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Preamble - This question is an offshoot from the following earlier questions here at MSE:

Can an odd perfect number be divisible by 825?

Can an odd perfect number be divisible by 165?

Odd perfect number divisors

My own question is the following:

Can an odd perfect number be divisible by $101$?

Here is my own (quick) attempt at a (partial) answer:

Let $N = {q^k}{n^2}$ be an odd perfect number given in Eulerian form (i.e. $q$ is prime with $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$). Also, let $$N = \prod_{i=1}^{\omega(N)}{{p_i}^{\alpha_i}}$$ be the canonical (i.e., prime) factorization of $N$.

If $q = 101$, then since $$(q+1)\mid\sigma(q^k) \mid \sigma(N)=2N,$$ then we have $$3\cdot17 = 51 = \frac{q+1}{2} \mid N.$$

Thus, it follows that $$3 \mid N \Longrightarrow 3^2 \mid N,$$ and $$17 \mid N \Longrightarrow {17}^2 \mid N.$$

Now test: (Note that $\alpha_i \geq \beta_i, \forall i$.) $$2 = I(N) \geq \prod_{i=1}^{\omega(N)}{\left(1 + \frac{1}{p_i} + \ldots + \frac{1}{{p_i}^{\beta_i}}\right)} \geq \frac{102}{101}\cdot\frac{13}{9}\cdot\frac{307}{289} = \frac{407082}{262701} \approx 1.549602019.$$

This is as far as I could go.

Of course, even after establishing $q \neq 101$, we still have to consider the remaining case $101 \mid n \mid N$, whence it follows that ${101}^2 \mid n^2 \mid N$.

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Cohen and Sorli has a paper in Integers (published 2012) titled "On Odd Perfect Numbers and Even 3-Perfect Numbers" where they list down some necessary conditions for a prime $q \equiv 1 \pmod 4$ to be the Euler prime of an odd perfect number $N = {q^k}{n^2}$. –  Jose Arnaldo Dris Aug 15 '13 at 6:59
    
(Pure pun intended here: If you wish to [quickly] get to the page in Cohen and Sorli's paper where a list of possible Euler primes is presented, just hit CTRL-F and search for $101$.) =) –  Jose Arnaldo Dris Aug 15 '13 at 7:00
1  
What's the pun? –  ShreevatsaR Aug 15 '13 at 7:08
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