For odd p the k=5 divisor sum factors into the two cyclotomic terms

For every odd natural number $p$, $2 \cdot (p^2+p+1) \cdot \left(\frac{p+1}{2}\right)(p^2-p+1) = \sigma(p^5)$. This is the exact factorisation of the first Dris equation on the $k=5$ branch: writing $U = p^2+p+1$ and $V = \frac{p+1}{2}(p^2-p+1)$ gives $\sigma(p^5) = 2UV$, so that equation reads $m^2 = d^2 \cdot (qUV)$ once the explicit square factor is cancelled. The identity is the polynomial one $\sigma(p^5) = 1+p+\dots+p^5 = (1+p+p^2)(1+p^3)$, and the oddness hypothesis is exactly what makes the Nat division $(p+1)/2$ exact rather than a truncation. The hypothesis is not cosmetic: for even $p$ the Nat quotient truncates and the identity fails, for instance $p=4$ gives $1092 \neq 1365$.