Find all integer triplets $(x, y, z)$ such that $x^{2}+y^{2}+z^{2}-2 x y z=0$.
First, since $x^{2}+y^{2}+z^{2}$ is even, $x, y, z$ cannot all be odd: at least one of them is even, which implies that $2 x y z=x^{2}+y^{2}+z^{2}$ is divisible by 4. But modulo 4, a square can only be congruent to 0 or 1. Therefore, for $x^{2}+y^{2}+z^{2} \equiv 0(\bmod 4)$, it is necessary that $x^{2}, y^{2}$ and $z^{2}$ are all $\equiv 0(\bmod 4)$, hence $x, y, z$ are all even.

Let's prove by induction that for all $n, x, y, z$ must all be multiples of $2^{n}$. This is true, as we have just seen, for $n=1$: the induction is initialized. Suppose this is true for $n \geqslant 1$, and let $x=2^{n} x_{n}, y=2^{n} y_{n}$ and $z=2^{n} z_{n}$. The equation $s'$ writes: $2^{2 n}\left(x_{n}^{2}+y_{n}^{2}+z_{n}^{2}\right)=$ $2^{3 n+1} x_{n} y_{n} z_{n}$, or $x_{n}^{2}+y_{n}^{2}+z_{n}^{2}=2^{n+1} x_{n} y_{n} z_{n}$. Since $n \geqslant 1, x_{n}^{2}+y_{n}^{2}+z_{n}^{2}$ is divisible by 4, so according to the reasoning above, $x_{n}, y_{n}$ and $z_{n}$ are all even, which means that $x, y, z$ are all divisible by $2^{n+1}$.

However, a non-zero integer cannot be divisible by all powers of 2. If $k$ is the exponent of 2 in the prime factorization of $x$, for $n>k, x$ is not divisible by $2^{n}$. The induction above therefore implies that $x, y$ and $z$ must all be zero. The only solution to the equation is $(0,0,0)$.