Question

Tìm các số nguyên dương a,b để thoải mãn phương trình:\(a^2+b^2-13\left(a-b\right)=0\)

Answer 1

\(\Leftrightarrow4a^2-52a+169+4b^2+52b+169=338\)

\(\Leftrightarrow\left(2a-13\right)^2+\left(2b+13\right)^2=338=7^2+17^2+13^2+13^2\)

\(\left[{}\begin{matrix}2a-13=7\\2b+13=17\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}a=10\\b=2\end{matrix}\right.\)

\(\left[{}\begin{matrix}2a-13=13\\2b+13=13\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}a=13\\b=0\end{matrix}\right.\) (loại)

\(\left[{}\begin{matrix}2a-13=17\\2b+13=7\end{matrix}\right.\) \(\Rightarrow b=-3< 0\left(l\right)\)