First, assume that \(2^n-1\) is prime. Then the factors of \(2^{n-1}\left(2^n-1\right)\) are coprime, so \[\sigma\left(2^{n-1}\left(2^n-1\right)\right)=\sigma\left(2^{n-1}\right)\sigma\left(2^n-1\right)=\left(2^n-1\right)\left(2^n-1+1\right)\] The steps are because of multiplicativity and the formulas we had above for \(\sigma\) of powers of two and primes. But then \[\left(2^n-1\right)\left(2^n-1+1\right)=2^n\left(2^n-1\right)=2\left[2^{n-1}\left(2^n-1\right)\right]\] so that the sum of divisors is exactly twice the original number.

Now for the converse, which is somewhat longer.

If we have an even perfect number, then it is divisible by some power of two. Let us (looking ahead!) call this the \(n-1\)th power! So that our even perfect number is \(2^{n-1}q\), where the \(q\) is the (odd) quotient. Let's divide the rest of the proof into steps.

• We know that this number is perfect, so \[\sigma\left(2^{n-1}q\right)=2^nq\]
• We also know how to compute \(\sigma\), so \[\sigma\left(2^{n-1}q\right)=\sigma\left(2^{n-1}\right)\sigma(q)=\left(2^n-1\right)\sigma(q)\]
• We can combine these to see that \[2^nq=\left(2^n-1\right)\sigma(q)\] Note that this means \(2^n-1\mid q\), since \(q\) is the only odd part of the left-hand side. Let's write \[\left(2^n-1\right)m=q\; .\]
• But then \[2^n\left(2^n-1\right)m=\left(2^n-1\right)\sigma(q)\Rightarrow 2^nm=\sigma(q)\]
• Notice that since \(m\) and \(q\) both divide \(q\), by definition of \(\sigma\) we have \[\sigma(q)\geq q+m=\left(2^n-1\right)m+m=2^nm\] as well.
• Since these two divisors alone add up to \(\sigma(q)\), it must be true that \(q\) has two divisors, so it is prime, and \(m=1\), and \(q=2^n-1\), and so the perfect number is \(2^{n-1}\left(2^n-1\right)\). Great!