# can't find the proof for the form of odd perfect numbers

I am currently researching perfect numbers and on the wikipedia page, as well as this paper:

http://www.math.dartmouth.edu/~jvoight/articles/opn-mass-rev-060211.pdf

it states that any odd perfect number N must be of the form: $$N=q^{\alpha} p_1^{2e_1} \cdots p_k^{2e_k},$$ where:

• q, p1, ..., pk are distinct primes
• q ≡ α ≡ 1 (mod 4)

however I can not find the proof of this anywhere (allegedly proven by Euler), does anybody know it/ can find it?

• Well, at least I haven't seen an odd perfect number not of this form yet :) – Hagen von Eitzen May 6 '14 at 20:31

Let $N=\prod_k p_k^{m_k}$ with all $p_k$ odd. Then $2N=\sigma(N)=\prod_k(1+p_k+p_k^2+\ldots +p_k^{m_k})$. Note that $$1+p+\ldots +p^{m}\equiv\begin{cases}m+1\pmod 4& p\equiv 1\pmod 4\\ 0\pmod 4&2\not\mid m,p\equiv-1\pmod 4\\ 1\pmod 4&2\mid m,p\equiv-1\pmod 4\end{cases}$$ Since $2N$ is not a multiple of $4$, there must not be any factor that os $\equiv 0\pmod 4$ and exactly one that is $\equiv 2\pmod 2$. Thus the cases $2\not\mid m$, $p\equiv -1\pmod 4$ and $m\equiv 3\pmod 4, p\equiv 1\pmod 4$ are forbidden; there must be exactly one case of $m\equiv 1\pmod 4$, $p\equiv 1\pmod 4$. The remaining cases, i.e. $m\equiv 0\pmod 2$, $p\equiv 1\pmod 4$ and $2\mid m$, $p\equiv 1\pmod 4$ can occur arbitrarily often.