9.3. Find all prime numbers $p$ for which $p^{2}+200$ is a perfect square of an integer.
Answer: $p=5$ or $p=23$. Solution. Let $p^{2}+200=a^{2}$, where $a$ is a natural number. Since $p=2$ does not satisfy the equation, the sought $p$, and thus $a$, are odd numbers. From the equality $(a-p)(a+p)=200$, considering that $a+p>a-p$ and both factors are even numbers, we get three cases: 1) $a+p=100, a-p=2$; 2) $a+p=50, a-p=4$; 3) $a+p=20, a-p=10$. By finding $p$ in each of these cases, specifically, 1) $p=49$; 2) $p=23$; and 3) $p=5$, we obtain two prime numbers: $p=$ 5 and $p=23$.