Question

Tìm tất cả cặp số nguyên dương x;y thỏa mãn

\(2x^2+4y^2-4xy+2x+3=2019\)

Answer 1

\(\Leftrightarrow\left(x^2-4xy+4y^2\right)+\left(x^2+2x+1\right)=2017\)

\(\Leftrightarrow\left(x-2y\right)^2+\left(x+1\right)^2=2017=9^2+44^2\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\left(x-2y\right)^2=9^2\\\left(x+1\right)^2=44^2\end{matrix}\right.\\\left\{{}\begin{matrix}\left(x-2y\right)^2=44^2\\\left(x+1\right)^2=9^2\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left(x;y\right)=\left(43;17\right);\left(43;26\right);\left(8;26\right)\)