If an odd perfect number exists, does it have exactly one prime factor of the form $4a+1$?

Question

I know that if an odd perfect number exists it must be of the form $p^kQ^2$ with $\gcd(p, Q) =1$ and $p \equiv k \equiv 1 \mod 4$.

Reading the book The man who only loved numbers by Paul Hoffman, it is stated in the second chapter that if there exists an odd perfect number, then it must have exactly one prime divisor of the form $4a+1$.

I haven't found anything about this fact online, nor a way of proving it by myself.

Is it a true statement or a mistake in the book?

Answer 1

If $N = p^k m^2$ is an odd perfect number with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$, and if we let $$m = \prod_{i=1}^{\omega(m)}{{r_i}^{\alpha_i}}$$ where the $r_i$'s are primes and the $\alpha_i$'s are positive integers, and where $\omega(m)$ is the number of distinct prime factors of $m$, then it is currently unknown whether $r_i \equiv 3 \pmod 4$, $\forall i$.

It is known however, that if $N = p^k m^2$ is an odd perfect number, $m$ has the prime factorization above, and $r_i \equiv 3 \pmod 4$ for all $i$, then $\sigma(p^k)/2$ is a composite number (P. Starni, On the Euler’s factor of an odd perfect number, J. Number Theory, 37 (1991), 366-369).

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Edit: Added January 09, 2023 - 13:29 PM (Manila time)

Chen and Luo proved in Odd Multiperfect Numbers that if $N = p^k m^2$ is an odd perfect number with special prime $p$, and if we write $$m^2 = \prod_{i}{{p_i}^{2\beta_i}}\prod_{j}{{q_j}^{2\gamma_j}}$$ where $p_i \equiv 1 \pmod 4$ and $q_j \equiv 3 \pmod 4$, then $$\sigma\left({q_j}^{2\gamma_j}\right) \equiv 1 \pmod 4.$$ Hence, we have $$\sigma(m^2)=\prod_{i}{\sigma\left({p_i}^{2\beta_i}\right)}\prod_{j}{\sigma\left({q_j}^{2\gamma_j}\right)} \equiv \prod_{i}{\left(2\beta_i + 1\right)} \pmod 4.$$ Clearly, if $\beta_i = 0$ for all $i$, that is $$m^2 = \prod_{j}{{q_j}^{2\gamma_j}},$$ then $\sigma(m^2) \equiv 1 \pmod 4$, and this implies that $p \equiv k \pmod 8$.