For $p \equiv 5 \pmod{12}$ the prime $3$ is a quadratic nonresidue modulo $p$

If $p$ is a prime with $p \equiv 1 \pmod 4$ and $p \equiv 2 \pmod 3$, equivalently $p \equiv 5 \pmod{12}$, then $3$ is a quadratic nonresidue modulo $p$.

By quadratic reciprocity, since $p \equiv 1 \pmod 4$ the symbol is symmetric, so $\left(\frac{3}{p}\right) = \left(\frac{p}{3}\right)$. Euler's criterion gives $\left(\frac{p}{3}\right) = p^{(3-1)/2} = p \bmod 3 = -1$, because $p \equiv 2 \equiv -1 \pmod 3$.

Consequence for the $k=5$ two-prime residual. The proved theorem sigma_source_of_p_is_fourth_power_residue says that any prime $t$ supplying $p$ in the second Dris equation satisfies $t^{(p-1)/4} = 1$ in $\mathbb{Z}/p$, so $t$ is in particular a quadratic residue modulo $p$. Taking $t = 3$ contradicts this theorem, and therefore

This matters because the first Dris equation forces $3 \mid m$ in the normalised two-prime branch, so without this exclusion the prime $3$ would look like a candidate supplier of the $p$-part of $\sigma(m^2) = p^5 s$. It removes the prime $3$ from the incoming $p$-valuation budget entirely, without ever claiming that $3 \nmid m$.