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,
$$P'(x) = 4046x(x^2-1)^{2022}, R'(x) = 34\left(2-\frac{2}{x^2}\right)\left(2x+1+\frac{2}{x}\right)^{33}$$
so obviously $P'(\pm 1) = R'(\pm 1) = 0$. 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
$$Q'(x) = 14 \cdot 2 \cdot (2x+1)^{13}.$$
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
$$P(x) \equiv Q(x) \mod f(x).$$
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
$$Q(x) = (2x+1)^{14} = (4x^2+4x+1)^7 \equiv (-3)^7 \pmod{f(x)},$$
$$R(x) = \left(\frac{2(x^2+1)}{x} + 1\right)^{34} \equiv (-2+1)^{34} \equiv 1 \pmod{f(x)}.$$
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 $x^6 \equiv 1 \pmod{f(x)}$ so
$$(x^2 - 1)^2 \equiv (-x - 2)^2 \equiv x^2 + 4x + 4 \equiv x^2 - 4x^2 \equiv -3x^2,$$
$$(x^2 - 1)^6 \equiv (-3x^2)^3 = -27x^6 \equiv -27 \pmod{f(x)}.$$
So that
$$P(x) \equiv (x^2 - 1)^{2022} \cdot (x^2 - 1) \equiv (-27)^{337} \cdot (-x - 2) \\ = 27^{337}(x + 2) \neq \text{const mod } f(x).$$
The problem is completely solved. □