Show $x=0$ and $x=1$ are the only integer solutions I'm trying to show that the only solutions to $x^2-x+1=y^2$ are when $x=0$ and $x=1$. All I can think of is completing the square gives $x^2-x+1=(x-\frac{1}{2})^2+\frac{3}{4}$, which is clearly not a perfect square. Does this suffice? Should I use induction? 

$$x^2-x+1=y^2\\
4x^2-4x+4=4y^2 \\
(2y)^2-(2x-1)^2=3 \\
\left( 2y-2x+1\right) \left(2y+2x-1 \right)=3
$$
