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Given that $p$ is a prime, and the sum of all positive divisors of $p^4$ is a perfect square, find the possible number of primes $p$ .

What I Tried: I know that the only divisors of $p^4$ will be $(1 , p , p^2 , p^3 , p^4)$ . From here I can conclude :- $$\rightarrow 1 + p + p^2 + p^3 + p^4 = k^2$$ For some positive integer $k$ . Now this is where I get stuck, I can write it as :- $$\rightarrow \frac{(1 - p^5)}{1 - p} = k^2$$ But I didn't understand how I can find all values of $p$. Next what I did is :- $$\rightarrow p(1 + p + p^2 + p^3) = (k + 1)(k - 1)$$ And I am stuck at the same problem.

Can anyone help me here?

nonuser
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Anonymous
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2 Answers2

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Just some thoughts. Still editing...

Note that the sum $1+p+p^2+p^3+p^4$ is odd, so $k^2\equiv1\bmod 8$, and hence, $p(1+p+p^2+p^3)$ is a multiple of $8$.

If $p=2$, this isn't the case, so $p$ must be odd, and hence, $1+p+p^2+p^3$ must be a multiple of $8$.

mod $8$, $1+p+p^2+p^3=1+p+1+p$, so $2+2p$ is a multiple of $8$, so $1+p$ is a multiple of $4$. So, $p\equiv3\bmod 4$.

Through similar logic $\bmod{12}$, we have $p\equiv-1\bmod{12}$, or $p=3$.

Rushabh Mehta
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  • I don't understand your last step. How did you exclude cases like $p=11, 23$, etc.? – Neat Math Jan 06 '21 at 16:10
  • @NeatMath The last steps was a bunch of steps in $1$. $\bmod 12$, our prime can be $3,7,11$. which, when plugged into the formula mod $12$, give you $1,5,1$. Hence, we can't be congruent to $7\bmod 12$. And, we can't be congruent to $3\bmod 12$ unless $p=3$, so the only possibility is $11\bmod12$. – Rushabh Mehta Jan 06 '21 at 16:12
  • I understand that $p\equiv -1 \pmod{12}$. I'm asking why you wrote $p=3$ in your last step? – Neat Math Jan 06 '21 at 16:14
  • That's because that the only way that $p\equiv3\bmod{12}$ is if $p=3$, since $p$ would be a multiple of $3$. – Rushabh Mehta Jan 06 '21 at 16:15
  • OK I see. Your "or" does mean "or", not "or, equivalently" which I hate to see other people using. – Neat Math Jan 06 '21 at 16:19
  • Hi, did you mean $p \equiv 3 (\mod 12)$ ? – Anonymous Jan 06 '21 at 16:21
  • @Anonymous Don Thousand said this is some "thoughts", not a complete answer. If you haven't fully understood the comments why did you accept it as the answer? – Neat Math Jan 06 '21 at 16:23
  • @NeatMath I have understood, that's why I did. What he did was basically consider the terms in mod 8 and then p comes to be in mod 4, then consider p in mod 12 and I suppose in that way the only solution would be when p = 3 . – Anonymous Jan 06 '21 at 16:47
  • @Anonymous Don's conclusion is either $p=3$ or $p\equiv -1 \pmod{12}$. Cases when $p=11, 23$ etc. have not been excluded yet. – Neat Math Jan 06 '21 at 17:10
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Notice that if $p>3$ we get $$ \Big(p^2+{p-1\over 2}\Big)^2<\underbrace{p^4+p^3+p^2+p+1}_{k^2}<\Big(p^2+{p+1\over 2}\Big)^2$$ implies $$p^2+{p-1\over 2}<k<p^2+{p+1\over 2}$$

So we are left to check if $p=2$ or $p=3$ works...

nonuser
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