📂Number Theory

Euclid's Derivation of the Perfect Number Formula

Formula ¹

If $2^{p}-1$ is a prime number, then $2^{p-1}(2^{p} - 1)$ is a perfect number.

Explanation

It is not certain that all perfect numbers will have this form, but this form is definitely a perfect number.

For example, for the prime number $(2^2 -1) = 3$, $2^{2-1}(2^2 -1) = 6$ is a perfect number.The fact that perfect numbers and Mersenne primes have this relationship could be somewhat inferred from the geometric series expansion of Mersenne primes.

Derivation

Since $2^{p}-1$ is a prime, the divisors of $2^{p-1}(2^{p} - 1)$ $$ 1,2, \cdots , 2^{p-1} \\ (2^{p}-1), 2 (2^{p}-1), \cdots , 2^{p-2} (2^{p}-1) $$ are divided into two categories. According to the sum of geometric series formula $$ 1 + 2 + \cdots + 2^{p-1} = { {2^{p}-1} \over {2 - 1} } = 2^{p}-1 $$ Similarly, $$ (2^{p}-1) + 2 (2^{p}-1) + \cdots + 2^{p-2} (2^{p}-1) = ( 2^{p-1} - 1 ) (2^{p}-1) $$ Adding the two, $$ 2^{p}-1 + ( 2^{p-1} - 1 ) (2^{p}-1) = 2^{p-1}(2^{p} - 1) $$ Therefore, $2^{p-1}(2^{p} - 1)$ is a perfect number.

■