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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.
Would appreciate your help:)

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

2 Answers 2

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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 $(-1,0)$.

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Rewrite is as$$(x-2y)(x+2y)=1$$ Then, there will be two cases $$\begin{cases} x-2y = 1 \\ x+2y = 1 \end{cases} \qquad \text{or}\qquad \begin{cases}x-2y = -1 \\ x+2y = -1 \end{cases}$$

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