Welcome back to Numberholic! I’m your host, Sejari, and it’s always a pleasure to have you here. Today, we’re diving into the eleventh episode of Proven Yet Hard to Believe, a series where we uncover the fascinating stories and directly explore the core ideas behind mathematical discoveries that have stood the test of time.
Today’s topic is perfect numbers!
A perfect number is a positive integer that is equal to the sum of its proper divisors, excluding itself. For example, 6 is a perfect number because the sum of its proper divisors (1, 2, and 3) equals 6. The second perfect number is 28, as 1 + 2 + 4 + 7 + 14 = 28.
If we include the number itself, a perfect number is defined as a number whose total sum of divisors equals exactly twice the number itself.
The Jewish philosopher Philo of the 1st century CE mentioned perfect numbers in his work “On the Creation,” emphasizing, “The world was created in six days, Six is the first perfect one, being made equal to its parts, and being made complete by them.”
Greek mathematicians had been interested in perfect numbers for a long time. Euclid, in Proposition 36 of Book 9 of The Elements, provided a proof regarding perfect numbers:
“Starting with unity, multiply by 2 repeatedly to form a sequence of numbers. If the sum of these numbers is prime, then the product of the sum and the last number in the sequence is a perfect number.”
In other words, if 1 + 2 + 4 + … + 2^n = 2^(n+1) – 1 is prime, then 2^n(2^(n+1) – 1) is a perfect number. Summing all divisors of this number results exactly in 2^(n+1)(2^(n+1) – 1), which is twice the number itself.
The complete proof of even perfect numbers was later established by the mathematician Euler in 1747, more than 2,000 years after The Elements was written.
Euler’s proof states:
All even perfect numbers must be of the form 2^nM, where M = 2^(n+1) – 1 is prime.
To establish this, we first factorize a natural number N into its even and odd factors and examine the sum of its divisors. Let the divisors of an odd number M be 1, d_1, d_2, …, M. Then, the sum of divisors is calculated as follows:
This sum must be equal to 2N = 2^(n+1)M.
Since (2^(n+1) – 1) is an odd number greater than or equal to 3, M must have a divisor y such that:
(2^(n+1) – 1)y = M.
Applying this to the sum of divisors, we can obtain an inequality.
For this inequality to become an equality, M must have only two divisors: y and M. This means y = 1 and M must be prime. Therefore, N must be of the form 2^nM, where M = 2^(n+1) – 1 is prime.
As for whether odd perfect numbers exist, the question remains unsolved. Mathematics has many unsolved problems, and there are two major open questions related to perfect numbers:
Are there infinitely many even perfect numbers? The key to this question lies in whether numbers of the form M = 2^p – 1 are prime. Such primes are known as Mersenne primes. In other words, do infinitely many Mersenne primes exist? Mathematicians strongly suspect so, but no formal proof has been found.
Do odd perfect numbers exist? While many negative indications suggest they might not, there is no definitive proof either way.
That’s all for today’s episode of Proven Yet Hard to Believe! If you enjoyed this deep dive, don’t forget to subscribe and stay tuned for our next journey into the world of mathematics. Until next time, keep exploring, keep questioning, and stay Numberholic!