\(\Leftrightarrow4x^2+4x+1=4y^4+4y^3+4y^2+4y+1\) (1)
\(\Leftrightarrow\left(2x+1\right)^2=\left(2y^2+y\right)^2+\left(y+1\right)\left(3y+1\right)\ge\left(2y^2+y\right)^2\)
Mặt khác
\(4y^4+4y^3+4y^2+4y+1=\left(2y^2+y\right)^2+4\left(2y^2+y\right)+4-\left(5y^2+3\right)< \left(2y^2+y+2\right)^2\)
\(\Rightarrow\left(2y^2+y\right)^2\le\left(2x+1\right)^2< \left(2y^2+y+2\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}\left(2x+1\right)^2=\left(2y^2+y\right)^2\\\left(2x+1\right)^2=\left(2y^2+y+1\right)^2\end{matrix}\right.\) (2)
Thay ngược (2) vào (1) tìm y, bạn tự giải nốt
