<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 $4$ to obtain $$(2x-1)^2+3=4y^2$$ or $$4y^2-(2x-1)^2=3$$ or $$(2y-2x+1)(2y+2x-1)=3$$
And the factors are in some order either $3,1$ or $-3, -1$
The sum of the two factors is $4y=4, -4$ respectively, so $y=\pm 1$
The difference between the two factors is $4x-2=\pm 2$ so $x=0,1$
All four combinations actually work.
</s>