Exercise 6. The strictly positive integers $x, y$, and $z$ satisfy the following two equations: $x+2y=z$ and $x^2-4y^2+z^2=310$. Find all possible values of the product $xyz$.
Solution to exercise 6 We have $310=x^{2}+z^{2}-(2 y)^{2}=x^{2}+z^{2}-(x-z)^{2}=2 z x$ so $z x=155=5 \times 31$. Therefore, $z$ is one of the values $155,31,5,1$ and $x$ takes respectively one of the values $1,5,31,155$. Since $z=x+2 y>x$, the last two cases are impossible. In the first two cases, we have
$y=\frac{z-x}{2}$, and the values of $x y z$ that we obtain are respectively $x y z=1 \times 77 \times 155=11935$ and $x y z=5 \times 13 \times 31=2015$.