$2019^{2n}$ can be expressed as a sum of ten different positive squares, for every positive integer $n$. Show that $2019^{2n}$ can be expressed as a sum of ten different positive squares, for every positive integer $n$. 
I have tried to solve this problem considering first particular cases and then expressing 2019 as the sum of two integers and trying to use and apply Newton's binomial, but I have not taken what is wanted. I would like to know if there is a less cumbersome test using perhaps a simpler method.

Hint: if $2019^{2m} = a_1^2 +\ldots + a_{10}^2$, then
$2019^{2m+2} = (2019 a_1)^2 + \ldots + (2019 a_{10})^2$.  So you really only
need to do it for $m=1$. 
