Remainder of 1?

Sum of squares of two consecutive integers can be written as

$n^2+(n+1)^2=2n^2+2n+1=2n(n+1)+1$

$n(n+1)$ is divisible by $2$ , since either $n$ or $n+1$ has to be even.

So $2n(n+1)$ is divisible by $4$ and $2n(n+1)+1$ therefore leaves a remainder of $1$ when divided by $4$ .

A pretty straightforward (maybe not well-written) solution: Of any 2 consecutive integers, 1 must be odd and the other must be even. Even squared is even while odd squared is odd. The result follows.