Odd Perfect Numbers

Problem

The divisors of a positive integer, excluding the number itself, are called the proper divisors . If the sum of proper divisors is equal to the number we call the number perfect. For example, the divisors of 28 are 1, 2, 4, 7, 14, and 28, so the sum of proper divisors is 1 + 2 + 4 + 7 + 14 = 28.

The first eight perfect numbers are 6, 28, 496, 8128, 33550336, 8589869056, 137438691328, 2305843008139952128.

It is well known that P is an even perfect number iff it is of the form $2^{n-1}(2^n - 1)$ where $2^n - 1$ is prime.

No one has yet discovered an odd perfect number, and the existence of them is in doubt.

However, if $Z$ is an odd perfect, prove that it must be of the form $c^2 q^{4k+1}$ where $q \equiv 1 \mod 4$ is prime.

Solution

We shall work modulo 4 throughout this proof, and we begin by writing $Z$ as a product of distinct prime factors, noting that as $Z$ is odd, all prime factors must be odd:

Let $Z = p_1^{a_1} p_2^{a_2} ... p_m^{a_m} q_1^{b_1} q_2^{b_2} ... q_m^{b_m}$, where $p \equiv 3$ and $q \equiv 1$.

If $a$ is even then $p^a \equiv 1$.
If $a$ is odd then $p^a \equiv 3$.
If $b$ is even or $b$ is odd then $q^b \equiv 1$.

So $Z \equiv 1,3$ and it follows that $2Z \equiv 2$.

If $Z$ is perfect then the sum of divisors, $\sigma(Z) = 2Z \equiv 2$.

Using the multiplicative property of the function, $\sigma(ab) = \sigma(a) \sigma(b)$, we need consider the possible values of each $\sigma(p^a)$ and $\sigma(q^b)$.

By definition, $\sigma(p^a) = 1 + p + ... + p^a$.

If $a$ is even then $\sigma(p^a) \equiv 1$ (sum of an odd number of odds).
If $a$ is odd then $\sigma(p^a) \equiv 0$ (sum of an even number of odds).

But if $a$ is odd then the product of these residuals will be zero and we know that $\sigma(Z) \equiv 2$. Hence $a$ must be even and the product of all $\sigma(p^a)$ terms will be congruent with 1.

We shall now consider $\sigma(q^b) = 1 + q + ... + q^b$ and recall that $q^b \equiv 1$ for all values of $b$.

If $b \equiv 0$ then $\sigma(q^b) \equiv 1$.
If $b \equiv 1$ then $\sigma(q^b) \equiv 2$.
If $b \equiv 2$ then $\sigma(q^b) \equiv 3$.
If $b \equiv 3$ then $\sigma(q^b) \equiv 0$.

In the same way that $a$ cannot be odd, $b$ cannot be of the form $4k + 3$.

If $b$ is even, that is, $b = 4k$ or $b = 4k + 2$, then the product of those $\sigma(q^b)$ terms will be congruent with 1 or 3.

Therefore $Z$ can comprise any number of primes, $p$ and $q$ raised to an even power, which will form a square number.

But it is necessary to have exactly one sigma term with $b = 4k + 1$ so that the product of all sigma terms will be congruent with 2.

Hence if $Z$ is an odd perfect then it must be of the form $c^2 q^{4k+1}$ where $q \equiv 1$ is prime. Q. E. D.

Problem ID: 328 (05 Jul 2007)     Difficulty: 4 Star

Only Show Problem