\(25-8\left(x-2016^2\right)=\left(y-1\right)^2.\)

\(Nx:\)\(8\left(x-2016\right)^2\ge0;\left(y-1\right)^2\ge0\)

\(\Rightarrow VT=\left(y-1\right)^2\Leftrightarrow8\left(x-2016\right)^2\le25\Rightarrow\left(x-2016\right)^2\le\frac{25}{8}\Rightarrow\left(x-2016\right)^2\le3\)

Mà \(\left(x-2016\right)^2\)là số chính phương \(\Rightarrow\orbr{\begin{cases}\left(x-2016\right)^2=1\\\left(x-2016\right)^2=0\end{cases}}\)

\(\left(x-2016\right)^2=1\Leftrightarrow\orbr{\begin{cases}x-2016=-1\Leftrightarrow x=2015\\x-2016=1\Leftrightarrow x=2017\end{cases}}\)

\(\left(x-2016\right)^2=0\Leftrightarrow x-2016=0\Leftrightarrow x=2016\)

\(Th1\left(x=2015;x=2017\right)\)

\(25-8=\left(y-1\right)^2\Leftrightarrow\left(y-1\right)^2=17\Leftrightarrow y-1=\sqrt{17}\Leftrightarrow y=\sqrt{17}+1\left(loại\right)\)

\(Th2\left(x=2016\right)\)

\(25-0=\left(y-1\right)^2\Leftrightarrow\left(y-1\right)=5\Leftrightarrow y=6\)

Vậy x = 2016 và y = 6