1. If $x$ is an integer, $3<x<200$, and $x^{2}+(x+1)^{2}$ is a perfect square, find the integer value of $x$.
$(2007$, I Love Mathematics Junior High School Summer Camp Mathematics Competition)
Let $x^{2}+(x+1)^{2}=v^{2}$. Then
$$
(2 x+1)^{2}+1=2 v^{2} \text {. }
$$

Let $u=2 x+1$. Then
$$
u^{2}-2 v^{2}=-1 \text {. }
$$

This is a Pell's equation.
It is easy to see that $\left(u_{0}, v_{0}\right)=(1,1)$ is a solution to equation (1), and the solution formula for Pell's equation (1) is
$$
u_{n}+\sqrt{2} v_{n}=\left(u_{0}+\sqrt{2} v_{0}\right)^{2 n+1} \text {. }
$$

Thus, we can obtain the solutions
$$
\begin{array}{l}
u_{1}+\sqrt{2} v_{1}=(1+\sqrt{2})^{3}=7+5 \sqrt{2}, \\
u_{2}+\sqrt{2} v_{2}=(1+\sqrt{2})^{5}=41+29 \sqrt{2}, \\
u_{3}+\sqrt{2} v_{3}=(1+\sqrt{2})^{7} \\
=(41+29 \sqrt{2})(1+\sqrt{2})^{2}=239+169 \sqrt{2}, \\
u_{4}+\sqrt{2} v_{4}=(1+\sqrt{2})^{9} \\
=(239+169 \sqrt{2})(1+\sqrt{2})^{2} \\
=1393+985 \sqrt{2} .
\end{array}
$$

Therefore, $u_{1}=7, u_{2}=41, u_{3}=239, u_{4}=1393$.
From $u=2 x+1$, we get $x=3,20,119,696$.
Given $3<x<200$, the values of $x$ are 20 and 119.