Euclid–Euler theorem

Statement

Let $n$ be an even positive integer. Then $n$ is a perfect number if and only if $n = 2^{p-1} (2^p - 1)$, where $2^p - 1$ is a Mersenne prime.

Remarks

The primes $2^p - 1$ for $p = 2, 3, 5, 7$ give the perfect numbers $6$, $28$, $496$ and $8128$. So the even perfect numbers correspond one to one to the Mersenne primes. The proof uses the multiplicativity of the sum of divisors: $\sigma(2^{p-1} q) = (2^p - 1)(q + 1)$ for an odd prime $q$.

History

Euclid proved in the Elements that $2^{p-1}(2^p - 1)$ is perfect when $2^p - 1$ is prime. Ibn al-Haytham conjectured the converse around 1000 CE. Euler proved the converse in the 18th century, in a paper published after his death.

Explanations

L1 human

Euclid found a recipe for perfect numbers: take a prime of the form $2^p - 1$ and multiply it by $2^{p-1}$. Euler showed that the recipe gives every even perfect number. Whether an odd perfect number exists is still unknown.