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 $p$ .

Question

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?

Answer 1

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 $p=3$.

Answer 2

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...