Question

[Nguyễn Đức Anh]

Tìm x;y\(\in\)Z:

x²-2y²=1

Answer 1

[Kudo Shinichi]

x²-2y²=1

=> x²-2y²=1²

=> x²-1²=2y²

=> ( x.x ) - ( 1.1 ) = 2y²

+) ( x.x ) - ( 1.1 ) = 2y²

^{=>\(\begin{cases}x-1=2\\x+1=2y\end{cases}\)}

=>\(\begin{cases}x=3\\x+1=y^2\end{cases}\)

Do x=3

=> x + 1 = y²

=>3 + 1 = y²

=> y = 2

+) (x - 1).(y + 1)= 2.y.y

=> x - 1 = y; x + 1= 2.y

=>\(\begin{cases}x=y+1\\x+1=2.y\end{cases}\)

Do x = y + 1

=> x + 1 = 2.y

=> (y + 1) + 1 = 2 . y

=> y + 2 = 2 . y

=> 2 . y - y = 2

Vậy (x; y) \(\in\) {3; 2}