2. Find all natural numbers $n$ for which there exists a prime number $p$ such that the number $p^{2}+7^{n}$ is a perfect square.
IV /2. Let $p^{2}+7^{n}=m^{2}$. Then $7^{n}=m^{2}-p^{2}=(m-p)(m+p)$. We consider two possibilities. If $m-p=1$ and $m+p=7^{n}$, it follows that $2 p=7^{n}-1$. Suppose $n \geq 2$. Then $2 p=7^{n}-1=(7-1)\left(7^{n-1}+7^{n-2}+\ldots+7+1\right)$, so we get $p=3\left(7^{n-1}+7^{n-2}+\ldots+7+1\right)$, which is not possible, as the expression in parentheses is greater than 1. If $n=1$, we get $p=3$.

The second possibility is that $m-p=7^{k}$ and $m+p=7^{n-k}$, where $k$ is some natural number. Clearly, $n-k>k$. Then $2 p=7^{n-k}-7^{k}=7^{k}\left(7^{n-2 k}-1\right)$. The expression in parentheses is an even number greater than 2, so we have written $2 p$ as the product of at least three prime numbers, which is not possible.

Thus, $n=1$ is the only natural number for which there exists a prime number $p$ such that $p^{2}+7^{n}$ is a perfect square. Such a prime number is $p=3$, and the perfect square is $3^{2}+7=16=4^{2}$.
Factorization $7^{n}=(m-p)(m+p)$ ..... 1 point
Writing $m-p=7^{k}$ and $m+p=7^{n-k}$ with the estimate $n-k>k$ ..... 1 point
In the case $k=0$, factorization $2 p=6\left(7^{n-1}+7^{n-2}+\ldots+7+1\right)$ ..... 1 point
In the case $k=0$, conclusion that $p=3$ ..... 1 point
In the case $k>0$, factorization $2 p=7^{k}\left(7^{n-2 k}-1\right)$ (or the observation that $7 \mid p$) ..... 1 point
In the case $k>0$, conclusion that $7^{k}\left(7^{n-2 k}-1\right)$ is the product of at least 3 primes (or the observation that $p=7$ is not a solution) ..... 1 point
Writing the solution: $p=3, n=1$ ..... 1 point