Question

Find two consecutive whole numbers such that the sum of their squares is 221.

Answer 1

$n^2+(n+1)^2=2n^2+2n+1=221$
$2n^2+2n=220$
$n^2+n=110$
$n^2+n-110=0$
$(n-10)(n+11)=0\implies n=10,n=-11$

But since $n$ must be a whole number, we ignore $n=-11$. So the two integers are $n=10$ and $n+1=11$.