10. If $x$ is an integer, $3<x<200$, and $x^{2}+(x+1)^{2}$ is a perfect square, then the value of the integer $x$ is $\qquad$.
10.20 or 119.

Let $x^{2}+(x+1)^{2}=v^{2}$, then $(2 x+1)^{2}=2 v^{2}-1$. Let $u=2 x+1$, then $u^{2}-2 v^{2}=-1$. This is a Pell's equation, its fundamental solution is $\left(u_{0}, v_{0}\right)=(1,1)$.
All positive integer solutions can be obtained by
$$
u_{n}+v_{n} \sqrt{2}=\left(u_{0}+v_{0} \sqrt{2}\right)^{2 n+1}
$$

where, $\left(u_{1}, v_{1}\right)=(7,5),\left(u_{2}, v_{2}\right)=(41$, 29), $\left(u_{3}, v_{3}\right)=(239,169), u_{4}>400$.
Therefore, $x=20$ or 119.