38. Find such real numbers $a, b, p, q$ so that the equality

$$
(2 x-1)^{20}-(a x+b)^{20}=\left(x^{2}+p x+q\right)^{10}
$$

is satisfied for any $x$.
38. Answer: $\quad a_{1,2}= \pm \sqrt[20]{2^{20}-1} ; \quad b_{1,2}=\mp \sqrt[20]{2^{20}-1} / 2$, $p=-1, q=1 / 4$ (two sets of coefficients in total).

Substituting $x=1 / 2$, we get

$$
\left(\frac{a}{2}+b\right)^{20}+\left(\frac{1}{4}+\frac{p}{2}+q\right)^{10}=0
$$

from which $a=-2 b$.

Now the identity takes the form

$$
(2 x-1)^{20}=(-2 b x+b)^{20}+\left(x^{2}+p x+q\right)^{10} .
$$

Calculating the coefficient of $x^{20}$ on both sides, we get

$$
2^{20}=2^{20} b^{20}+1 ; \quad b= \pm \sqrt[20]{2^{20}-1} / 2
$$

Now the identity takes the form

$$
(x-1 / 2)^{20}=\left(x^{2}+p x+q\right)^{10}
$$

from which $x^{2}+p x+q=(x-1 / 2)^{2}$, i.e., $p=-1, q=1 / 4$.