Question

tìm các nguyện tự nhiên (x,y) của phương trình:

\(\left(x^2+4y^2+28\right)^2=17\left(x^4+y^4+14y^2+49\right)\)

Answer 1

Answer 2

\(\Leftrightarrow-\left(4x^2-y^2-7\right)^2=0\)

SURPRISE MOTHERFUKA !!

Answer 3

\(\left(x^2+4y^2+28\right)^2=17\left(x^4+y^4+14y^2+49\right)\)

đặt \(z=y^2+7;y\in N;z\in Z;z\ge7\)

\(\Leftrightarrow\left(x^2+4z\right)^2=17\left(x^4+z^2\right)\)

\(\Leftrightarrow x^4+8x^2z+16z^2=17x^4+17z^2\)

\(\Leftrightarrow16x^4-8x^2z+z^2=0\Leftrightarrow\left(4x-z\right)^2=0\)

\(\Leftrightarrow4x^2=z\Leftrightarrow y^2+7=4x^2\Leftrightarrow\left(2x\right)^2-y^2=7\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x\right)^2=16\\y^2=9\end{matrix}\right.\) \(\Leftrightarrow x,y,\in N;\left(x;y\right)=\left(2;3\right)\)