+288 Find all \(n\) for which \(n!+(n+1)!+(n+2)!\) is a perfect square.
Thanks in advance for any help!
A short computer code could only find one such n:
When n=1. 1! + 2! + 3! = 9
Sqrt(9) = 3 - which is a perfect square.
+26367 Find all \(n\) for which \(n!+(n+1)!+(n+2)!\) is a perfect square.
\(\begin{array}{|rcll|} \hline && n!+(n+1)!+(n+2)! \\ &=& n! + n!(n+1)+n!(n+1)(n+2) \\ &=& n!\Big(1+(n+1)+(n+1)(n+2)\Big) \\ &=& n!\Big(n^2+4n+4\Big) \\ &=& \mathbf{n!(n+2)^2} \\ \hline \end{array} \)
\(n!+(n+1)!+(n+2)!\) is a perfect square, if \(n!\) is a perfect square!
But n! is only a perfect square, if \(n=0 ~(0!=1=1^2)\) or \(n = 1~ (1!=1=1^2)\)
\(\begin{array}{|lrcll|} \hline n=0: & 0!+1!+2! = 1+1+2 = 4 = 2^2 \\ n=1: & 1!+2!+3! =1+2+6 = 9 = 3^2 \\ \hline \end{array}\)
