Question

[Trần ngô hạ uyên]

Tìm các số nguyên x,y thỏa mãn phương trình: \(xy^2+2xy+x=32y\)

Answer 1

[Phùng Gia Bảo]

\(xy^2+2xy+x=32y\)

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

\(\Rightarrow x=\frac{32y}{\left(y+1\right)^2}\)

Vì \(\left(y,\left(y+1\right)^2\right)=1\)và \(x\inℤ\)\(\Rightarrow\left(y+1\right)^2\inƯ\left(32\right)=Ư\left(2^5\right)=\left\{2^2;2^4\right\}\)

\(Khi\left(y+1\right)^2=2^2=4\Rightarrow\orbr{\begin{cases}y+1=2\\y+1=-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}y=1\\y=-3\end{cases}}\)

\(\cdot y=1\Rightarrow x=\frac{32.1}{4}=8\)        

\(\cdot y=-3\Rightarrow x=\frac{32.\left(-3\right)}{4}=-24\)

\(Khi\left(y+1\right)^2=2^4=16\Rightarrow\orbr{\begin{cases}y+1=4\\y+1=-4\end{cases}\Leftrightarrow\orbr{\begin{cases}y=3\\y=-5\end{cases}}}\)

\(\cdot y=3\Rightarrow x=\frac{32.3}{16}=6\)

\(\cdot y=-5\Rightarrow x=\frac{32.\left(-5\right)}{16}=-10\)

Vậy nghiệm phương trình \(\left(x;y\right)=\left(8;1\right);\left(-24;-3\right);\left(6;3\right);\left(-10;-5\right)\)