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  "body": "\n## Proof\n\nLet $\\sigma$ be the [[sum of divisors|009QCP5T3Z]]. A positive integer $n$ is [[perfect|0SB7Q8SYSF]] exactly when $\\sigma(n) = 2n$.\n\n1. For $k \\ge 0$, the divisors of $2^k$ are $1, 2, \\dots, 2^k$, by the [[fundamental theorem of arithmetic|084HH1NDCB]]. So $\\sigma(2^k) = 1 + 2 + \\dots + 2^k = 2^{k+1} - 1$.\n2. For a [[prime|0SN0MAEKYX]] $q$, the divisors of $q$ are $1$ and $q$, so $\\sigma(q) = q + 1$.\n3. An odd number and a power of $2$ have no common prime factor, so they are [[coprime|0GFKP8F8AG]]. Then [[multiplicativity of $\\sigma$|04WC36XKV6]] gives $\\sigma(2^k m) = \\sigma(2^k) \\sigma(m)$ for odd $m$.\n\n**If $n = 2^{p-1}(2^p - 1)$ with $q = 2^p - 1$ a [[Mersenne prime|06GPJF9V7M]], then $n$ is perfect.**\n\n4. Since $q$ is prime, $q \\ge 2$, so $p \\ge 2$. Also $q$ is odd.\n5. By steps 1 to 3, $\\sigma(n) = \\sigma(2^{p-1}) \\sigma(q) = (2^p - 1)(q + 1) = q \\cdot 2^p = 2n$. So $n$ is perfect.\n\n**If $n$ is an even perfect number, then $n$ has this form.**\n\n6. Let $n$ be even and perfect. Write $n = 2^k m$ with $k \\ge 1$ and $m$ odd.\n7. By step 3 and step 1, $\\sigma(n) = (2^{k+1} - 1)\\sigma(m)$. Since $\\sigma(n) = 2n$, we get $(2^{k+1} - 1)\\sigma(m) = 2^{k+1} m$.\n8. The numbers $2^{k+1} - 1$ and $2^{k+1}$ are coprime, since a common divisor divides their difference $1$.\n9. So $2^{k+1} - 1$ divides $m$, by [[the generalized Euclid's lemma|03BAREW1Z5]]. Write $m = (2^{k+1} - 1) d$ with $d$ a positive integer.\n10. Then step 7 gives $(2^{k+1} - 1)\\sigma(m) = 2^{k+1}(2^{k+1} - 1) d$, so $\\sigma(m) = 2^{k+1} d = m + d$.\n11. Since $k \\ge 1$, we have $2^{k+1} - 1 \\ge 3$, so $d \u003c m$. Thus $m$ and $d$ are two distinct divisors of $m$.\n12. The sum $\\sigma(m)$ of all divisors of $m$ is $m + d$. So every divisor of $m$ is $m$ or $d$, since any other divisor would add a positive term.\n13. The number $1$ divides $m$, and $1 \\ne m$ because $m \\ge 3$. So $d = 1$, and the only divisors of $m$ are $1$ and $m$.\n14. Hence $m = 2^{k+1} - 1$ is prime. Put $p = k + 1$. Then $n = 2^{p-1}(2^p - 1)$, and $2^p - 1$ is a Mersenne prime.\n\n## Idea\n\nThe sum of divisors is multiplicative, so it splits over the power of $2$ and the odd part. For an even perfect number this forces $2^{k+1} - 1$ to divide the odd part $m$, and then $\\sigma(m) = m + d$. A number whose divisors add up to only two of them must be prime.\n",
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