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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.

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Mirrana
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    All fine. For completeness you should say why that fraction is not $p^n$. – André Nicolas Sep 24 '13 at 00:25
  • I'm not quite sure how to do that... – Mirrana Sep 24 '13 at 00:26
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    Find some integer which is a divisor of one and not the other. – Zvi Rosen Sep 24 '13 at 00:27
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    Another hint is to observe that $p^{n+1} +1 > 2p^n $. – Calvin Lin Sep 24 '13 at 00:30
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    An alternative way of proving that $p^n$ cannot be perfect is to note that the sum of divisors of $p^n$ must necessarily be congruent to 1 mod $p$, while $p^n$ itself is clearly congruent to 0. – Dennis Meng Sep 24 '13 at 00:40
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    Another quick way through is to note that $\frac{p^n-1}{p-1}\leq p^n-1\lt p^n$, since $p\geq 2$. – Steven Stadnicki Sep 24 '13 at 00:45
  • @StevenStadnicki Thanks - I can actually follow the inequality quite easily. It's a hell of a lot shorter than the proposed proof by contradiction. – Mirrana Sep 24 '13 at 00:48
  • I've updated the solution with the suggested proof of inequality. Thanks for the advice. – Mirrana Sep 24 '13 at 00:50
  • If x is a perfect number, then the sum of divisors of x is exactly 2x. In many cases, like this one, you can show that the sum of divisors is < 2x. In other cases, you can show that the sum of divisors is > 2x (for example if x=6k, k ≠ 1, then it has divisors 6k, 3k, 2k, k, and 1, which add up to 2x+1 making it not perfect). – gnasher729 Apr 13 '15 at 15:19

3 Answers3

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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.

Jose Arnaldo Bebita Dris
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As André Nicolas your reasoning is good an it's enough to prove the that no perfect number is a prime of power, but you should prove that $p^n \neq \frac{p^n - 1}{p-1}$ to complete the answer, because it's not so obivous and something we take for granted. Here's some help.

Try to prove using contradiction. Assume that:

$p^n = \frac{p^n - 1}{p-1}$

$p-1$ is obviously not 0, so we multiply by it.

$$p^n(p-1) = p^n - 1$$ $$p^{n+1} - p^n - p^n = -1$$ $$p^{n+1} - 2p^n = -1$$ $$p^n(p-2) = -1$$

Because both terms are integers, that means that we have two separate cases:

$$ \left\{\begin{aligned} &p^n = 1\\ &p-2 = -1 \end{aligned} \right.$$

This implies that $p=1$, but because p is a prime, it can't be 1.

$$ \left\{\begin{aligned} &p^n = -1\\ &p-2 = 1 \end{aligned} \right.$$

The second equation implies that $p=3$, but $3^n = -1$ isn't possible in any case. So because we exhausted all the possibilites and we didn't find a solution, that means that our initial assumption is wrong.

Q.E.D.

Stefan4024
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    This is the long way around. Much easier to go $p\geq 2$ so $p-1\geq 1$ and $\frac{p^n-1}{p-1}\leq p^n-1\lt p^n$. – Steven Stadnicki Sep 24 '13 at 00:42
  • I first thought of this and I didn't try thinking about another easier solution. – Stefan4024 Sep 24 '13 at 00:45
  • @DennisMeng As an alternate solution, could you show the proof of that? I know what congruence is, but I'm not quite following your reasoning why that is. – Mirrana Sep 24 '13 at 00:45
  • Sure. The basic idea is that since the sum of the divisors is $1 + p + p^2 + ... + p^n$, it has to leave a remainder of 1 when divided by $p$. $p^n$ on the other hand, leaves a remainder of 0. Since the two leave different remainders, they can't be equal (since if two numbers are equal, they must have the same remainder when divided by $p$). – Dennis Meng Sep 24 '13 at 00:49
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    An easy way to finish is to write $p^n(p-1)=p^n-1$. If $n\gt 0$, $p$ divides the left but not the right. We need to deal separately with $n=0$ anyway. – André Nicolas Sep 24 '13 at 01:07
  • Or just consider your $p^{n + 1} - 2 p^n = -1$ modulo $p$, which gives the ridiculous $0 \equiv -1 \pmod{p}$. – vonbrand Oct 25 '15 at 00:19
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An easier proof involving no inequalities may go as follows:

Let p be a prime number. Assume p^n is a perfect number where n is some natural number.

This means the sum of the factors of p^n can be written as p^n

As the top proof says, the sum of the factors of p^n can be shown as the sum of the powers of p below it all the way down to the power of zero(otherwise known as 1).

Consider the sum of the factors of p^n other than 1. One can see intuitively that it is some quantity of p, but it can also be shown with the converse of the distributive property.

That is, {p^(n-1)+p^(n-2)+...+p^2+p} can also be written as

p(p^(n-2)+p^(n-3)+...+p+1)

With this one can see this sum is divisible by p, for p is one of its factors.

Now consider the 1 we left out. If we include it in the sum, we can now say the sum of the factors of p^n are indivisible by p, for there is now a remainder of 1.

But recall the assumption that the sum of the factors of p^n can be shown to be p^n.

With the same converse as before we can show this to be equal to

p(p^(n-1))

This shows that it is also divisible by p, for it has p as a factor

But this is a contradiction, as it was just shown that the sum of the factors of p^n are not divisible by p, and therefore the original assumption was incorrect. A power of a prime cannot be perfect, exactly as it was meant to be shown.

ThoughtOfGod
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