How to solve in integers the equation $2x^2+x+2x^2y-y^2+y=1$?

Question

Any hints? I could've reduced it to Pell's equation and solved it, if there wasn't $2x^2y$ part.

This is a part of a bigger problem, and that's all I have left to solve the main one.

Answer 1

\begin{align*} 2x^2 &+x+2x^2y-y^2+y=1 \\ \\ 2(y+1)x^2 &+(1)x-(y^2-y+1)=0\\ x &= \dfrac{\sqrt{8 y^3 + 9} - 1}{4 (y + 1)} \\ \\ y^2& - (2x^2+1)y -(2x^2+x-1)=0 \\ y &= \dfrac{2 x^2 + 1+ \sqrt{4 x^4 + 12 x^2 + 4 x - 3}}{2} \end{align*} Solving for $\space x\space$ suggests that $\space y\ne-1,\space$ but solving for $\space y\space$ and viewing a graph of this function shows that this is not necessarily so. The graph of this function suggests that solutions are not far from the origin.

A spreadsheet may not show everything but, for $\space -100 \le x \le 100,\space$ the integer solutions are $\quad (-1,0),\space (-1,3),\space (3,-1), \space (3,20).$