I have been thinking about odd perfect numbers and came upon an argument which must be invalid because it is too simple.
I would be very interested in understanding what is wrong with this argument.
Let:
- $s(x)$ be the aliquot sum for $x$.
- $p$ be an odd prime.
(1) If $p | x$, then $s(px) = s(x) + ps(x)$
(2) If $p \nmid x$, then $s(px) = s(x) + ps(x) + x$
(3) Assume that $px$ is an odd perfect number such that $s(px) = px$
(4) If $p | x$, then $px = s(x)(p+1)$ which is impossible since $p+1$ is even but $px$ is odd.
(5) If $p \nmid x$, then $px = s(x) + ps(x) + x$ and $x(p-1) = s(x)(p+1)$
(6) There exists integers $a$ and $b$ such that $2^a | (p-1)$ but $2^{a+1} \nmid (p-1)$ and $2^b | (p+1)$ but $2^{b+1} \nmid (p+1)$
(7) $a \ne b$ since $4 | (p-1)(p+1)$ but $4 | (p-1)$ if and only if $4 \nmid (p+1)$ since $4 \nmid (4a-2)$ and $4 \nmid (4a+2)$
(8) $a > b$ since $x$ is odd.
(9) So, $s(x)$ is even
(10) $s(x)$ is only even if $x$ is not squarefree since:
- Base Case: if $x$ is a prime then $s(x)=1$
- Assume that if $x$ is squarefree, then $s(x)$ is odd.
- Inductive Case: $s(px) = s(x) + ps(x) + x = $ odd + odd + odd = odd.
(11) If $x$ is not squarefree, there exists an prime $q$ where $q^2 | x$.
(12) But this is impossible, since we would have the case where $px = s\left(q(p\dfrac{x}{q}\right) = s\left(p\dfrac{x}{q}\right) + qs\left(p\dfrac{x}{q}\right) = s\left(p\dfrac{x}{q}\right)(q + 1)$ where $qx$ is odd but $(q+1)$ is even.
Assumptions (1) fails in general. Pick $x=p^2$. Then $s(x)=1+p$, $s(px)=1+p+p^2$, so $s(xp)\neq s(p)+ps(x)$. I think it only holds if $x$ divides $p$ exactly once.