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How do I prove:

Let $p$ be a prime, and $n$ be a positive integer. Then $p^n$ is not a perfect number.

One example is when $p = 2$ and $n = 3$, the question is to show $8$ is not a perfect number. And I know that out of $1, 2, 3, 4, 5, 6, 7, 8$, the proper divisors of $8$ are $1, 2,$ and $4$, with $1 + 2 + 4 = 7 \ne 8$, so $8$ is not a perfect number.

But how do I show this for any $n$ and $p$?

Anonymous - a group
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Joshua Myers
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2 Answers2

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The divisors of $p^n$ are $1,p,p^2,\ldots,p^n$. The proper divisors are all but the last one. The sum of those is $$ 1+p+p^2+p^3+\cdots+p^{n-1}. $$ This is a geometric series. Apply the standard formula for the sum of a finite geometric series and see if you get $p^n$.

Milo Brandt
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Michael Hardy
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  • I'm still having some trouble understanding this. Perhaps can you go further? – Joshua Myers Jan 28 '15 at 02:11
  • Suppose one has $\text{sum}=1+p+p^2+p^3+p^4+p^5$. Then $p\times\text{sum}=p+p^2+p^3+p^4+p^5+p^6$. Now subtract: $(p\times\text{sum}) - \text{sum}$. All of the terms cancel except $p^6$ and $1$, so you get $p^6-1$. But $(p\times\text{sum}) - \text{sum} = (p-1)\times\text{sum}$. Consequently $(p-1)\times\text{sum}=p^6-1$, so $\text{sum}=\dfrac{p^6-1}{p-1}$. The question is now whether that is equal to $p^6$. ${}\qquad{}$ – Michael Hardy Jan 28 '15 at 02:15
  • It isn't, right? – Joshua Myers Jan 28 '15 at 02:19
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    We don't need a formula for the sum of a GP. Note that $p$ divides every term except $1$, so if we have equality then $p$ must divide $1$, not true. – André Nicolas Jan 28 '15 at 02:31
  • @AndréNicolas So basically if I write what you wrote in the comments for my actual question, then I should be fine? – Joshua Myers Jan 28 '15 at 02:33
  • @JoshuaMyers : Correct. The sum cannot be divisible by $p$ unless the numerator is divisible by $p$, and the numerator differs by $1$ from something divisible by $p$. But the comment by André Nicolas also suffices. ${}\qquad{}$ – Michael Hardy Jan 28 '15 at 02:40
  • @MichaelHardy I'll give it to you though...Thanks to you and Andre – Joshua Myers Jan 28 '15 at 02:45
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    @JoshuaMyers: The solution of Michael Hardy exploits another frequently useful idea, note that $\frac{p^n-1}{p-1}$ is "too small." – André Nicolas Jan 28 '15 at 03:18
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$p$ prime then, $1 +p + p^2 + \cdots + p^{n-1} \neq p^n$.

Loreno Heer
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