For prime p, sigma(p^5) = 2(p^2+p+1)((p+1)/2)(p^2-p+1)

For an odd prime $p$, the divisor sum of $p^5$ factors exactly as $\sigma(p^5) = 2UV$ with $U = p^2+p+1$ and $V = \frac{p+1}{2}(p^2-p+1)$. Since $p$ is an odd prime, $(p+1)/2$ is an exact integer and $\sigma(p^5) = (p^6-1)/(p-1) = (p+1)(p^2+p+1)(p^2-p+1)$, so the displayed identity holds. This is the correct statement of the factorisation: it requires $p$ prime. The primality hypothesis is necessary, since for a composite base such as $p = 9$ one has $\sigma(9^5) = 88573 \ne 2UV = 66430$, because $p^5$ then has divisors that are not powers of $p$.