2. a) Calculate

$$
(a-b)\left(a^{4}+a^{3} b+a^{2} b^{2}+a b^{3}+b^{4}\right)
$$

b) Using the equality obtained in a), determine a prime number $p$ such that $2 p^{2}+1=k^{5}$, where $k$ is a natural number.
Solution. a) We have:

$$
\begin{aligned}
& (a-b)\left(a^{4}+a^{3} b+a^{2} b^{2}+a b^{3}+b^{4}\right)= \\
& \quad=a^{5}+a^{4} b+a^{5} b^{2}+a^{2} b^{3}+a b^{4}-a^{4} b-a^{3} b^{2}-a^{2} b^{3}-a b^{4}-b^{5} \\
& \quad=a^{5}-b^{5}
\end{aligned}
$$

b) From $2 p^{2}+1=k^{5}$ we get $2 p^{2}=k^{5}-1$, so according to a) we have

$$
2 p^{2}=(k-1)\left(k^{4}+k^{3}+k^{2}+k+1\right)
$$

Since $2 p^{2}=k^{5}-1$, we get that $k^{5}$ is an odd number, which means that $k$ is an odd number, for example $k=2 n+1$ and $k>1$, so $n \in \mathbb{N}$. Now by substitution in (1) and after canceling with 2 we have $p^{2}=n\left(k^{4}+k^{3}+k^{2}+k+1\right)$. Since $p$ is a prime number, the last equality is possible only if $n=1$ or $n=k^{4}+k^{3}+k^{2}+k+1$. Since $k=2 n+1$, i.e., $k>n$, it is not possible that $n=k^{4}+k^{3}+k^{2}+k+1$. Therefore, $n=1$, and from there $k=3$. By substituting $k=3$ we get $p^{2}=3^{4}+3^{3}+3^{2}+3+1=121$. The required number is $p=11$.