Odd abundant number question

[hv at crypt.org]

Earlier I wrote:
:Hmm, now I'm not so sure. Note that the abundance of n is (sigma(n) - 2n).
:Now given m pseudoperfect, n prime to m:
:  sigma(mn) - 2mn = sigma(m) . sigma(n) - 2mn
:      = n . (sigma(m) - 2m) + sigma(m) . (sigma(n) - n)
:
:Since m is pseudoperfect, (sigma(m) - 2m) is expressible as a sum of
:the proper divisors of m, and (sigma(n) - n) is precisely the sum of
:the proper divisors of n, so mn is also necessarily pseudoperfect (and
:so not weird).

Silly me, I missed the forest for the trees.

If {a_1, a_2, ... a_k} are proper divisors of m that sum to m, then
{na_1, na_2, ... na_k} are proper divisors of mn that sum to mn.
So any multiple of a pseudoperfect number is trivially pseudoperfect.

Hugo