$5 \cdot 37$ Find real numbers $a, b, p, q$, such that the equation
$$
(2 x-1)^{20}-(a x+b)^{20}=\left(x^{2}+p x+q\right)^{10}
$$

holds for any $x$.
【Sol】Let $x=\frac{1}{2}$, we get
$$
\left(\frac{a}{2}+b\right)^{20}+\left(\frac{1}{4}+\frac{p}{2}+q\right)^{10}=0 .
$$

From this, we obtain
$$
\frac{a}{2}+b=0,
$$

which means
$$
a=-2 b .
$$

Thus, the original equation can be transformed into
$$
(2 x-1)^{20}-(-2 b x+b)^{20}=\left(x^{2}+p x+q\right)^{10} \text {, }
$$

which is
$$
\begin{array}{l}
2^{20}\left(x-\frac{1}{2}\right)^{20}-(2 b)^{20}\left(x-\frac{1}{2}\right)^{20}=\left(x^{2}+p x+q\right)^{10}, \\
{\left[2^{20}-(2 b)^{20}\right]\left(x-\frac{1}{2}\right)^{20}=\left(x^{2}+p x+q\right)^{10} .}
\end{array}
$$

By comparing the coefficients of $x^{20}$ on both sides of the equation, we get
$$
2^{20}-(2 b)^{20}=1 \text {, }
$$

Therefore,
$$
\left(x-\frac{1}{2}\right)^{20}=\left(x^{2}+p x+q\right)^{10} \text {, }
$$

which means $\left(x-\frac{1}{2}\right)^{2}=x^{2}+p x+q$,
hence $p=-1, q=\frac{1}{4}$.
At the same time, from (1) we can solve
$$
b= \pm \frac{1}{2} \cdot \sqrt[20]{2^{20}-1}
$$

Thus, $a=\mp \sqrt[20]{2^{20}-1}$.