Euler equation for an odd perfect number with special exponent $k = 5$ has no solution

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

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

Since $p^{5}$ and $m^{2}$ are coprime and $\sigma$ is multiplicative, the displayed equation says exactly that $N = p^{5}m^{2}$ 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 $k = 5$. The statement is therefore the special-exponent-$5$ case of the odd perfect number conjecture, isolated from the general case $k \equiv 1 \pmod 4$, $k \ge 5$.

Formalization Note Perfection is expressed directly through the divisor-sum equation $\sigma(p^{5})\sigma(m^{2}) = 2p^{5}m^{2}$ rather than through Nat.Perfect; the two are equivalent here because $p^{5}$ and $m^{2}$ are coprime and $N = p^{5}m^{2} > 0$ is odd. The sum-of-divisors function is written as an explicit sum over Nat.divisors.