Recreatii Matematice 1/2016, Problem VIII.201

Problem:
Prove that the number $2(n^4-n^2+1)$ is the sum of two perfect squares for all $n \in \mathbb{N}$.

Proposed by Alessandro Ventullo, Milan, Italy

Solution:We have $$\begin{array}{lll}2(n^4-n^2+1)&=&2n^2(n^2-1)+2\\&=&2n\cdot(n+1)n(n-1)+2\\&=&[(n+2)+(n-2)](n+1)n(n-1)+2\\&=&(n+2)(n+1)n(n-1)+1+(n+1)n(n-1)(n-2)+1 \\&=&[(n+2)(n-1)(n+1)n+1]+[(n+1)(n-2)n(n-1)+1]\\&=&[(n^2+n-2)(n^2+n)+1]+[(n^2-n-2)(n^2-n)+1]\\&=&(n^2+n-1)^2+(n^2-n-1)^2. \end{array}$$