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.
 A: One way is to see if you can find them.  Complete the square when dealing with a quadratic equation such as this:
$x^2+x=y^2-1 \to \cdots \to (2y)^2-(2x+1)^2=3$
So the revelation is that it's a difference of squares equation.  
For any integer that's not divisible by $2$ only once (example $2,6,10,14, \ldots$), one can always find at least one pair of integers whose difference of squares equals that number.  
If $3$ can be broken up into two divisors in the following ways:  $(3,1)$ and $(-3,-1)$, you can call the first divisor $d_1$ and the second $d_2$ and use the convenient fact that:
$$\left[\frac{d_1+d_2}{2}\right]^2-\left[\frac{d_1-d_2}{2}\right]^2=d_1\cdot d_2$$
Thus you can equate the following:
$$
\begin{cases}
2y&=\frac{d_1+d_2}{2} \\
2x+1&=\frac{d_1-d_2}{2}
\end{cases}
$$
And then plug in the divisor-pairs, $(d_1,d_2)$, of $3$ from above...you should get $(x,y)$ values $(0,0)$ and $(-1,-1)$.
You have only one answer simply because $3$ can only be decomposed into two factors in one way, cuz it's prime.  But suppose your equation instead $x^2+x=y^2-7$.  You'd have the same difference of squares, just now equaling $27$.  Yet $27=27\cdot 1=9\cdot 3$, so here you'd have one more solution. 
A: Let $x,y$ be such that $x$ is a positive integer and $x^2+x+1=y^2$.  $$x^2<x^2+x+1<x^2+2x+1\\x^2<y^2<(x+1)^2$$ 
So clearly $y$ cannot be also be a positive integer since there are no positive integers between $x$ and $x+1$.
A: $x^2+x=y^2-1 \Rightarrow x(x+1)=(y-1)(y+1)$. In other words, two consecutive numbers must have the same product as two numbers that differ by $2$. 
It should be clear that this is impossible for integers $x,y>0$. If both $(y-1)$ and $(y+1)$ are either larger than $x+1$, or smaller than $x$, the product $y^2-1$ will be respectively either larger or smaller than $x^2+x$. 
So one of $(y-1)$ or $(y+1)$ must be equal to one of $x$ or $(x+1)$. But for whichever one is equal, the other one is not, so the two products cannot be equal. The sole exception is the case in which one of the terms in $y$ and one of the terms in $x$ are both $0$, because then the products are both $0$ even though the non-zero terms in $x$ and $y$ are not equal.
A: Assume that $y=mx$. Since $x,y$ are integers you have that $m \in \mathbb Q$. Now plug $y$ into your equation. You get $$(1-m^2)x^2+x+1=0.$$ A necessary condition in order to have integer values of $x$ from this equation is that its discriminant must be zero. Since $$\Delta=4m^2-3$$ this leads to irrational values of $m$ and this is a contradiction.
A: 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
