Exercise 4.2.19 (Every even perfect number has the form $2^{n-1} q$, where $q = 2^n - 1$ is prime)

Proof. Let $m$ be an even perfect number. Since $m\ge 1$, we can write $m$ in the form $m=2^{n-1}q$, where $q$ is odd, and $n>1$ because $m$ is even.

We know that $\sigma (q)\ge q$ for all positive integer $q$, so $\sigma (q)=q+k$ for some $k\ge 0$.

By definition of a perfect number, $\sigma (m)=2m$, so

Since $\sigma$ is multiplicative, and $q$ is odd, by Theorem 4.5,

The comparison of (1) and (2) gives $2^{n}q=(2^n-1)(q+k)$, therefore

This shows that $k\mid q$ and $k<q$, because $n>1$. By Problem 18, we obtain that $k=1$, so $\sigma (q)=q+1$, and by Problem 17, $q$ is a prime.

Thus $m=2^{n-1}q$ where $q=2^n-1$ is a prime, thus $m$ has the form given by Euclid. □