1. Determine all prime numbers of the form $1+2^{p}+3^{p}+\ldots+p^{p}$ where $p$ is a prime number.
Solution. First method. For $p=2$, we get the number $1+2^{2}=5$, which is a prime number. Let $p>2$, that is, let $p$ be an odd number. Then the number

$$
1+2^{p}+\ldots+p^{p}=\left(1^{p}+(p-1)^{p}\right)+\left(2^{p}+(p-2)^{p}\right)+\ldots+\left(\left(\frac{p-1}{2}\right)^{p}+\left(\frac{p+1}{2}\right)^{p}\right)+p^{p}
$$

is divisible by $p$, since each of the numbers in parentheses is divisible by $p$. Indeed, from

$$
k^{p} \equiv k^{p}(\bmod p), p-k \equiv-k(\bmod p) \text { and }(p-k)^{p} \equiv(-k)^{p} \equiv-k^{p}(\bmod p)
$$

it follows that

$$
k^{p}+(p-k)^{p} \equiv k^{p}-k^{p} \equiv 0(\bmod p)
$$

According to this, for any odd prime number $p$, a number of the form $1+2^{p}+3^{p}+\ldots+p^{p}$ is a composite number.

Finally, the only solution to the problem is the number 5, which is obtained for $p=2$.

Second method. From Fermat's Little Theorem, it follows that $a^{p} \equiv a(\bmod p)$, for any natural number $a$ and any prime number $p$. Therefore,

$$
1+2^{p}+3^{p}+\ldots+p^{p} \equiv 1+2+\ldots+p=\frac{p(p+1)}{2}(\bmod p)
$$

If $p$ is an odd number, then $p+1$ is even, from which it follows that $\frac{p(p+1)}{2} \equiv 0(\bmod p)$, meaning that in this case the problem has no solution. For $p=2$, we get $1+2^{2}=5$, so the only solution to the problem is the number 5.