Suppose $n$ is an odd perfect number then it exists $p$ such that $\frac{n}{p}$ is a square

elementary-number-theoryperfect numbers

Suppose $n$ is an odd perfect number.

How to show that it exists a prime number $p$ such that $\frac{n}{p}$ is a square number?

My idea was:

$n$ is perfect if $\sigma(n)=2n$.

Let $n=2k+1, \ k \in \mathbb{N_0}$.

Then it's $\sigma(n)=\sigma(2k+1)=\sum \limits_{d \vert n}d=1+…+d_{n-1}+d_n=2n=2(2k+1)$ with $d_1,…,d_n$ odd.

So $\frac{1+…+d_{n-1}}{p}
\Leftrightarrow\frac{2n}{2p}$.

But I don't know how to continue to show that such a $p$ exists, such that $\frac{n}{p}$ is a square number.

How can it be shown?

Best Answer

Let $n$ be an odd perfect number. We write it in the prime factor decomposition $$ n= \prod_{1\le k\le s}p_k^{a(k)}\ , $$ where there are $s$ prime numbers involved, $p_1,p_2,\dots ,p_s$, respectively to the powers $a(1),a(2),\dots,a(s)$. We build the sum of the divisors to get $$ \begin{aligned} 2n&= \sigma(n)\\ &= \sigma\left(\prod_{1\le k\le s}p_k^{a(k)}\right) \\ &= \prod_{1\le k\le s}\sigma\left(p_k^{a(k)}\right) \\ &= \prod_{1\le k\le s}\left(1+p_k+\dots+p_k^{a(k)}\right) %\\ %&= %\prod_{1\le k\le s}\frac{p_k^{a(k)-1}}{p_k-1} \ . \end{aligned} $$ What can we say about the factors in the last product. Exactly one of them is even. (And this one factor is not divisible by $4$.) So we have to examine what happens with the expression $$ 1+p+\dots+p^a $$ for an odd prime $p$, and a power $a>0$.

  • If $a$ is even, we have an odd number of terms, so the result is an odd number.
  • If $a$ is odd, we have an even number of terms, so the result is an even number.

In our case we have exactly one even such factor, so we can already get the needed information for the OP, namely that all powers among $a(1), a(2),\dots,a(s)$ are even numbers, except for one of them, which is odd. This odd prime is the $p$ in the OP.

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