$5 \cdot 37$ 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 $x$.
[Solution] 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 $\square$
$$\frac{a}{2}+b=0$$

That is
$$a=-2 b$$

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

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

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

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

That is $\left(x-\frac{1}{2}\right)^{2}=x^{2}+p x+q$,
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 $a=\mp \sqrt[20]{2^{20}-1}$.