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If $\sigma(n)$ is the sum of all divisors of $n$ , we want to find integers $r$ such that $\sigma(pr) = 2pr + 992$ for all primes $p$ greater than $500$ . If $r$ is such a number then, by choosing $p$ large enough, we can assume that $p$ and $r$ are coprime. Since $\sigma$ is a multiplicative function, it follows that $(p+1)\sigma(r) \; = \; \sigma(p)\sigma(r) \; = \; \sigma(pr) \; = \; 2pr + 992$ and hence $p(\sigma(r) - 2r) \; = \; 992 - \sigma(r) \;.$ Since this identity must be true for all large enough $p$ , and so certainly for $p$ greater than $|992 - \sigma(r)|$ , we deduce that $\sigma(r) - 2r = 0$ and $992 - \sigma(r) = 0$ . Since $496$ is perfect, $r=496$ is the only solution of these equations, and so is such that $\sigma(pr) = 2pr + 992$ for all primes $p$ coprime to $r=496$ .

Since any prime $p$ greater than $500$ must be coprime to $496$ , we see that $496$ does indeed satisfy the equation, and so is the only solution. The sum of all solutions is, therefore, $496$ .

Let $r_1, r_2, r_3 \ldots$ be the factors of $r$ less than $r$ . Since $p$ is prime,

$S(rp)=r_1p+r_2p+r_3p+ \ldots +r+r_1+r_2+r_3+ \ldots$

$S(rp)-rp=(r_1p+r_2p+r_3p+ \ldots -rp) +(r+r_1+r_2+r_3+ \ldots)$

$S(rp)-rp=p(r_1+r_2+r_3+ \ldots -r) +(r+r_1+r_2+r_3+ \ldots)$

This gives us $r_1+r_2+r_3+ \ldots =r$

Therefore, $S(rp)-rp=p((r_1+r_2+r_3+ \ldots) -r) +(r+r_1+r_2+r_3+ \ldots)$

$S(rp)-rp=p(r-r) +(r+r_1+r_2+r_3+ \ldots)$

$S(rp)-rp=(r+(r_1+r_2+r_3+ \ldots))$

$S(rp)-rp=r+r$

$S(rp)-rp=2r$

$992=2r$

$r=496$