I need to prove the following, though I'm not 100% certain I understand the definition of a perfect number.
Prove that no perfect number is a power of a prime.
First of all, I'm assuming that the question is asking me to prove that for any prime $p$ and for all natural numbers\positive integers $n$, $p^n$ is not a perfect number. Am I correct in this understanding of the problem?
Based on this, I've come up with the following to prove this theorem…
Let $p$ be a prime number. Assume that $p^n$ is a perfect number for some $n\in\mathbb{N}$. Therefore, $$p^n=\underbrace{p\cdot p\cdot p\cdots p}_{n\text{ prime numbers}}=\underbrace{1+p+p^2+p^3+\cdots +p^{n-1}}_{\text{sum of all divisors of $p^n$ except itself}}=\frac{p^n-1}{p-1}.$$
As $\frac{p^n-1}{p-1}\leq p^n-1<p^n$ for all $p\geq 2$, this is a contradiction, thus proving that no power of a prime can be a perfect number.
Without elaborating too much, I'm assuming that my proof ends here, because the definition I was given for a perfect number is that it is equal to the sum of all of its divisors except itself. Since the only valid divisors of a number of the form $p^n$ are 1 and all powers of $p$ from $1$ to $n-1$, this is what I come up with. And since $1$ is not a prime number by convention, this seems to hold.
Note: I used the identity $1+x+x^2+x^3+\cdots +x^{n-1}=\frac{x^n-1}{x-1}$ because it was conveniently proven in my textbook.
Best Answer
A quick way to prove that no prime power is perfect is to notice that, if $p$ is a prime, then $p \geq 2$, so that
$$\frac{\sigma(p^k)}{p^k} = \frac{1 + p + \ldots + p^k}{p^k} = \frac{p^{k+1} - 1}{p^k(p - 1)} < \frac{p^{k+1}}{p^k(p - 1)} = \frac{p}{p - 1}.$$
Now, since $p \geq 2$, we get that
$$\frac{1}{p} \leq \frac{1}{2} \Longrightarrow -\frac{1}{p} \geq -\frac{1}{2} \Longrightarrow 1 - \frac{1}{p} \geq 1 - \frac{1}{2} = \frac{1}{2}.$$
Consequently, we have:
$$\frac{p - 1}{p} \geq \frac{1}{2} \Longrightarrow \frac{p}{p - 1} \leq 2.$$
We conclude that:
$$\frac{\sigma(p^k)}{p^k} < \frac{p}{p - 1} \leq 2.$$
In fact, this inequality shows that all prime powers are deficient. Hence, no prime power is perfect.