# Prove that $x^2 + x +1 = y^2$ has no positive integer solutions

I began the proof by factoring the LHS to be $$x(x+1)+1 = y^2$$. Since the LHS contains the product of two consecutive integers, that part is even. But since $$1$$ is added, it must be odd. That means $$y$$ must also be odd. How should I continue from here?

If $$x^2+x+1=y^2$$ with $$x$$ a positive integer, then $$x^2
• there are no perfect squares between $x^2$ and $(x+1)^2$, so $y$ can't be an integer; i.e., $y$ is strictly between consecutive integers $x$ and $x+1$, so $y$ is not an integer – J. W. Tanner Sep 11 '19 at 23:54
• Proof by contradiction: suppose the equation has a solution in the positive integers. However, using Tanner's analysis, you can see that $x < y < x + 1$, so $y$ cannot be an integer, and there's your contradiction. Done. – Sigma Sep 11 '19 at 23:58
$$x^2+x+1=y^2$$ implies $$4x^2+4x+4=(2y)^2$$ or $$(2y)^2-(2x+1)^2 = 3$$. The only integer squares that differ by $$3$$ are $$4$$ and $$1$$, so the only solutions are given by $$(x,y)\in\{(0,1),(0,-1),(-1,1),(-1,-1)\}$$.