Science:Math Exam Resources/Courses/MATH437/December 2011/Question 06/Solution 1

Let $\displaystyle n=2^{k}m$ with k at least 1. Further, let

$\displaystyle \sigma (n)=\sum _{d\mid n}d$

and recall that this function is multiplicative (the sum is over positive divisors of n). Since we are given that n is an even perfect number, we know that

$\displaystyle 2n=\sigma (n)$.

Next, we use the multiplicativity of $\displaystyle \sigma$ and see that

$\displaystyle 2^{k+1}m=2n=\sigma (n)=\sigma (2^{k}m)=\sigma (2^{k})\sigma (m)=(1+2+2^{2}+...+2^{k})\sigma (m)=(2^{k+1}-1)\sigma (m)$.

Since $\displaystyle \gcd(2^{k+1},2^{k+1}-1)=1$, we see that $\displaystyle (2^{k+1}-1)\mid m$. Let M be an integer such that $\displaystyle m=(2^{k+1}-1)M$. Substituting this into the above yields

$\displaystyle 2^{k+1}(2^{k+1}-1)M=2^{k+1}m=(2^{k+1}-1)\sigma (m)$.

Cancelling on both sides yields

$\displaystyle 2^{k+1}M=\sigma (m)$.

Using the fact that both m and M are divisors of m (and hence are terms inside the expansion of $\displaystyle \sigma (m)$, we have that

$\displaystyle 2^{k+1}M=\sigma (m)\geq m+M=(2^{k+1}-1)M+M=2^{k+1}M$.

Thus, the greater than sign is in fact an equality. This means that $\displaystyle \sigma (m)$ has only two factors, namely m and M. Since $\displaystyle m>M$, we must have that $\displaystyle M=1$ and so $\displaystyle m=(2^{k+1}-1)M=2^{k+1}-1$. As m has only two prime factors, we also know that $\displaystyle m=2^{k+1}-1$ is prime. This completes the proof.