Ta có:
`x^4 + 4x^3 + 6x^2+ 4x = y^2 `
`⇔ x^4 + 4x^3 + 6x^2 + 4x + 1 - y^2 = 1`
`⇔ (x+1)^4 – y^2 = 1`
`⇔ [(x+1)^2 –y] [(x+1)^2+y]= 1`
`⇔` $\displaystyle \left\{ \begin{array}{l}(x+1)_{{}}^{2}-y=1\\(x+1)_{{}}^{2}+y=1\end{array} \right.$
Hoặc
`⇔` $\displaystyle \left\{ \begin{array}{l}(x+1)_{{}}^{2}-y=-1\\(x+1)_{{}}^{2}+y=-1\end{array} \right.$
`⇒` $\displaystyle \left[ \begin{array}{l}1+y=1-y\\-1+y=-1-y\end{array} \right.$
`⇒ y = 0`
`⇒ (x+1)^2 = 1`
`⇔ x+1 = ±1`
`⇒ x = 0` hoặc `x = -2`
Vậy `( x, y ) = ( 0, 0 ); ( – 2, 0 )`
Xin hay nhất !
