Question

[Bùi Đức Anh]

Giải pt nghiệm nguyên \(xy^2+2xy+x=32y\)

Answer 1

[Trang Triệu]

Ta có: \(xy^2+2xy+x=32y \)

⇔ \(x\left(y^2+2y+1\right)=32y\)

⇔\(x=\dfrac{32y}{\left(y+1\right)^2}\) 

⇔\(x=\dfrac{32y-32+32}{\left(y+1\right)^2}\)

⇔\(x=\dfrac{32\left(y+1\right)}{\left(y+1\right)^2}-\dfrac{32}{\left(y+1\right)^2}\)

⇔\(x=\dfrac{32}{y+1}-\dfrac{32}{\left(y+1\right)^2}\)

Để x là số dương ⇒ \(\left(y+1\right)^2\)∈ \(U_{\left(32\right)}\)={-32 ;-16;-8;-4;-2;-1;1;2;4;8;16;32}

Nhưng \(\left(y+1\right)^2\)là số chính phương ⇒ \(\left(y+1\right)^2\)∈ {1;4;16}

⇒\(\left[{}\begin{matrix}\left(y+1\right)^2=1\\\left(y+1\right)^2=4\\\left(y+1\right)^2=16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y+1=1\\y+1=2\\y+1=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=0\\y=1\\y=3\end{matrix}\right.\)  

Thay :

y = 0 ⇒ x = 0

y = 1 ⇒ x = 8

y = 3 ⇒ x = 6

Vậy x;y = ( 0;0) ; ( 8;1) ; ( 6;3)