Tìm nghiệm nguyên của pt: $x^{3}-y^{3}-2y^{2}-3y-1=0$ - Số học - Diễn đàn Toán học

Ta có:

\(x^3-y^3-y^2-3y-1=0\)

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

Dễ dàng thấy:

\(\left(y-1\right)^3< y^3+2y^2+3y+1\le\left(y+1\right)^3\)

\(\Leftrightarrow y^3+2y^2+3y+1=\left[\left(y^3\right);\left(y+1\right)^3\right]\)

Làm tiếp nhé

a)

\(\Leftrightarrow yz=z^2+2z+3\Leftrightarrow z\left(y-2-z\right)=3\)

\(\hept{\begin{cases}z=\left\{-3,-1,1,3\right\}\\y-2-z=\left\{-1,-3,3,1\right\}\end{cases}\Rightarrow\hept{\begin{cases}x=\left\{-2,0,2,4\right\}\\y=\left\{-2,-4,6,6\right\}\end{cases}}}\)

\(4x^2=4y^6-4y^3\)

\(\Leftrightarrow4y^6-4y^3+1-4x^2=1\)

\(\Leftrightarrow\left(2y^3-1\right)^2-4x^2=1\)

\(\Leftrightarrow\left(2y^3-1-2x\right)\left(2y^3-1+2x\right)=1\)

\(x^3+y^3=5+x^2y+xy^2\Rightarrow x^3+y^3-\left(x^2y+xy^2\right)=5\)

\(\Rightarrow\left(x+y\right)\left(x^2-xy+y^2\right)-xy\left(x+y\right)=5\)

\(\Rightarrow\left(x+y\right)\left(x-y\right)^2=5\)

Vì \(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0\\5>0\end{matrix}\right.\Rightarrow x+y>0\)

Lại có \(\left\{{}\begin{matrix}\left(x-y\right)^2\in N\\\left(x-y\right)^2< 5\end{matrix}\right.\) và \(\left(x-y\right)^2\) là số chính phương

\(\Rightarrow\left(x-y\right)^2=1\Rightarrow\left\{{}\begin{matrix}x+y=5\\x-y=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\)