Question

Giải phương trình nghiệm nguyên

\(x^4+2x^3+3x^2+2x=y^2-y\)

Answer 1

\(\Leftrightarrow4x^4+8x^3+12x^2+8x+1=\left(2y-1\right)^2\)

Ta có:

\(VT=\left(2x^2+2x+2\right)^2-3< \left(2x^2+2x+2\right)^2\)

\(VT=\left(2x^2+2x+1\right)^2+4x\left(x+1\right)\)

Do \(x\in Z\Rightarrow x\left(x+1\right)\ge0\Rightarrow VT\ge\left(2x^2+2x+1\right)^2\)

\(\Rightarrow\left(2x^2+2x+1\right)^2\le\left(2y-1\right)^2< \left(2x^2+2x+2\right)^2\)

\(\Rightarrow\left(2y-1\right)^2=\left(2x^2+2x+1\right)^2\)

\(\Rightarrow\left(2x^2+2x+1\right)^2+4x\left(x+1\right)=\left(2x^2+2x+1\right)^2\)

\(\Rightarrow x\left(x+1\right)=0\Rightarrow\left[{}\begin{matrix}x=0\Rightarrow y=\left\{0;1\right\}\\x=-1\Rightarrow y=\left\{0;1\right\}\end{matrix}\right.\)