26. Let $p$ be an odd prime, and let $T_{p}^{*}(r)$ denote the number of solutions to the congruence equation
$$x_{1}^{2}+\cdots+x_{r}^{2} \equiv 0(\bmod p), \quad 1 \leqslant x_{1} \leq c$$
where $c = (p-1)/2$. When $r > 5$, we have
$$\begin{aligned} (3!) p T_{p}^{*}(3)= & \sum_{l=1}^{p} \sum_{x_{1}=1}^{c} \sum_{x_{2}=1}^{c} \sum_{x_{3}=1}^{c} \mathrm{e}^{2 \pi i l\left(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}\right) / p} \\ & -3 \sum_{l=1}^{p} \sum_{x_{1}=1}^{c} \sum_{x_{2}=1}^{c} \mathrm{e}^{2 \pi i l\left(x_{1}^{2}+2 x_{2}^{2}\right) / p}+2 \sum_{l=1}^{p} \sum_{x_{1}=1}^{c} \mathrm{e}^{2 \pi i l\left(3 x_{1}^{2}\right) / p}, \end{aligned}$$
This formula also holds for $p=3,5$.
(iii) Prove that $T_{3}^{*}(4)=T_{5}^{*}(4)=T_{i}^{*}(4)=0$. When $p>7$, we have
$$\begin{array}{l} \text { (4!) } p T_{p}^{*}(4)=\sum_{l=1}^{p} \sum_{x_{1}=1}^{c} \cdots \sum_{x_{4}=1}^{c} \mathrm{e}^{2 \pi i l\left(x_{1}^{2}+\cdots+x_{4}^{2}\right) / p} \\ \quad-6 \sum_{l=1}^{p} \sum_{x_{1}=1}^{c} \sum_{x_{2}=1}^{c} \sum_{x_{3}=1}^{c} \mathrm{e}^{2 \pi i l\left(x_{1}^{2}+x_{2}^{2}+2 x_{3}^{2}\right) / p} \\ \quad+3 \sum_{l=1}^{p} \sum_{x_{1}=1}^{c} \sum_{x_{2}=1}^{c} \mathrm{e}^{2 \pi i l\left(2 x_{1}^{2}+2 x_{2}^{2}\right) / p} \\ \quad+8 \sum_{l=1}^{p} \sum_{x_{1}=1}^{c} \sum_{x_{2}=1}^{c} \mathrm{e}^{2 \pi i l\left(x_{1}^{2}+3 x_{2}^{2}\right) / p} \end{array}$$
This formula also holds for $p=3,5,7$.
(iv) Using the method of Example 4 in §4 and equation (63), find the expressions for $T_{p}^{*}(3)$ and $T_{p}^{*}(4)$.