so on. Thus $x, y, u, v$ can be divided by $3^{n}$ for any integer $n$, so they are all zero.

The only solution is then $(0,0,0,0)$.