Tìm x để `x^2-2x-4` là số chính phương

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Đặt `x^2-2x-4=y^2`

Ta có: `x^2-2x-4=y^2`

`<=>` `x^2-2x-4-y^2=0`

`<=>` `x^2-2x+1-5-y^2=0`

`<=>` `(x^2-2x+1)-y^2=5`

`<=>` `(x-1)^2-y^2=5`

`<=>` `(x-1-y)(x-1+y)=5`

Ta có bảng sau:

\begin{array}{|c|c|c|}\hline \text{x-1-y}&\text{1}&\text{5}&\text{-1}&\text{-5}\\\hline \text{x-1+y}&\text{5}&\text{1}&\text{-5}&\text{-1}\\\hline \text{x}&\text{4}&\text{4}&\text{-2}&\text{-2}\\\hline \text{y}&\text{-2}&\text{2}&\text{2}&\text{-2}\\\hline\end{array}

Vậy `x=4;x=-2`