5. Find all integers $x, y$ and $z$ with $2 \leq x \leq y \leq z$ such that
$$
x y \equiv 1 \quad(\bmod z), x z \equiv 1 \quad(\bmod y), y z \equiv 1 \quad(\bmod x) .
$$
5. The only solution is $(x, y, z)=(2,3,5)$.

First of all, observe that $(x, y)=(x, z)=(y, z)=1$. Then $2 \leq x1
$$

Since $x<y<z$, it follows that
$$
1<\frac{1}{x}+\frac{1}{y}+\frac{1}{z}<\frac{3}{x}
$$
and this gives $x=2$ as the only value. In this case, the inequalities give
$$
\frac{1}{2}<\frac{1}{y}+\frac{1}{z}<\frac{2}{y},
$$
which implies that $y=3$. It follows that the only possible values of $z$ are 4 and 5 . Hence, for $2 \leq x<y<z$, the solutions are $(x, y, z)=(2,3,4)$ and $(2,3,5)$. Since 2 and 4 are not relatively prime, the only solution is $(x, y, z)=(2,3,5)$.