Gazeta Matematica 6-7-8/2017, Problem 27406

Problem:
Prove that the number $$N=1+(1!)^2\cdot1\cdot3+\ldots+(n!)^2\cdot n\cdot(n+2)$$ is a perfect square for all $n \in \mathbb{N}^*$.

Proposed by Alessandro Ventullo, Milan, Italy

Solution:
Observe that for all $k=1,2,\ldots,n$, we have $$\begin{array}{lll} (k!)^2\cdot k\cdot(k+2)&=&(k!)^2[(k^2+2k+1)-1]\\&=&(k!)^2[(k+1)^2-1]\\&=&[(k+1)!]^2-(k!)^2. \end{array}$$ Adding all these equalities, we get $$\begin{array}{lll} N&=&1+[(2!)^2-(1!)^2]+\ldots+[((n+1)!)^2-(n!)^2]\\&=&1+[(n+1)!]^2-(1!)^2\\&=&[(n+1)!]^2. \end{array}$$