19. Determine the number of sets of solutions $(x, y, z)$, where $x, y$ and $z$ are integers, of the the equation $x^{2}+y^{2}+z^{2}=x^{2} y^{2}$.
19. Answer: 1
Solution. Note that $x=0, y=0$ and $z=0$ is a solution of this equation. We shall show that this is its only integer solution by proving that if $x, y, z$ is a solution of this equation and whenever $x, y, z$ are divisible by $2^{k}$, they are also divisible by $2^{k+1}$ for any $k \geq 0$.
Let $x=2^{k} x^{\prime}, y=2^{k} y^{\prime}$ and $z=2^{k} z^{\prime}$. Then $x^{2}+y^{2}+z^{2}=x^{2} y^{2}$ is changed to
$$
x^{\prime 2}+y^{\prime 2}+z^{\prime 2}=2^{k} x^{\prime 2} y^{\prime 2}
$$

It is easy to verify that only when $x^{\prime}, y^{\prime}, z^{\prime}$ are all even, $x^{\prime 2}+y^{\prime 2}+z^{\prime 2}$ and $2^{2 k} x^{\prime 2} y^{\prime 2}$ have the same remainder when divided by 4 . Thus $x, y, z$ are divisible by $2^{k+1}$.