\(y=\sqrt{x^2+4x+5}\left(y>1\right)\)
<=> \(y^2=x^2+4x+5=\left(x+2\right)^2+1\)
<=>\(y^2-\left(x+2\right)^2=1\)
<=>\(\left(y-x-2\right)\left(y+x+2\right)=1\) (1)
Do x,y nguyên => \(\left\{{}\begin{matrix}y-x-2\in Z\\y+x+2\in Z\end{matrix}\right.\)
(1) => \(\left\{{}\begin{matrix}y-x-2\inƯ\left(1\right)\\y+x+2\inƯ\left(1\right)\end{matrix}\right.\)
Ta có bảng:
y-x-2 | 1 | -1 |
y+x+2 | 1 | -1 |
y-x | 3 | 1 |
y+x | -1 | -3 |
y | 1 | -1 |
x | -2 | -2 |
K/luận | Tm | Ktm |
Vậy (x,y)=(-2,1)