Solve the following equation on the set of real numbers:

$$
x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+\cdots+x_{n}^{2}-x_{n+1}=\sqrt{x_{1}+x_{2}+\cdots+x_{n}-x_{n+1}}-\frac{n+1}{4}
$$
Let $a=\sqrt{x_{1}+x_{2}+\cdots+x_{n}-x_{n+1}}$; with this substitution, the equation transforms as follows:

$$
\begin{gathered}
x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}-\left(x_{1}+\cdots+x_{n}-a^{2}\right)=a-\frac{n+1}{4} \\
\left(x_{1}^{2}-x_{1}+\frac{1}{4}\right)+\left(x_{2}^{2}-x_{2}+\frac{1}{4}\right)+\ldots\left(x_{n}^{2}-x_{n}+\frac{1}{4}\right)+\left(a^{2}-a+\frac{1}{4}\right)=0 \\
\left(x_{1}-\frac{1}{2}\right)^{2}+\cdots+\left(x_{n}-\frac{1}{2}\right)^{2}+\left(a-\frac{1}{2}\right)^{2}=0
\end{gathered}
$$

This holds precisely when $x_{1}=\cdots=x_{n}=a=\frac{1}{2}$, in which case $x_{n+1}=x_{1}+\cdots+x_{n}-a^{2}=\frac{2 n-1}{4}$. With these values, both sides of (1) equal $\frac{1}{2}$.