5. Given $x=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}, y=$ $\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}-\sqrt{n}}$ ( $n$ is a positive integer) is a solution to the equation $2 x^{2}+$ $207 x y+2 y^{2}=2003$. Then the value of $n$ is ( ).
(A) 7
(B) 8
(C) 9
(D) 10
5.A.

It is easy to get $x y=1, x+y=4 n+2$.
Since $2003=2 x^{2}+207 x y+2 y^{2}=2(x+y)^{2}+203$ $=2(4 n+2)^{2}+203$, therefore, $(4 n+2)^{2}=900, n=7$.