Prove that there exist infinitely many pairs $(a,b)$ of integers such that $x^{2020}=ax+b$ has among its solutions two distinct real numbers whose product is 1.
I tried to construct $a$ and $b$ by taking two real roots whose product is one and the rest of the roots as complex roots of unity but still wasn't working for me.
Can this be extended for any $n$ instead of $2020$? Maybe once I have the proof then it will be clear.
On
The polynomial $x^{2020} - ax - b$ has a pair of roots multiplying to $1$ if and only if it is divisible by a quadratic of the form $x^2 - px + 1$. To enforce these roots be real and distinct, we further insist on discriminant of this quadratic being positive, i.e. $p^2 - 4 > 0$.
So, we are assuming that $x$ is a real number satisfying $x^2 = px - 1$ where $p^2 > 4$ and we wish to have $x^{2020} = ax + b$. Let us do some investigation as to the powers of $x$. The fact that $x^2$ can be expressed as $px - 1$ allows any power of $x$ to be expressed in terms of $x$ and $1$. Let's say that we have sequences $c_n, d_n$ such that $$x^n = c_n x + d_n$$ Then $$x^{n+1} = x(c_n x + d_n) = c_n x^2 + d_n x = c_n (px - 1) + d_n x = (pc_n + d_n)x - c_n.$$ If we restrict our attention to integers $p \ge 3$, then $x^n$ can be expressed as an integer combination of $x$ and $1$. We will get $x^{2020} = c_{2020}x + d_{2020}$, where $c_{2020}, d_{2020} \in \Bbb{Z}$. Plus, $x^2 - px + 1$ will divide the polynomial, and hence there will be two distinct real roots whose product is $1$. So, simply by choosing $p$ to be one of the infinitely many integers greater than or equal to $3$, we obtain corresponding $a = c_{2020}$ and $b = d_{2020}$ so that the roots of $x^2 - px + 1$ are roots of $x^{2020} - ax - b$.
There's a slight potential kink here, which is that different values of $p$ may not produce different values of $a$ and $b$. However, if distinct $p_1, \ldots, p_n \ge 3$ produce the same $a, b$ then that says both $x^2 - p_ix + 1$ divides the polynomial $x^{2020} - ax - b$ for all $i = 1, \ldots, n$. Since each $p_i$ is distinct, and produces a quadratic with completely different positive roots, the quadratics $x^2 - p_ix + 1$ and $x^2 - p_j x + 1$ share no common factors, and so we can conclude $$\prod_{i=1}^n (x^2 - p_ix + 1) \mid x^{2020} - ax - b.$$ But obviously this poses a problem if $n > 1010$. Therefore, while it is possible for distinct $p \ge 3$ to produce the same $a, b$, it still must produce infinitely many distinct $a, b$ pairs.
Notice that two roots of product $p=1$ and sum $s$ verify the equation $x^2-sx+p=0$
$x^3=x^2x=(sx-1)x=sx^2-x=s(sx-1)-x=(s^2-1)x-s$
You can prove by induction that $x^n=f_n(s)x+g_n(s)$ with $\begin{cases}f_{n+1}(s)=sf_n(s)+g_n(s)\\g_{n+1}(s)=-f_n(s)\end{cases}$
For for these two roots $x,y$ you get $\begin{cases}x^{2020}=f_{2020}(s)x+g_{2020}(s)=ax+b\\y^{2020}=f_{2020}(s)y+g_{2020}(s)=ay+b\end{cases}$
By difference and since $x-y\neq 0$ as the problem specifies explicitely two distinct real numbers
we get: $\begin{cases}a=f_{2020}(s)\\b=g_{2020}(s)\end{cases}$
As long as $s$ is an integer, it follows that $f_n(s)$ and $g_n(s)$ are also integers.
Therefore you can generate $(a,b)$ integers and $(x,y)$ real roots of product $1$ that answer the problem by starting from $x^2-sx+1=0$ and varying $s\in\mathbb Z$.
But notice that $f_n(s)=\dfrac{x^n-y^n}{x-y}$ and that $xy=1\implies |x|>1\text{ and }|y|<1\quad$
(or vice-versa, WLOG let assume $|x|>|y|$)
It ensues that $|y|^{2020}\ll 1$ and that $a=\operatorname{round}(\frac{x^{2020}}{x-y})\ge\operatorname{round}(\frac 12x^{2019})$
And since $x=\frac{s+\sqrt{s^2-4}}{2}\nearrow$ when $s\nearrow$ we have also $a\nearrow$, and strict monotonicity gives injectivity.
Edit:
Note that for $s$ large, $x\sim s$ and $a\sim s^{n-1}$, so you are pretty much ensured of injectivity for any exponent instead of $2020$ as long as your $s$ is large enough, which is not an issue since $s$ can be taken arbitrarily in $\mathbb Z$.