The Problems of Ukrainian Authors

For $n\ge 2$ let $q_1(x)=q_2(x)=...=q_n(x)=0$ and $p_2(x)=p_3(x)=...=p_n(x)=0$, $q_2(x)=-\frac{x+1}{3}$, $q_1(x)=\frac{x-2}{3}$, $p_1(x)=\frac{1}{\sqrt{3}}(x+1)$. Then

$$\sum _{i=1}^n(p_i^2(x)+q_i^3(x))=x,(1)$$ and changing $x$ by $g(x)$ for all polynomials $p_i(x)$ and $q_i(x)$, any polynomial $g(x)$ can be obtained:

$$q_2(x)=-\frac{g(x)+1}{3},q_1(x)=\frac{g(x)-2}{3},p_1(x)=\frac{1}{\sqrt{3}}(g(x)+1),$$ from (1) it is easy to see that we get what is required.

Lets show that $n=1$ does not suit. Suppose that there exist such polynomials $p(x)$ and $q(x)$ that $$p^2(x)+q^3(x)=x.(2)$$ Changing $x$ by $x^2$ in (2), we get: $(x-p(x^2))(x+p(x^2))=q^3(x^2)$. Suppose that both polynomials $x-p(x^2)$ and $x+p(x^2)$ have common (possibly complex) root $\alpha$, then $\alpha =p({\alpha }^2)=-p({\alpha }^2)$, hence $\alpha =0$ and $p(0)=p({\alpha }^2)=\alpha =0$, $q(0)=0$. But in such case the left-hand side of (2) is divisible by $x^2$, and the right-hand side is not, which is

impossible. So $x-p(x^2)$ and $x+p(x^2)$ have no common roots. Then each of them has to be a cube of some polynomial (with, possibly, complex coefficients). I.e. $x-p(x^2)=q_1^3(x)$ and $x+p(x^2)=q_2^3(x)$. Adding up these equalities we get $$2x=q_1^3(x)+q_2^3(x)=(q_1(x)+q_2(x))(q_1^2(x)-q_1(x)q_2(x)+q_2^2(x)),$$ and 2 cases are possible: 1) $q_1(x)+q_2(x)=c$; 2) $q_1(x)+q_2(x)=2cx$. For the first case, $\frac{2x}{c}=q_1^2(x)-q_1(x)q_2(x)+q_2^2(x)=c^2-3q_1(x)(c-q_1(x))$, which is impossible because the degree of the left-hand side is odd and of the right-hand side is even. For the second case $\frac{1}{c}=q_1^2(x)-q_1(x)q_2(x)+q_2^2(x)=4c^2{x}^2-3q_1(x)(2cx-q_1(x))$, thus $\mathrm{deg}{q}_1(x)=1$, i.e. $q_1(x)=ax+b$ ($a\ne 0$). Then the equality $p(x^2)=x-q_1^3(x)$ is impossible, because the right-hand side is the polynomial of the third degree, and the left-hand side is the polynomial of even degree.