# Characterizing all even perfect numbers

Over 2300 years ago Euclid proved that If 2* ^{k}*-1 is a prime
number (it would be a Mersenne
prime), then 2

^{k-1}(2

*-1) is a perfect number. A few hundred years ago Euler proved the converse (that every even perfect number has this form). It is still unknown if there are any odd perfect numbers (but if there are, they are large and have many prime factors).*

^{k}**Theorem:**- If 2
-1 is a prime number, then 2^{k}^{k-1}(2-1) is a perfect number^{k}*and*every even perfect number has this form. **Proof:**-
Suppose first that

*p*= 2-1 is a prime number, and set^{k}*n*= 2^{k-1}(2-1). To show^{k}*n*is perfect we need only show σ(*n*) = 2*n*. Since σ is multiplicative and σ(*p*) =*p*+1 = 2, we know^{k}σ(

*n*) = σ(2^{k-1})σ(^{.}*p*) = (2-1)2^{k}= 2^{k}*n*.This shows that

*n*is a perfect number.On the other hand, suppose

*n*is any even perfect number and write*n*as 2^{k-1}*m*where*m*is an odd integer and*k*Again σ is multiplicative so__>__2.σ(2

^{k-1}*m*) = σ(2^{k-1})^{.}σ(*m*) = (2-1)^{k}^{.}σ(*m*).Since

*n*is perfect we also know thatσ(

*n*) = 2*n*= 2^{k}m.Together these two criteria give

2

(2^{k}m =-1)^{k}^{.}σ(*m*),so 2

-1 divides 2^{k}hence 2^{k}m-1 divides^{k}*m*, say*m =*(2-1)^{k}*M*. Now substitute this back into the equation above and divide by 2-1 to get 2^{k}^{k}M*=*σ(*m*). Since m and M are both divisors of m we know that2

= σ(^{k}M*m*)__>__*m*+*M*= 2,^{k}Mso σ(

*m*) =*m*+*M*. This means that*m*is prime and its only two divisors are itself (*m*) and one (*M*). Thus*m*= 2-1 is a prime and we have prove that the number^{k}*n*has the prescribed form. ∎