📂Number Theory

Euler's Proof of the 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 might seem like the inverse of the Euclid’s perfect number formula, but the difference is that it only mentions even numbers.

However, this theorem tells almost everything about perfect numbers, mainly because no odd perfect number has been discovered yet. The only known fact about odd perfect numbers up to now is that ‘if they exist, they must be very large’.

Proof

Part 1.

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

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

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

For some odd $m$, if we set $n = 2^{k} m$, $$\begin{align*} 2^{k+1} m =& 2n \\ =& \sigma (n) \\ =& \sigma (2^{k} m) \\ =& \sigma (2^{k} ) \sigma ( m) \\ =& (2^{k+1} - 1) \sigma (m) \end{align*}$$ It simplifies to $(2^{k+1} - 1) \sigma (m) = 2^{k+1} m$, but since $2^{k+1} -1$ is odd, $\sigma (m)$ is a multiple of $2^{k+1}$. That is, for some $c$, the following holds. $$\begin{align*} \sigma (m) =& 2^{k+1} c \\ m =& (2^{k+1} - 1) c \end{align*}$$

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Part 2.

Assuming $c \ge 1$, $$\sigma (m) \ge 1 + c + m = 1 + c + (2^{k+1} -1) c = 1 + 2^{k+1}c$$ However, since $\sigma (m) = 2^{k+1}c$, it contradicts $2^{k+1}c \ge 1 + 2^{k+1} c$, so $c=1$.

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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.

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