9. Find an integer $x$ such that $\left(1+\frac{1}{x}\right)^{x+1}=\left(1+\frac{1}{2003}\right)^{2003}$
9. 求一個整數 $x$, 使得 $\left(1+\frac{1}{x}\right)^{x+1}=\left(1+\frac{1}{2003}\right)^{2003}$ 。
9. -2004
9. Setting $y=x+1$, the equation becomes $\left(1+\frac{1}{y-1}\right)^{y}=\left(1+\frac{1}{2003}\right)^{2003}$. Setting $z=-y$, we get
$$
\begin{aligned}
\left(1+\frac{1}{-z-1}\right)^{-z} & =\left(1+\frac{1}{2003}\right)^{2003} \\
\left(\frac{z}{z+1}\right)^{-z} & =\left(1+\frac{1}{2003}\right)^{2003} \\
\left(\frac{z+1}{z}\right)^{z} & =\left(1+\frac{1}{2003}\right)^{2003}
\end{aligned}
$$

Hence one possible integer solution is $z=2003$, which corresponds to $y=-2003$ and $x=-2004$.
[From the above solution we also see that the solution for $z$, and hence the solution for $x$, is unique. Indeed, the left hand side of $\left(^{*}\right)$ is strictly increasing in $z$ on the natural numbers, so there is no positive integer solution for $z$ other than 2003. It is clear that $z$ is not equal to 0 or -1 . If $z$ is smaller than -1 , then $x$ is a positive integer. Looking at the original equation, we readily see that this is not possible because on one hand $x$ has to be divisible by 2003 , and on the other hand the power $x+1$ has to be equal to 2003 by counting the exponent of 2003.]