5. Let $m=\left|n^{2}-100\right|$.

We have $m=\left|n^{2}-100\right|=|(n-10)(n+10)|$.

The number $m$ is prime if and only if one of the numbers $|n-10|$ and $|n+10|$ is equal to 1, and the other is a prime number.

$1 \text{ POINT}$

Notice that there are no natural numbers $n$ for which $|n+10|=1$.

Let's find all natural numbers $n$ for which $|n-10|=1$.

If $|n-10|=1$, then $n=9$ or $n=11$.

For $n=9$, $m=|(n-10)(n+10)|=19$, which is a prime number.

For $n=11$, $m=|(n-10)(n+10)|=21$, which is not a prime number.

Therefore, there is only one natural number $n$ for which $\left|n^{2}-100\right|$ is a prime number, and that is $n=9$.
5. Determine all natural numbers $n$ for which the number $\left|n^{2}-100\right|$ is prime.

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