I assume [tex]n\in\mathbb{Z}[/tex] (or at least [tex]n\in\mathbb{N}[/tex]).
[tex]n^2+(n+1)^2=n^2+n^2+2n+1=2n^2+2n+1=2(n^2+n)+1[/tex]
The above is always odd, because one of the factors of [tex]2(n^2+n)[/tex] is even, which makes the product even as well, and 1 is odd, and the sum of an even number and an odd number is an odd number.
