Parity bridge: the cofactor $m$ of the packaged Euler configuration is odd

Euler's form of a hypothetical odd perfect number $N = p^k m^2$ is frequently packaged into the three identities

together with the Euler congruences $p \equiv k \equiv 1 \pmod 4$. Many statements about this configuration also carry the hypothesis that the cofactor $m$ is odd. This theorem says that this hypothesis is redundant: as soon as $m \neq 0$, the three identities and the two congruences already force $m$ to be odd.

The argument is $2$-adic. The divisor sum of a nonzero perfect square is always odd (for an even square the $2$-part contributes $2^{2a+1}-1$, and every odd prime contributes a sum of an odd number of odd terms). Since $p \equiv 1 \pmod 4$ we have $\sigma(p^k) \equiv k+1 \equiv 2 \pmod 4$, so $t$ is odd. Multiplying the last two identities gives $\sigma(m^2)\,t = p^k m^2$, whose left-hand side is a product of two odd numbers; as $p$ is odd, the right-hand side can be odd only if $m$ is odd.