Question

Tìm nghiệm nguyên của pt

1+x+\(x^2\)+\(x^3\)=\(y^3\)

Answer 1

Ta có:\(y^3=x^3+x^2+x+1=x^3+\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>x^3\)(1)

Lại có:\(\left(x+2\right)^3-y^3=x^3+6x^2+12x+8-1-x-x^2-x^3=5x^2+11x+7=\left(\sqrt{5}x+\dfrac{11}{2\sqrt{5}}\right)^2+\dfrac{19}{20}>0\)

\(\Rightarrow\left(x+2\right)^3>y^3\left(2\right)\)

Từ (1),(2) và yEZ\(\Rightarrow y^3=\left(x+1\right)^3\)

\(\Rightarrow1+x+x^2+x^3=x^3+3x^2+3x+1\)

\(\Leftrightarrow2x^2+2x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}y=1\\y=0\end{matrix}\right.\)