$\sigma(p^k) \ne 2m^2$ when $6 \mid k+1$

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

Let $p$ be an odd prime and let $k$ be a natural number with $6 \mid k+1$. Then the equation

has no solution in natural numbers $m$.

Equivalently, $\tfrac{1}{2}\sigma(p^{k}) = \tfrac{1}{2}\bigl(1+p+\cdots+p^{k}\bigr)$ is never a perfect square when $6 \mid k+1$.

Why this is relevant to odd perfect numbers. If $N = p^{k}m^{2}$ is an odd perfect number in Euler form ($p$ prime, $p \equiv k \equiv 1 \pmod 4$, $p \nmid m$), then multiplicativity of $\sigma$ gives the Euler equation $\sigma(p^{k})\sigma(m^{2}) = 2p^{k}m^{2}$. Since $p \nmid \sigma(p^{k})$ and $k$ is odd, one gets $\tfrac{1}{2}\sigma(p^{k}) \mid m^{2}$, and writing $2m^{2} = \sigma(p^{k})\,s$, $\sigma(m^{2}) = p^{k} s$ shows that $m^{2} \le p^{k}$ forces $s = 1$, i.e. $2m^{2} = \sigma(p^{k})$. The statement above therefore rules out $m^{2} \le p^{k}$ whenever $6 \mid k+1$; for the special exponents $k \equiv 1 \pmod 4$ with $k \ge 5$ this covers $k \equiv 5 \pmod{12}$, in particular the first case $k = 5$.

Proof idea. Writing $k+1 = 6w$ and $y = p^{w}$, the geometric series identity factors $\sigma(p^{k}) = \sigma(p^{2w-1})\,(y^{4}+y^{2}+1)$, where $\sigma(p^{2w-1})$ is even and divides $y^{2}-1$. Hence $2m^{2} = \sigma(p^k)$ becomes

The two quadratic factors are coprime, and each has greatest common divisor with $y^{2}-1$ dividing $3$; since $3$ cannot divide both, one of them is coprime to the remaining product and must be a perfect square. That is impossible because $y^{2} < y^{2}+y+1 < (y+1)^{2}$ and $(y-1)^{2} < y^{2}-y+1 < y^{2}$.