Team selection tests

First of all, from $P(x),Q(x)$, after all of the operations, we get a polynomial variable $x$, in which $R(x)$ is not a polynomial, so it is not possible to have $$P,Q\to R.$$ Next, we will show that $R,P\to Q$ cannot be obtained. Indeed, $$P^{\prime }(x)=4046x(x^2-1{)}^{2022},R^{\prime }(x)=34\left(2-\frac{2}{x^2}\right){\left(2x+1+\frac{2}{x}\right)}^{33}$$ so obviously $P^{\prime }(\pm 1)=R^{\prime }(\pm 1)=0$. The given operations will generate functions that are composite functions of $P(x),R(x)$ so for each of those functions, if taking the derivative, they still have a solution of $\pm 1$. However, $Q(x)$ does not satisfy this because $$Q^{\prime }(x)=14\cdot 2\cdot (2x+1{)}^{13}.$$ Finally, we will show that $R,Q\to P$ cannot be obtained using polynomial congruence, namely by considering modulo $f(x)=x^2+x+1$. Next, if $P(x)-Q(x)$ is divisible by $f(x)$, we denote $$P(x)\equiv Q(x)\mathrm{mod}f(x).$$ It is easy to check that the properties of the integer congruence are still true here. From this, we can also define congruence for fractions. Back to the problem, we have $$Q(x)=(2x+1{)}^{14}=(4x^2+4x+1{)}^7\equiv (-3{)}^7(\mathrm{mod}f(x)),$$ $$R(x)={\left(\frac{2(x^2+1)}{x}+1\right)}^{34}\equiv (-2+1{)}^{34}\equiv 1(\mathrm{mod}f(x)).$$ Therefore, all functions generated by $Q,R$, when taking congruent modulo $f(x)$ are constant. We show that $P(x)$ does not have that property. Indeed, notice that $x^6\equiv 1(\mathrm{mod}f(x))$ so $$(x^2-1{)}^2\equiv (-x-2{)}^2\equiv x^2+4x+4\equiv x^2-4x^2\equiv -3x^2,$$ $$(x^2-1{)}^6\equiv (-3x^2{)}^3=-27x^6\equiv -27(\mathrm{mod}f(x)).$$ So that $$\begin{gathered} P(x)\equiv (x^2-1{)}^{2022}\cdot (x^2-1)\equiv (-27{)}^{337}\cdot (-x-2) \\ =27^{337}(x+2)\ne \text{const mod }f(x). \end{gathered}$$ The problem is completely solved. □