4. Given real numbers $a, b, c$ simultaneously satisfy $a-7b+8c=4$ and $8a+4b-c=7$. Then, $a^{2}-b^{2}+c^{2}=$ $\qquad$ .
4.1.

According to the conditions $a+8c=4+7b, 8a-c=7-4b$. Squaring both sides of the two equations and then adding them yields
$$
\begin{array}{l}
(a+8c)^{2}+(8a-c)^{2} \\
=(7+4b)^{2}+(7-4b)^{2} .
\end{array}
$$

Simplifying and organizing, we get $65\left(a^{2}+c^{2}\right)=65\left(1+b^{2}\right)$.
Therefore, $a^{2}-b^{2}+c^{2}=1$.