To let $ sigma (x) $ denote the *Sum of divisors* the positive integer $ x $, The *frequency Index* from $ x $ is then given by the formula $ I (x) = sigma (x) / x $,

A number $ N in mathbb {N} $ it is said that *Perfect* if $ sigma (N) = 2N $, Equivalent, if $ N $ is perfect $ I (N) = 2 $,

Euler showed that one *odd perfect number* $ M $, if available, must necessarily have the form $ M = p ^ k m ^ 2 $, from where $ p $ is the *special prime number* satisfying $ p equiv k equiv 1 pmod 4 $ and $ gcd (p, m) = 1 $,

Here is my question:

What makes it so difficult to prove that? $ 3 $ does not share an odd perfect number?

**MY ATTEMPT**

So leave it now $ M = p ^ k m ^ 2 $ be an odd perfect number with a special prime $ p $, Let's assume the opposite $ 3 mid $,

Since $ 3 equiv 3 pmod $ 4, then $ p neq 3 $, (Actually, $ p $ is prime with $ p equiv 1 pmod 4 $ implies that $ p geq $ 5).

So that means that if $ 3 mid $, then $ 3 ^ 2 mid n ^ 2 mid $, Accept that $ 3 ^ 2 || M $ (It means that $ 3 ^ 2 mid $ but $ 3 ^ 4 nmid M $). This implies that $ 13 = 1 + 3 + 3 ^ 2 = sigma (3 ^ 2) mid sigma (M) = 2M $, which means, that $ 13 mid $, This implies that

$$ I (3 ^ 2) I (13) leq I (M), $$

provided that $ p = 13 $, or

$$ 1.60673 approx frac {9507} {5917} = I (3 ^ 2) I ({13} ^ 2) I ({61} ^ 2) I ({97} ^ 2) leq I (M) = 2 $$

assumed $ {13} ^ 2 || M $ implies that

$$ 3 cdot {61} = 183 = 1 + 13 + {13} ^ 2 = sigma ({13 ^ 2}) mid sigma (M) = 2M $$

and

$$ 3 cdot {13} cdot {97} = 3783 = 1 + 61 + {61} ^ 2 = sigma ({61} ^ 2) mid sigma (M) = 2M, $$

where is there no contradiction.

**postscript**

I dare assume that an odd perfect number is not divisible *everything* the odd primes below $ 100 (except for $ 3 $. $ 5 $, and $ 7 $ – since we already know that $ 105 nmid M $). However, I have no proof. (Of course, I know that all it takes is a simple calculation.)