Odd Perfect Number

A perfect number is a positive integer, the sum of whose factors(except the number itself) add upto the number.

Example: 28 = 14 + 7 + 4 + 2 + 1

There are about 48 known perfect numbers at the time of writing this article.But none of them are odd. Computers have checked upto $10^{300}$ but could never find one. So I tried my luck disproving their existence.I had a little experience proving this (pefect squares have odd number of factors). But rather than disproving their existence I ended up finding constraints on their existence.

Let $x$ be a odd perfect number.

Then $x$ can be denoted as $x = p^a.q^b.r^c ..….$ where $p, q, r.....$ are prime factors.

Theorem 1: One and only one of the prime factors have an odd power i.e. only one of $a, b, c....$ is odd and all others are even.

Proof: 

All the combination of factors of $x$ can be denoted by varying the powers of the prime factors.

The sum of all the factors of $x$ can then be written as

$s = (p^0 + p^1 + p^2 + ....p^a)(q^0 + ..... q^b)(r^0 + ...… r^c).....$

By defenition of perfect number we have

$x = (p^0 + p^1 + p^2+...p^a)(q^0 + ... q^b)(r^0+...r^c)...-p^a.q^b.r^c....$

$x = (1 + p + p^2 + ....p^a)(1 + ..... q^b)(1 + ...… r^c)..... - x$

$2x = (1 + p + p^2 + ....p^a)(1 + ..... q^b)(1 + ...… r^c).....$

The left side of the equation has a 2 multiplied by a odd number. So the right should have one and only one even multiple.

Prime number is a odd number. An odd number raised to any positive integer power is also odd (3x3=9).

This implies if $a$ is odd then $(1 + p + p^2 + ....p^a)$ is even. Thus all other terms are odd and $q, r, .....$ are even.

Theorem 2: The only odd power is of the form $4m+1$ and the prime factor is of the form $4k+1$ where $m$ is a non negative integer and $k$ a positive integer

Proof:

From the previous proof we have 

$2x = (1 + p + p^2 + ....p^a)(1 + ..... q^b)(1 + ...… r^c).....$

$2x = (1 + p + p^2 + ....p^a).odd$

Since x is also an odd number $(1 + p + p^2 + ....p^a)$ is only once divisible by 2 and should be of the form $2(2n+1)$ where n is a non negative integer.

$2(2n+1) = (1 + p + p^2 + ....p^a)$

$2(2n+1) = \frac{p^{a+1}-1}{p-1}$

$2(2n+1) = \frac{(1+p-1)^{a+1}-1}{p-1}$

$2(2n+1) = \frac{\binom{a+1}{0}+\binom{a+1}{1}(p-1)+...+\binom{a+1}{a+1}(p-1)^{a+1}-1}{p-1}$

$2(2n+1) = \frac{(a+1)(p-1)+\binom{a+1}{2}(p-1)^2...+(p-1)^{a+1}}{p-1}$

$2(2n+1) = (a+1)+\binom{a+1}{2}(p-1)...+(p-1)^a$

Since $(p-1)$ is even all the terms in the right except the 1st and the 2nd has at least $2^2$ as factor.

So for rhs to be of the form $2(2n+1)$

$(a+1)=2(2m+1) => a = 4m+1$

and

$(p-1)=4k => p=4k+1$

To show the above statement true

$rhs = 2(2m+1 + (2m+1)a(2k) + even)$

$= 2(1 + even)$

$= 2(2n+1)$
$= lhs$

Conclusion:
$x=(4k+1)^{4m+1}.O^2$
Where O is a odd positive integer, m a whole number and k a natural number.

Please comment for any clarification.

Update: This result though in a different form has already been found out by Euler.