Sum of Consecutive Perfect Squares If $n\in\mathbb{Z}$ and $2n+1$ is a perfect square, then is it true that $n+1$ is a sum of two consecutive perfect squares?

Since $2n + 1$ is odd, if it is a perfect square it is the square of an odd number.  So write $2n+1 = (2k+1)^2$.  Then we can solve to get $n+1 = \frac{(2k+1)^2 - 1}{2} + 1$, and further simplify this to $2k^2 + 2k + 1 = k^2 + k^2 + 2k + 1 = k^2 + (k+1)^2$, showing that $n+1$ is indeed the sum of two consecutive squares.
