Tìm các số dương x, y thỏa x(y+1)^2=32y

Ta có:\(32⋮y\Rightarrow x\left(y+1\right)^2⋮y\) . Mà \(\left(y,y+1\right)=1\Rightarrow\left(y+1\right)^2\) \(⋮̸y\Rightarrow x⋮y\)

Đặt x=yt. Ta có: \(x\left(y+1\right)^2=32y\)

\(\Rightarrow yt\left(y+1\right)^2=32y\)

\(\Rightarrow t\left(y+1\right)^2=32\)

\(\Rightarrow\left(y+1\right)^2\) là Ư chính phương của 32

TH1\(\)\(\left\{\begin{matrix}t=32\\\left(y+1\right)^2=1\end{matrix}\right.\)

\(\Rightarrow\left\{\begin{matrix}t=32\\y+1=1\end{matrix}\right.\Rightarrow\left\{\begin{matrix}t=32\\y=0\end{matrix}\right.\)(loại vì \(y\in\) N*)

TH2\(\left\{\begin{matrix}t=2\\\left(y+1\right)^2=16\end{matrix}\right.\)

\(\Rightarrow\left\{\begin{matrix}t=2\\y+1=4\end{matrix}\right.\)\(\Rightarrow\left\{\begin{matrix}t=2\\y+1=4\end{matrix}\right.\Rightarrow\left\{\begin{matrix}t=2\\y=3\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=6\\y=3\end{matrix}\right.\)

TH3\(\left\{\begin{matrix}t=8\\\left(y+1\right)^2=4\end{matrix}\right.\)

\(\Rightarrow\left\{\begin{matrix}t=8\\y+1=2\end{matrix}\right.\)\(\Rightarrow\left\{\begin{matrix}t=8\\y+1=2\end{matrix}\right.\Rightarrow\left\{\begin{matrix}t=8\\y=1\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=8\\y=1\end{matrix}\right.\)

Vậy có 2 cặp số x,y. Đó là (x=6,y=3) và (x=8,y=1)