<s>[INST] How to check if $ x^2 + x = y^2 - 1$ has positive integer solutions? How to check if $ x^2 + x = y^2 - 1$ has positive integer solutions?
I've tried using different modulos to simplify problem, but nothing has worked yet.
 [/INST] As AmateurMathPirate said, we have $(2y)^{2} - (2x+1)^{2} = 3$
Thus $(2y - 2x - 1)(2y + 2x + 1) = 3$ 
Then $(2y - 2x - 1)|3$ and $(2y + 2x +1)|3$
So $(2y - 2x -1) = 1, -1, 3,$ or $-3$
i) if $(2y - 2x - 1) = 3 \implies (2y + 2x + 1) = 1 \implies 2y + 2x = 0 \implies y = -x \implies x\leq 0$ or $y\leq 0$ then it can't be
ii) if $(2y - 2x - 1) = -3 \implies (2y + 2x + 1) = -1 \implies 2y + 2x = 2 \implies x + y = 1$, but $x,y\geq 1$ then $x + y \geq 2$ then it can't be
iii) if $(2y - 2x -1) = 1 \implies (2y + 2x + 1) = 3$ but $x,y\geq 1$ then $2y + 2x + 1 \geq 5$ then it can't be
iv) if $(2y - 2x - 1) = -1 \implies 2y+ 2x + 1 = -3$ but $2x + 2y +1 >0$ then it can't be
So there is not exist that positive intengers
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