Problem 8.6. A natural number $n$ is such that the value of the expression $n^{2}+492$ is a perfect square.
(a) (2 points) Indicate any possible value of $n$.

(b) (2 points) What can $n$ be equal to? Indicate all possible options.
Answer: $n=38 ; 122$.

Solution. Let $n^{2}+492=k^{2}$ for some natural $k>n$. Then

$$
2^{2} \cdot 3 \cdot 41=492=k^{2}-n^{2}=(k-n)(k+n)
$$

The numbers $k-n$ and $k+n$ have the same parity, so they must both be even. Since $k-n<k+n$, we have only two possible cases.

- Let $k-n=2$ and $k+n=246$. Then $k=\frac{246+2}{2}=124$ and $n=\frac{246-2}{2}=122$.
- Let $k-n=6$ and $k+n=82$. Then $k=\frac{82+6}{2}=44$ and $n=\frac{82-6}{2}=38$.

It is easy to verify that both of these pairs satisfy the condition.