Theorem of Even Perfect Numbers/Necessary Condition

Theorem

Let $a \in \N$ be an even perfect number.

Then $a$ is in the form:

$2^{n - 1} \paren {2^n - 1}$

where $2^n - 1$ is prime.

Proof

Let $a \in \N$ be an even perfect number.

We can extract the highest power of $2$ out of $a$ that we can, and write $a$ in the form:

$a = m 2^{n - 1}$

where $n \ge 2$ and $m$ is odd.

Since $a$ is perfect and therefore $\map {\sigma_1} a = 2 a$:

So:

$\map {\sigma_1} m = \dfrac {m 2^n} {2^n - 1}$

But $\map {\sigma_1} m$ is an integer and so $2^n - 1$ divides $m 2^n$.

From Consecutive Integers are Coprime, $2^n$ and $2^n - 1$ are coprime.

So from Euclid's Lemma $2^n - 1$ divides $m$.

Thus $\dfrac m {2^n - 1}$ divides $m$.

Since $2^n - 1 \ge 3$ it follows that:

$\dfrac m {2^n - 1} < m$

Now we can express $\map {\sigma_1} m$ as:

$\map {\sigma_1} m = \dfrac {m 2^n} {2^n - 1} = m + \dfrac m {2^n - 1}$

This means that the sum of all the divisors of $m$ is equal to $m$ itself plus one other divisor of $m$.

Hence $m$ must have exactly two divisors, so it must be prime by definition.

This means that the other divisor of $m$, apart from $m$ itself, must be $1$.

That is:

$\dfrac m {2^n - 1} = 1$

Hence the result.

$\blacksquare$

Historical Note

René Descartes stated in $1638$ that he had a proof that every even perfect number is of the form $2^{p - 1} \paren {2^p - 1}$, but failed to actually produce it.

The first actual published proof was made by Leonhard Paul Euler.

Sources

• 1971: George E. Andrews: Number Theory ... (previous) ... (next): $\text {3-5}$ The Use of Computers in Number Theory: Conjecture $2$
• 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $28$
• 1992: George F. Simmons: Calculus Gems ... (previous) ... (next): Chapter $\text {B}.2$: More about Numbers: Irrationals, Perfect Numbers and Mersenne Primes
• 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $28$