We leave many details to Exercise 19.6.24.

• An odd perfect number cannot be a prime power. This is easy; \(2=\frac{\sigma(n)}{n}<\frac{p}{p-1}\) isn't possible, even for \(p=2\); since we are looking for an odd perfect number, it definitely won't be possible!

• An odd perfect number cannot be a product of exactly two prime powers. Use the same idea, but now with the biggest possible values for odd primes.

• An odd perfect number cannot be a product of exactly three prime powers unless the first two are \(3^e\) and \(5^f\). This proof is slightly longer.

  • Suppose that \(3\) is not the smallest prime involved. Then the biggest that \begin{equation*}\frac{p_1}{p_1-1}\frac{p_2}{p_2-1}\frac{p_3}{p_3-1}\end{equation*} can be is \begin{equation*}\frac{5}{4}\frac{7}{6}\frac{11}{10}=\frac{77}{48}<2\; .\end{equation*}

  • Suppose that \(5\) is not the second-smallest prime involved (assuming \(3\) is the smallest). We again get a contradiction.

Assume this works for some \(N\). Then \(3\sigma(N)=5N\).

Let's look at divisors. First, \(3\mid N\). So if \(N\) is even, then \(6\mid N\), so by Fact 19.4.11, \begin{equation*}\sigma_{-1}(N)\geq \sigma_{-1}(6)=2>\frac{5}{3}\; ,\end{equation*} which is impossible. If \(N\) is not even, then \(N\) is odd, so \(3\sigma(N)=5N\) is odd, which implies \(\sigma(N)\) itself is odd.

Since \(3\mid N\) and using Lemma 19.5.3, we see that we must have that \(3^2\mid N\).

Let's return to the divisors. We know that \(5\nmid N\), because otherwise \begin{equation*}\sigma_{-1}(N)\geq \sigma_{-1}\left(3^2\cdot 5\right)=\frac{26}{15}>\frac{5}{3}\end{equation*} which is again impossible.

Now we can compute directly that \begin{equation*}\sigma_{-1}(5N)=\sigma_{-1}(5)\sigma_{-1}(N)=\frac{6}{5}\frac{5}{3}=2\; !\end{equation*}