ta có: 49 - y2 = 12(x - 2001)2
=> \(12\left(x-2001\right)^2\le49\\ \Rightarrow\left(x-2001\right)^2\le\frac{49}{12}\approx4\)
mà (x - 2001)2 là số chính phương
=> \(\left(x-2001\right)^2=\left\{0;1;4\right\}\)
nếu (x - 2001)2 = 0
=> x - 2001 = 0 => x = 2001
=> 49 - y2 = 0 => y2 = 49 \(\Rightarrow\left[\begin{matrix}y=7\\y=-7\left(loại\right)\end{matrix}\right.\)
nếu (x - 2001)2 = 1
\(\left\{\begin{matrix}x-2001=1\\x-2001=-1\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=2002\\x=2000\end{matrix}\right.\)
\(\Rightarrow49-y^2=12\Rightarrow y^2=37\left(loại\right)\)
nếu (x - 2001)2 = 4
\(\Rightarrow\left\{\begin{matrix}x-2001=2\\x-2001=-2\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=2003\\x=1999\end{matrix}\right.\)
\(\Rightarrow49-y^2=12.4=48\Rightarrow y^2=1\Rightarrow\left\{\begin{matrix}y=1\\y=-1\left(loại\right)\end{matrix}\right.\)
vậy ta có các cặp (x;y) là (2001;7), (2003;1), (1999;1)
