Solve the equation $x^{2}+y^{2}=z-16$ in the set of positive prime numbers.
Solution. $x$ and $y$ are primes, so $x^{2}+y^{2}>0$, which means $z-16>0$ also holds. Since $z>16$ and is prime, it is odd, and thus $z-16$ is also odd. Therefore, $x^{2}+y^{2}$ is odd. Since there is only one even prime number, which is 2, one of $x$ and $y$ must be 2, and the other must be odd.

For example, if $y=2$, then $x^{2}+4=z-16$, from which $x^{2}=z-20$. Clearly, $z>20$. If $x=3$, then $z=29$.

If $y=2$ and $x$ were a prime number larger than 3, then $x=3k \pm 1(k \in \mathbb{Z}$ and $k>1)$, so $x^{2}=9k^{2} \pm 6k+1$ and $z=9k^{2} \pm 6k+21=3(3k^{2} \pm 2k+7)$, which means $z$ would also be divisible by 3, and thus could not be a prime number.

Therefore, if $y=2$, then $x=3$ and $z=29$.

Since the roles of $x$ and $y$ in the equation are interchangeable, the other solution to the equation is: $x=2, y=3$ and $z=29$.