Determine all triples $(x,y,z)$ of positive integers such that $$(x+y)^2 + 3x + y + 1 = z^2$$
Assume $x\geq y$ then
$$(x+x)^2 + 3x + x + 1 = 4x^2 + 4x + 1\geq z^2$$
Taking this as a quadratic in $x$ we have $\Delta = 4^2 - 4(4)(1-z^2) \leq 0 \implies 16z^2 \leq 0 \implies z = 0$, but $z\in \mathbb{Z^{+}}$
Contradiction.
However the problem does actually have solutions of the form $k,k,2k+1$