For prime $p$ do we have $p^3+p^2+p+1=n^2$ infinitely often?

Question

This is a question to ponder about the occurrence of prime $p$ giving $p^3 + p^2 + p +1=n^2$ as is true with $7$ giving $400=20^2$. Do you think this will ever happen again?

Factor, $(p+1)(p^2+1)$ should be a square. For $p > 2$ the $\gcd$ of the factors is $2$, so $p+1 = 2k^2$ and $p^2+1 = 2m^2$. From $p^2 - 2m^2 = - 1$, you get a quick recurrence for candidates. Which of those are primes such that $(p+1)/2$ is a square …? — Daniel Fischer, Jul 17 '18 at 18:58
posed first by Fermat in 1657. See volume I of Dickson's History, pages 54-58. I think $7$ turns out to be the only prime with $\sigma(x^3) = y^2$ — Will Jagy, Jul 17 '18 at 19:02
Does anyone know up to what p was this examined? One also has 50 as twice 5 squared minus 1 is seven squared. Does this ever happen again? — J. M. Bergot, Jul 17 '18 at 19:07
One also has 18^2 + 18 + 1 = 343=7^3. Can another composite like 18 produce a prime cubed? — J. M. Bergot, Jul 17 '18 at 19:09
https://books.google.com/books?id=9agqAwAAQBAJ&pg=PA54&lpg=PA54&dq=fermat+and+wallis+on+sums+of+divisors&source=bl&ots=uTgxFw7t7b&sig=dmHYkM1JEZnhua8cpjy30tNguN0&hl=en&sa=X&ved=0ahUKEwiC6K7S6qbcAhWj0FQKHW5dCscQ6AEIODAD#v=onepage&q=fermat%20and%20wallis%20on%20sums%20of%20divisors&f=false — Will Jagy, Jul 17 '18 at 19:09
Special case of Find all integer solutions of $1+x+x^2+x^3=y^2$ — Sil, Jul 17 '18 at 19:43

score 10 · Answer 1 · answered Jul 17 '18 at 19:26

There are no more positive prime solution other than $p = 7$.

First, we known $p = 2$ doesn't work, we can assume $p$ is an odd prime.
Let's say $p$ is an odd prime such that

$$p^3 + p^2 + p + 1 = n^2\quad\iff\quad(p^2+1)(p+1) = n^2$$

Since $\gcd(p+1,p^2+1) = \gcd(p+1,2) = 2$, we find for some $k, m > 0$,

$$\begin{cases} p^2 + 1 &= 2m^2\\ p + 1 &= 2k^2\\ \end{cases} \quad\iff\quad \begin{cases} m = \sqrt{\frac{p^2+1}{2}}\\k = \sqrt{\frac{p+1}{2}}\end{cases}$$

Subtracting the two equations on the left, we obtain

$$p(p-1) = 2(m+k)(m-k)$$

Since $0 < m - k < m = \sqrt{\frac{p^2+1}{2}} < \sqrt{\frac{p^2+p^2}{2}} = p$, both $2$ and $m-k$ are coprime to $p$.
Since $p$ is a prime, we obtain $p | (m+k)$.
Since $m+k < 2m < 2p$, this forces $p = m + k$. As a result,

$$\begin{align} & 2(p^4 + m^4 + k^4) - (p^2 + m^2 + k^2)^2 = (p+m+k)(p+m-k)(p-m+k)(p-m-k) = 0\\ \iff & p^4 - 6p^3 - 7p^2 = (p-7)(p+1)p^2 = 0\\ \implies & p = 7\end{align} $$

In short, the equation

$$p^3 + p^2 + p + 1 = n^2$$ has one and only one positive prime solution. Namely, $p = 7$.

For prime $p$ do we have $p^3+p^2+p+1=n^2$ infinitely often?

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