Home > Number theory, Problem Solving > IMC 2014 Day 1 Problem 4

## IMC 2014 Day 1 Problem 4

Let ${n > 6}$ be a perfect number and ${n = p_1^{e_1}...p_k^{e_k}}$ be its prime factorization with ${1. Prove that ${e_1}$ is an even number.

A number ${n}$ is perfect if ${s(n)=2n}$, where ${s(n)}$ is the sum of the divisors of ${n}$.

IMC 2014 Day 1 Problem 4

If $\p_1=2$, then assuming $\e_1$ is odd yields that $\p_2 = 3$. However, if $n$ is divisible by 6, then it is easy to see that $\sigma(n)>2n$, by using simple inequality argument.
In the other case, assuming $e_1$ odd would imply that $n$ has a prime divisor that divides $p_1+1$, which contradicts the fact that $p_1$ is the smallest.
If the smallest prime divisor of n is 2 and if $e_1$ is odd, then it would imply that $n$ is also divisible by 6. However, it is easy to see that no perfect number is divisible by 6 except 6 itself.