# Find all the integer solutions of $x^2-4y^2=1$

Find all the integer solutions of the equation: $$x^2-4y^2=1$$.

I know I can't solve it like a PELL equation because d is a square in this case.

• In such cases you can factorise using difference of squares Jan 5, 2021 at 8:56
• What is the problem? $(x-2y)(x+2y) = 1$ 1.1 =1 and -1.-1=1
– user822140
Jan 5, 2021 at 9:00

We have $$1=x^{2}-4y^{2}=x^{2}-(2y)^{2}=(x-2y)(x+2y)$$, and since we are working over the integers, both of these must be integers which implies that both are either $$1$$ or $$-1$$.
Hence, we must have $$x-2y=x+2y \implies -2y=2y \implies -2 = 2$$ for $$y \neq 0$$, and so there are no solutions where $$y \neq 0$$.
If $$y = 0$$, then $$x^{2} = 1$$ $$\implies x= 1$$ and $$x = -1$$ are the only solutions.
Therefore, the solutions to $$x^{2}-4y^{2}=1$$ are $$(1,0)$$, and $$(-1,0)$$.
Rewrite is as$$(x-2y)(x+2y)=1$$ Then, there will be two cases $$\begin{cases} x-2y = 1 \\ x+2y = 1 \end{cases} \qquad \text{or}\qquad \begin{cases}x-2y = -1 \\ x+2y = -1 \end{cases}$$