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Find all \(n\) for which \(n!+(n+1)!+(n+2)!\) is a perfect square.

 

Thanks in advance for any help!

 Feb 21, 2020
 #1
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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.

 Feb 21, 2020
 #2
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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 \(n = 1~ (1!=1=1^2)\)


\(\begin{array}{|lrcll|} \hline n=0: & 0!+1!+2! = 1+1+2 = 4 = 2^2 \\ n=1: & 1!+2!+3! =1+2+6 = 9 = 3^2 \\ \hline \end{array}\)

 

laugh

 Feb 21, 2020

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