Problem 4. Given a quadratic trinomial $g(x)=x^{2}+a x+b$, which has exactly one root. Find the coefficients $a$ and $b$, if it is known that the polynomial $g\left(x^{5}+2 x-1\right)+g\left(x^{5}+3 x+1\right)$ also has exactly one root.
Solution. Let $x_{0}$ be the unique root of the polynomial $g(x)$. Then the function $g(x)$ maintains its sign at all points $x \neq x_{0}$ and only $g\left(x_{0}\right)=0$. Therefore, the root of the polynomial $f(x)=g\left(x^{5}+2 x-1\right)+g\left(x^{5}+3 x+1\right)$ is only such a point $x_{1}$ that

$$
x_{1}^{5}+2 x_{1}-1=x_{1}^{5}+3 x_{1}+1=x_{0} .
$$

Simplifying the equation: $2 x_{1}-1=3 x_{1}+1$, we find $x_{1}=-2$. Therefore, $x_{0}=(-2)^{5}+2(-2)-1=-37$.

From the uniqueness of the root, it follows that $g(x)$ has the form

$$
g(x)=\left(x-x_{0}\right)^{2}=(x+37)^{2}=x^{2}+74 x+37^{2}
$$

from which $a=74, b=37^{2}=1369$.

Answer: $a=74, b=37^{2}=1369$.