<s>[INST] 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? 
 [/INST] Multiply by the equation by $4$ and complete the square for $x$. We have 
\begin{eqnarray*}
(2x-1)^2+3=(2y)^2 \\
(2y+2x-1)(2y-2x+1)=3
\end{eqnarray*}
So Either 
\begin{eqnarray*}
\cases{2y+2x-1=3 \\
2y-2x+1=1} \text{or}
\cases{2y+2x-1=-3 \\
2y-2x+1=-1}
\end{eqnarray*}
or 
\begin{eqnarray*}
\cases{2y+2x-1=1 \\
2y-2x+1=3} \text{or} \cases{2y+2x-1=-1 \\
2y-2x+1=-3}
\end{eqnarray*}
so  $(x,y)$ has solutions $\color{blue}{(1, \pm 1)}$ and $\color{green}{(0, \pm 1)}$.
</s>