Gazeta Matematica 4/2014, Problem E:14645

Problem:
Show that for any integer $n > 3$, the number $a_n=10(n^2+1)$ can be written as the sum of four distinct perfect squares.

Proposed by Alessandro Ventullo

Solution:
We have \begin{eqnarray*} 10(n^2+1)&=&8n^2+2+2n^2+8\\&=&4n^2-4n+1+4n^2+4n+1+n^2-4n+4+n^2+4n+4\\&=&(2n-1)^2+(2n+1)^2+(n-2)^2+(n+2)^2, \end{eqnarray*} and the statement follows.