Euler equation for an odd perfect number with special exponent $k \\ge 9$ has no solution

Write $\sigma(n)=\sum_{d\mid n} d$ for the sum-of-divisors function.

Let $p$ be a prime with $p \equiv 1 \pmod 4$, let $k$ be a natural number with $k \equiv 1 \pmod 4$ and $k \ge 9$, and let $m$ be an odd natural number with $p \nmid m$. The assertion is that

Since $p^{k}$ and $m^{2}$ are coprime and $\sigma$ is multiplicative, the displayed equation says exactly that $N = p^{k}m^{2}$ satisfies $\sigma(N) = 2N$, i.e. that $N$ is an odd perfect number written in Euler's form with special (Euler) prime $p$ and special exponent $k$. Together with the case $k = 5$, this covers every special exponent $k \equiv 1 \pmod 4$ with $k \ge 5$, that is, the whole complement of the case $k = 1$ predicted by the Descartes-Frenicle-Sorli conjecture.

Formalization Note Perfection is expressed directly through the divisor-sum equation $\sigma(p^{k})\sigma(m^{2}) = 2p^{k}m^{2}$ rather than through Nat.Perfect; the two are equivalent here because $p^{k}$ and $m^{2}$ are coprime and $N = p^{k}m^{2} > 0$ is odd. The congruences are written as p % 4 = 1 and k % 4 = 1, and the sum-of-divisors function as an explicit sum over Nat.divisors.