Question

[Nguyễn Công Hiếu]

Tìm các số nguyên dương x, y, z thỏa mãn xz=y^2 và x^2+z^2+99=7y^2.

Answer 1

[Diệu Huyền]

\(xz=y^2\Rightarrow2xz=2y^2\)

\(x^2+z^2+99=7y^2\)

\(\Rightarrow x^2+z^2+2xz+99=7y^2+2y^2\)

\(\Rightarrow\left(x+z\right)^2+99=9y^2=\left(3y\right)^2\)

\(\Rightarrow\left(x+z\right)^2-\left(3y\right)^2=-99\)

\(\Rightarrow\left(x+z+3y\right)\left(x+z-3y\right)=-99=-\left(9.11\right)=-\left(3.33\right)=-\left(99.1\right)\)

Gọi: \(x+z=a;3y=b\)

\(\Rightarrow\left(a+b\right)\left(a-b\right)=-\left(99.1\right)=-\left(3.33\right)=-\left(99.1\right)\)

Trường hợp 1: \(\left(a+b\right)\left(a-b\right)=-\left(9.11\right)\)

\(\Rightarrow\)\(\left[{}\begin{matrix}\left\{{}\begin{matrix}a+b=11\\a-b=-9\end{matrix}\right.\\\left\{{}\begin{matrix}a+b=9\\a-b=-11\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=1\\b=10\end{matrix}\right.\\\left\{{}\begin{matrix}a=-1\\b=10\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+z=1\\3y=10\end{matrix}\right.\\\left\{{}\begin{matrix}x+z=-1\\3y=10\end{matrix}\right.\end{matrix}\right.\) \(\left(ktm\right)\)

Trường hợp 2: \(\left(a+b\right)\left(a-b\right)=-\left(9.11\right)\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a+b=33\\a-b=-3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=15\\b=18\end{matrix}\right.\\\Rightarrow\left\{{}\begin{matrix}x+z=15\\y=6\Rightarrow xz=6^2=36\end{matrix}\right.\\\left\{{}\begin{matrix}a+b=3\\a-b=-33\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+z=15\\3y=18\end{matrix}\right.\\\left\{{}\begin{matrix}x=12\\y=6\\z=3\end{matrix}\right.\\\left\{{}\begin{matrix}x+z=-15\\3y=18\end{matrix}\right.\end{matrix}\right.\)

Trường hợp 3: Không thỏa mãn

Vậy \(x=12;y=6;z=3\) hoặc \(x=3;y=6;z=12\)