## Frequently Asked Questions

#### What are perfect, abundant, and deficient numbers?

The divisors of a natural number, 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.

If the sum of proper divisors exceeds the number it is called abundant (for example, 12: 1 + 2 + 3 + 4 + 6 = 16), and if the sum of proper divisors are less than the number it is called deficient (for example, 14: 1 + 2 + 7 = 10).

In number theory it is more convenient to define a perfect number as a number for which the sum of all its divisors is twice the number. We use the sum of divisors function, σ(`n`), to represent the sum of divisors of the natural number, `n`. So if P is a perfect number, σ(P)=2P.

The ancient Greeks knew that all the perfect numbers they discovered were of the form `q`×2^{p−1}, where `p` and `q` are prime.

For example, 6=3×2^{2−1}, 28=7×2^{3−1}, 496=31×2^{5−1}, 8128=127×2^{7−1}.

Although we have produced perfect numbers for `p` = 2, 3, 5, 7, it is not perfect when `p` = 11; the next cases work for `p` = 13, 17, 19, and then 31.

Leonhard Euler (1707-1783) was able to prove the conjecture: if a perfect number is even it WILL be of the form `q`×2^{p−1}, where `q` is prime. What no one has been able to prove is whether or not any odd perfect numbers exist.

By utilising the sum of divisor function, σ(`n`), we shall being by showing that if `q` is a prime of the form, 2^{p}−1, it will always be perfect.

Given the even number, P = `q`×2^{p−1}, where `q` is prime, we can then write,

σ(P) = σ(`q`×2^{p−1}) = σ(`q`)×σ(2^{p−1}) = (`q`+1)(2^{p}−1).

But if `q` is a prime of the form 2^{p}−1, we get,

σ(P) = (2^{p}−1+1)(2^{p}−1) = 2^{p}(2^{p}−1) = 2×2^{p−1}(2^{p}−1) = 2×`q`×2^{p−1} = 2P.

Hence, if `q` is a prime of the form, 2^{p}−1, then the even number, P = `q`×2^{p−1}, will be perfect.

We shall now prove the converse...

**Theorem**

If P is an even perfect number it must be of the form P = 2^{k−1}(2^{k}−1).

**Proof**

Suppose that P is an even perfect number. All even numbers can be written in the form, `q`×2^{k−1}, where `q` is odd and `k` ³ 2.

Therefore, σ(P) = σ(`q`)×(2^{k}−1).

As P is perfect, σ(P) = 2P = `q`×2^{k}, and we can write, σ(`q`)×(2^{k}−1) = `q`×2^{k}. This leads to, σ(`q`) = `q`×2^{k}/(2^{k}−1).

Because σ(`q`) is integer, `q` must be divisible by 2^{k}−1, so let `q` = `r`(2^{k}−1), therefore, σ(`q`) = `r`×2^{k}.

As `q` is divisible by `r`, we know that its sum of divisors will at least be `q`+`r`. Therefore, σ(`q`) = `r`×2^{k} ³ `q`+`r` = `r`(2^{k}−1)+`r` = `r`(2^{k}−1+1) = `r`×2^{k}, which tells us that σ(`q`) = `q`+`r`. As `q` and `r` are the only divisors, `q` must be prime and `r` must be 1, hence `q` will be a prime of the form 2^{k}−1.

And so, if P is an even perfect, it will be of the form 2^{k−1}(2^{k}−1). **Q.E.D.**