📂Number Theory

Proof of Euler's Perfect Number Theorem

Theorem ¹

If an even number $n = 2^{p-1} (2^p - 1)$ is a perfect number, then $2^{p}-1$ is a Mersenne prime.

Explanation

At first glance it looks like the converse of the Euclid perfect number formula, but it differs in that it is stated only for even numbers.

However, this theorem tells us almost everything about perfect numbers, because in fact no odd perfect number has ever been found. To this day, essentially the only thing known about odd perfect numbers is that ‘if one exists, it must be very large’.

Proof

Part 1.

Since $n$ is a perfect number, $2n = \sigma (n)$.

Properties of the sigma function: For $\displaystyle \sigma (n) : = \sum_{d \mid n} d$, the following hold.

• [1]: For a prime $p$, $$\sigma ( p^k ) = {{p^{k+1} - 1} \over {p-1}}$$
• [2]: If $\gcd (n , m ) = 1$, then $$\sigma (nm) = \sigma (n) \sigma (m)$$

Setting $n = 2^{k} m$ for some odd number $m$, $$ \begin{align*} 2^{k+1} m =& 2n \\ =& \sigma (n) \\ =& \sigma (2^{k} m) \\ =& \sigma (2^{k} ) \sigma ( m) \\ =& (2^{k+1} - 1) \sigma (m) \end{align*} $$ Rearranging gives $(2^{k+1} - 1) \sigma (m) = 2^{k+1} m$, and since $2^{k+1} -1$ is odd, $\sigma (m)$ is a multiple of $2^{k+1}$. That is, for some $c$ the following holds. $$ \begin{align*} \sigma (m) =& 2^{k+1} c \\ m =& (2^{k+1} - 1) c \end{align*} $$

---

Part 2.

Assuming $c \ge 1$, $$ \sigma (m) \ge 1 + c + m = 1 + c + (2^{k+1} -1) c = 1 + 2^{k+1}c $$ holds. But since $\sigma (m) = 2^{k+1}c$, we would have $2^{k+1}c \ge 1 + 2^{k+1} c$, which is a contradiction, so $c=1$.

---

Part 3.

$$ \sigma (m) = 2^{k+1} = (2^{k+1} - 1) + 1 = m+1 $$

That $m$ has only itself and $1$ as divisors means that $m$ is prime. Since $m = 2^{k+1} - 1$, $m$ is a Mersenne prime.

■