Find the values of $n≥1$ for which

$$n!+(n+1)!+(n+2)!$$ is a perfect square.

This problem is from the book ELEMENTARY NUMBER THEORY by DAVID M BURTON under the chapter Primes and their distribution.Here is the link: https://books.google.com/books/about/Elementary_Number_Theory.html?id=2dI3AwAAQBAJ

Any hints of how to solve it are welcome.

  • $\begingroup$ $n=1$ works! ${}{}{}{}{}{}$ $\endgroup$ – mjw Feb 17 at 4:15
  • $\begingroup$ Is there any scope of improvement in my post? Feel free to outline them $\endgroup$ – Jordan Lawson Feb 17 at 4:15
  • $\begingroup$ @mjw The problem requires solutions other than 1 $\endgroup$ – Jordan Lawson Feb 17 at 4:17
  • $\begingroup$ Seems that $n=1$ is the only solution. $\endgroup$ – mjw Feb 17 at 4:28

You have

$$\begin{equation}\begin{aligned} n! + (n+1)! + (n+2)! & = (n!)(1 + (n+1) + (n+1)(n+2)) \\ & = (n!)(n + 2 + (n^2 + 3n + 2)) \\ & = (n!)(n^2 + 4n + 4) \\ & = (n!)(n + 2)^2 \end{aligned}\end{equation}\tag{1}\label{eq1A}$$

Also, Bertrand's postulate states that for any $m \gt 3$, there's a prime number $p$ such that

$$m \lt p \lt 2m - 2 \tag{2}\label{eq2A}$$

If $n \ge 6$ in \eqref{eq1A} is even, there's an $m \gt 3$ such that $n = 2m - 2$ so then $n + 2 = 2m$. From \eqref{eq2A}, there's a prime $p \gt m$, but $p \lt 2m - 2$, so it's a factor in \eqref{eq1A}. However, you also have that $2p \gt 2m$ and, as such, there's only one factor of it in \eqref{eq1A}. This means that \eqref{eq1A} can't be a perfect square.

If $n \ge 7$ in \eqref{eq1A} is odd, there's an $m \gt 3$ such that $n = 2m - 1$ so then $n + 2 = 2m + 1$. As before, there's a prime $p \gt m$, but $p \lt 2m - 1$, so it's a factor in \eqref{eq1A}. However, $2p \ge 2m + 2$, so also as before, there's only one factor in \eqref{eq1A}, meaning it can't be a perfect square.

This just leaves checking for $1 \le n \le 5$, which give $9$, $32$, $150$, $864$ and $5880$, respectively. Thus, as mjw's comment says, $n = 1$ is the only solution as $9 = 3^2$.

Update: As you stated in the comment below, the solution above could've been a bit simpler & shorter since, in \eqref{eq1A}, as $(n+2)^2$ is already a square, it only needed to check for $n!$ being a perfect square.

  • 2
    $\begingroup$ Very nice!${}{}{}{}{}{}$ $\endgroup$ – mjw Feb 17 at 4:33
  • $\begingroup$ @mjw Thanks. I made a small mistake originally, which I've fixed. Also, out of curiosity, I wanted to check on the other possibilities myself, so I finished the rest of the answer at the same time. $\endgroup$ – John Omielan Feb 17 at 4:40
  • 1
    $\begingroup$ I believe you could have just used the fact that $n!$ cannot be a perfect square $\endgroup$ – Jordan Lawson Feb 17 at 5:10
  • $\begingroup$ @HowardDickson You're right. That would have made the answer a bit shorter and simpler. $\endgroup$ – John Omielan Feb 17 at 5:12

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