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.*](https://en.wikipedia.org/wiki/Landau%27s_problems#Near-square_primes)
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.



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> **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.**
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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.



Number of distinct near-squares primes dividing an odd perfect number
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.


