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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

Community
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HyperLuminal
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  • @peter.petrov Sorry, please look at my edit. – HyperLuminal May 28 '15 at 22:13
  • The second time you denote by E,Q,R,S different things than the first time. But now it looks more or less OK. I understand it now. The LFH is an integer while the RHS is not unless n=1. Seems correct now, why do you think it's not correct? – peter.petrov May 28 '15 at 22:15
  • No, I think it's correct :P Is this like not exactly ground breaking or something? – HyperLuminal May 28 '15 at 22:19
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    I don't think there is anything wrong with your demonstration. In fact, Euler demonstrated that any odd perfect number $P$ must be of the form $p^{4m+1} Q^2$, where $\lambda$ is a prime of the form $4n+1$. This implies (among other things) that $P$ cannot be square-free, as you yourself have also shown. This is no mean accomplishment, mind you, but it is already known. – Brian Tung May 28 '15 at 22:22
  • You should continue thinking along these lines and see what else you can come up with! – Brian Tung May 28 '15 at 22:23
  • @BrianTung So this is like an extra proof? Also has anyone proven anything about a perfect cube odd perfect number? – HyperLuminal May 28 '15 at 22:24
  • I removed an earlier hasty response. But do look here for more information about what's been shown about odd perfect numbers (should they exist): http://mathworld.wolfram.com/OddPerfectNumber.html – Brian Tung May 28 '15 at 22:39
  • @JessePFrancis Yea I know. – HyperLuminal Jun 02 '15 at 22:56

2 Answers2

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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 $N+1$.

André Nicolas
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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.

peter.petrov
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