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Let $\sigma(x)$ be the sum of the divisors of $x$. So, for example, $\sigma(6) = 1 + 2 + 3 + 6 = 12$.

A positive integer $N$ is said to be perfect if $\sigma(N) = 2N$.

Consider the equation $$\sigma(N_1) + \sigma(N_2) = 2(N_1 + N_2).$$

Clearly, if the equation is true, then we have the biconditional

$N_1$ is perfect $\Longleftrightarrow$ $N_2$ is perfect.

This begs the question:

Do there exist perfect numbers $N_1$ and $N_2$ such that $$\sigma(N_1 + N_2) = \sigma(N_1) + \sigma(N_2)?$$

(This question was first considered by Antalan in On The Solutions of Two Sum of Divisor Equations.)

If there do not exist any perfect numbers $N_1$ and $N_2$ satisfying this last equation, is it possible to prove this rigorously?

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Well, let's see. Even perfect numbers are all of the form $2^{n-1}(2^n-1)$, where $2^n-1$ is a Mersenne prime. As such, they are completely determined by the highest power of 2 dividing them. This value in a sum of two such numbers will be the same as in the smallest of them. Really, let $N_1=2^{n_1-1}(2^{n_1}-1) < N_2=2^{n_2-1}(2^{n_2}-1),\;n_1<n_2$. Clearly, $N_1+N_2$ is divisible by $2^{n_1-1}$ and gives an odd number. Thus its highest power of 2 is the same as in $N_1$, but it is not $N_1$ itself, hence it can't be perfect.

As for the odd perfect numbers, I'd rather wait till I see one of them.

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    $\begingroup$ "This value in a sum of two such numbers will be the same as in the smallest of them." -- Care to elaborate a bit on that statement? $\endgroup$ Oct 7, 2015 at 15:33
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    $\begingroup$ Ohh okay. So for a concrete example: $N_1 = 6$ and $N_2 = 28$ are both (even) perfect numbers. $N_1 + N_2 = 34 = 2\cdot{17}.$ $\endgroup$ Oct 7, 2015 at 15:41
  • $\begingroup$ Can you include your last comment in your answer, so that I may be able to accept it? Thanks! $\endgroup$ Oct 7, 2015 at 15:41

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