HMMT February, 2020

Solution:
Let $P(x)$ have degree $n = 2020$ with roots $r_{1}, \ldots, r_{n}$. Let $R(x) = \prod_{i} \left(x - r_{i}^{2}\right)$. Then
$$\prod_{i} r_{i}^{n} P\left(\frac{x}{r_{i}}\right) = \prod_{i} \prod_{j} \left(x - r_{i} r_{j}\right) = Q(x)^{2} R(x)$$
Using $R\left(x^{2}\right) = (-1)^{n} P(x) P(-x)$ and Vieta, we obtain
$$P(x) P(-x) Q\left(x^{2}\right)^{2} = P(0)^{n} \prod_{i} P\left(\frac{x^{2}}{r_{i}}\right)$$
Plugging in $x = \alpha$, we use the facts that $P(\alpha) = 4$, $P(-\alpha) = 4 - 2 \alpha$, and also
$$P\left(\frac{\alpha^{2}}{r_{i}}\right) = \frac{\alpha^{4040}}{r_{i}^{2020}} + \frac{\alpha^{2}}{r_{i}} + 2 = -\frac{(\alpha-2)^{2}}{r_{i}+2} + \frac{\alpha^{2}}{r_{i}} + 2 = \frac{2\left(-\alpha - r_{i}\right)^{2}}{r_{i}\left(r_{i}+2\right)}.$$
This will give us
$$P(\alpha) P(-\alpha) Q\left(\alpha^{2}\right)^{2} = 2^{n} \prod_{i} \frac{2\left(-\alpha - r_{i}\right)^{2}}{r_{i}\left(r_{i}+2\right)} = 2^{n} \cdot \frac{2^{n} P(-\alpha)^{2}}{P(0) P(-2)}$$
Therefore,
$$\begin{aligned} Q\left(\alpha^{2}\right)^{2} & = \frac{4^{n} P(-\alpha)}{P(0) P(-2) P(\alpha)} \\ & = \frac{4^{n}(4-2 \alpha)}{2 \cdot 2^{2020} \cdot 4} \\ & = \frac{2^{4041}(2-\alpha)}{2^{2023}} \\ & = 2^{2018}(2-\alpha) \end{aligned}$$
We can check that $P(x) - 4$ has no double roots (e.g. by checking that it shares no roots with its derivative), which means that all possible $\alpha$ are distinct. Therefore, adding over all $\alpha$ gives $2020 \cdot 2^{2019}$, because the sum of the roots of $P(x) - 4$ is $0$.