Show $x=0$ and $x=1$ are the only integer solutions

Question

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?

Answer 1

Note that we have (for positive $x$): $$ (x-1)^2\leq x^2-x+1\leq x^2 $$ so your expression lies between two consecutive squares. Now you just have to check for equality on either side, and you will have found all cases.

Answer 2

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

Answer 3

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 $\color{green}{(0, \pm 1)}$.

Answer 4

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.