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I'm curious about if the following question is in the literature or what work can be done about it.

Denote the number of distinct primes dividing an odd perfect number $N$ with the arithmetic function $\omega(N)$. Wikipedia has this section dedicated to near-square primes from the article Landau's problems. Under the assumption that there exists an odd perfect number $N$, we denote the number of distinct near-squares primes dividing $N$ as $\omega_{\text{nsp}}(N)$, thus $\omega_{\text{nsp}}(N)\leq \omega(N)$.

Here I add as general reference that a near-square prime is a prime number of the form $p=n^2+1$, that's the sequence A002496 from The On-Line Encyclopedia of Integer Sequences.

Question. I would like to know if we can improve the previous lower bound and/or the upper bound, in this way $$\text{lower bound}<\omega_{\text{nsp}}(N)<\text{upper bound}\tag{1}$$ being these bounds functions of $N$, that's $\text{lower bound}=\text{lower bound}(N)$ and $\text{upper bound}$ denotes a function $\text{upper bound}(N)$. Many thanks.

I can to deduce obvious improvements for particular cases of Touchard's theorem, or when I consider Euler's theorem for odd perfect numbers. Also I know bounds for the radical of an odd perfect number, bounds for the Euler's totient function $\varphi(N)$, and I know the product formula representation for these arithmetic functions.

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  • $\begingroup$ I can not to deduce relevant inequalities of the form $(1)$, I hope that the question is interesting. In past I've edited the post Odd perfect numbers having as prime factors exclusively Mersenne primes and Fermat primes and identifier 362043 , thanking the excellent references that I cited from Mathematics Stack Exchange,, and the answer for this question on MathOverflow. $\endgroup$
    – user142929
    Aug 21, 2022 at 15:42
  • $\begingroup$ I've raised a flag in my question. I've edited the post on Meta Stack Exchange meta.stackexchange.com/questions/381823/… that was deleted by a moderator after of 1 minute. This is for your information. I add it here because I have no reputation to add it neither in the chat of Meta Stack Exchange nor in the chat of MathOverflow since I was suspended in the chat. My belief is that the moderator team of MathOverflow suspended to me the action to raise more flags in the post with identifier 429420 $\endgroup$
    – user142929
    Sep 6, 2022 at 13:21

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In general, very few prime factors in an odd perfect number can be of the form $n^2+1$.

In particular, if N is an odd perfect number then $\frac{\sigma(N)}{N}=2$, and for any $m$ (perfect or not), $\frac{\sigma(n)}{n} \leq \prod_{p|n}\frac{p}{p-1}$ with equality if and only if $n=1$.

Thus, for any perfect number, one needs $ \prod_{p|N}\frac{p}{p-1}>2$. However,

$$\prod_{p, p =i^2+1}\frac{p}{p-1} < \prod_{i=1}^{\infty} \frac{(2i)^2+1}{(2i)^2} < 1.4$$

So, only a small fraction of the primes can be of the form $n^2+1$. Turning this into a bound on $\omega(N)$ directly is going to be very difficult, because one cannot rule out that one has an odd perfect number with a few primes not of this form, and then a lot of very big primes of the form $n^2+1$ that contribute very little to the product.

Note that similar remarks would apply to almost any class of primes of the form $p=Q(n)$ for some polynomial $Q$. There's very little content here dealing with the fact that we're looking at $n^2+1$ at all.

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  • $\begingroup$ Many thanks professor, for your excellent answer. I think that I should to accept the answer in next few days after I've read your remark that turning your reasoning and computations into a bound for $\omega(N)$ is very difficult. $\endgroup$
    – user142929
    Nov 30, 2022 at 9:23

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