## Perfect Number & Mersenne Prime

Posted: December 25, 2018 in Mathematics

One of the interesting thing about Number Theory is that quite often a theorem or a conjecture is easy to be stated and yet it is hard to be proven. For example, a conjecture that there are infinitely many perfect numbers. To this date no one has ever proved it true or false. However it is believed to be true by many mathematicians. First of all, what is a perfect number? A perfect number is a positive integer such that the sum of all its proper divisors equal to the number itself. For instance, 6 is a perfect number since 1, 2, and 3 are the proper divisors of 6 and 6 = 1 + 2 + 3. The next perfect number would be 28 since 28 = 1 + 2 + 4 + 7 + 14. This has to be one of the oldest puzzles in the history of mathematics. Euclid (around 300 BC) had shown that when $2^k-1$ is prime, $2^{k-1}(2^k-1)$ is perfect. There is a special name for numbers in the form $2^k-1$. They are called the Mersenne Numbers, which is named after a French priest Marin Mersenne (1588 – 1648) who devoted a lot of his time and effort studying them. And whenever a Mersenne number is prime, it is called a Mersenne Prime. So there is a relationship between Mersenne Prime and Perfect Number. To be more accurate, when you find a Mersenne prime, you will find an even perfect number. It was proven by Euler around the 18th Century that all even perfect numbers must be in the form $2^{k-1}(2^k-1)$. Now, the proof is not really that difficult in today’s standard.

Euclid-Euler Theorem. For $k > 1$, if $2^k-1$ is prime, then $n = 2^{k-1}(2^k-1)$ is perfect. And every even perfect number is of this form.

Proof. Denote the sum of all divisors function as $\sigma$. Let $2^k-1$ be Mersenne Prime and $n = 2^{k-1}(2^k-1)$ be an integer. Since $(2^{k-1}, 2^k-1) = 1$ and $\sigma$ is multiplicative, we then have

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

Since the factors of $2^{k-1}$ are $1, 2, 2^2, \ldots, 2^{k-1}$ and $2^k-1$ is Mersenne prime,

$\sigma(2^{k-1}) = 2^k-1 \quad\text{and}\quad \sigma(2^k-1) = 2^k$

$\therefore \sigma(n) = (2^k-1)2^k = 2n$

And this is exactly the definition of a perfect number.

To prove the converse, assume n is an even perfect number such as $n = 2^{k-1}(2a-1)$ where a and k are positive integers and $k\geq2$. Since $(2^{k-1}, 2a-1) = 1$, we have

$\sigma(n) = \sigma(2^{k-1}(2a-1)) = \sigma(2^{k-1})\sigma(2a-1) = (2^k-1)\sigma(2a-1)$

Since n is perfect, then

$\sigma(n) = 2n = 2\cdot 2^{k-1}(2a-1) = 2^k(2a-1)$

$\therefore 2^k(2a-1) = (2^k-1)\sigma(2a-1) \qquad (1)$

Hence, $2^k-1 \;|\; 2^k(2a-1)$, but $(2^k-1, 2^k)=1$, so $2^k-1 \;|\; 2a-1$, which means $2a-1 = (2^k-1)b$ for some integer b. Then equation (1) becomes

$2^k(2^k-1)b = (2^k-1)\sigma(2a-1)$

$2^kb = \sigma(2a-1)$

Since $2a-1$ and $b$ are both divisors of $2a-1$, we have

$2^kb = \sigma(2a-1) \geq (2a-1) + b = (2^k-1)b + b = 2^kb$

This forces the inequality to equality,

$\sigma(2a-1) = (2a-1) + b$

And this implies that $2a-1$ has only two divisors and so $2a-1$ must be prime and $b=1$. Therefore $2a-1 = (2^k-1)b = 2^k-1$ is a prime number, which completes the proof. QED.

So the problem of finding even perfect numbers boils down to the hunt for Mersenne primes. And I remember the last time I wrote the article on the largest known prime, the last greatest Mersenne prime was $2^{77232917}-1$. But now the record has been broken again. On December 7, 2018 the largest known prime (or Mersenne prime) is found to be

$2^{82589933}-1$

which makes the largest known perfect number to be

$2^{82589932}(2^{82589933}-1)$

which is 49,724,095 digits long!