Find all integer solutions of the equation:

$$
x^{2}+y^{2}=3 z^{2}
$$
There are two methods, both based on the principle of infinite descent. The first method consists in noticing that modulo 4, since a square is congruent to 0 or 1, we have 4 divides $x^{2}, y^{2}$ and $z^{2}$, so $(x, y, z)$ solution $\Longrightarrow\left(\frac{x}{2}, \frac{y}{2}, \frac{z}{2}\right)$ solution. By the principle of infinite descent, we easily deduce that $(x, y, z)=(0,0,0)$ is the only solution.

The second method consists in noticing that modulo 3, a square being congruent to 0 or 1, 3 divides $x$ and $y$ so $x=3 x^{\prime}$ and $y=3 y^{\prime}$, so $3\left(x^{\prime 2}+y^{\prime 2}\right)=z^{2}$, so $z=3 z^{\prime}$, so $x^{\prime 2}+y^{\prime 2}=3 z^{\prime 2}$, so $\left(\frac{x}{3}, \frac{y}{3}, \frac{z}{3}\right)$ and we conclude as before that $(x, y, z)=(0,0,0)$ is the only solution.