Proof that odd perfect numbers cannot consist of single unique factors? I'm a high school student, so please point out my mistakes nicely :)
So we already know odd perfect numbers cannot be in the form of a square, but how about that they cannot be in this form: $$P=abcd...$$ where P, the odd perfect number, equals the product of unique single factors, i.e. a is not b, b is not c, etc. 28, having a prime factorization of $$1 * 2 * 2 * 7$$ is not only not an odd perfect number but has a non-unique factor 2.
So how do we prove that odd perfect numbers cannot have such factors? From the formula for finding the sum of all divisors of a number here, we can deduce the following: $$(a+1)(b+1)(c+1)(d+1)...=2P.$$ Now, are the factors of an odd number even? Of course not. Therefore the factors a, b, c, d, etc, must be odd. Which means we can get the following: $$(E)(Q)(R)(S)...=2P$$ where e, q, r, s,... are even numbers. If there are n factors in P, then: $$\frac{(E)(Q)(R)(S)}{2^n}=\frac{2P}{2^n}$$ therefore $$\frac{(E)(Q)(R)(S)}{2^{n-1}}=\frac{P}{2^{n-1}}$$ Evidently,
impossible.
Well, unless P only had one divisor, but... then P would equal 1. :P
So please point out any mistakes nicely, please. Thanks! :D
 A: I think you have the right idea. However, the exposition is insufficiently clear. Let $N$ be an odd perfect number. We show that $N$ is divisible by a perfect square greater than $1$.
Suppose to the contrary that 
$$N=p_1 p_2\cdots p_n,$$
where the $p_i$ are distinct primes.
Then
$$2N=(p_1+1)(p_2+1)\cdots (p_{n}+1).$$
This is impossible of $n\gt 1$. For $2^n$ divides the right-hand side, while the highest power of $2$ that divides $2N$ is $2^1$.
We conclude that $n=1$, that is, $N$ is prime. That is impossible, since the sum of the divisors of $N$ would then be $N+1$.
A: If there are n unique prime factors in P, then (as you noticed)
$(1)\ \ \ E_1 . E_2 . ... . E_n = 2P$,
where $E_i = p_i + 1$, and where $p_i$ is the i-th
(unique) prime factor of P (our perfect odd number).     
Now let's denote $M_i = E_i / 2$. The numbers $M_i$ are integers obviously.
From (1) after dividing by $2^n$, we get:
$(2)\ \ \ M_1 . M_2 . ... . M_n = P / (2^{n-1})$
OK, if n > 1, the RHS is not an integer (as P is odd) while the LHS is an integer.
The case n=1 can be considered entirely separately, it's trivial. 
