Mathzeus1024 wrote:

$\sum_{k=0}^{n}c_{k}(2a^{k}-0) = \sum_{k=0}^{n}c_{k}b^{k} \Rightarrow b^{k}=2a^{k} \Rightarrow \frac{b}{a} = 2^{1/k}$

which is a contradiction since a rational number $\neq$ an irrational number. Thus, no such $b \in \mathbb{Z}$ exists under these conditions.

You can't assume that $b^k=2a^k$ just because $\sum_{k=0}^{n}c_{k}(2a^{k}) = \sum_{k=0}^{n}c_{k}b^{k}.$ In fact, this clearly wrong when $k=0$ and $2^{\frac 1k}$ isn't irrational when $k=1$ , yet such $f,a,$ and $b$ exist as alexheinis pointed out. The rest of your solution is neat, its just this one step that makes an assumption. In fact, I believe the statement is false for all $f$ except constant and quadratic polynomials. Here is a proof:

The constant case is clear and the construction above proves the linear case. If $f$ has degree $n\geq 3$ , then take
$f(x)=g(x)x(x-1)(x-2)+x+2020,$ where $g$ has degree $n-3$ with integer coefficients. Then $f(0)=2020, f(1)=2021,$ and $f(2)=2022.$ (This type of construction is common when you have one polynomial behaving like a smaller degree polynomial, such as $x+2020$ , for a set of $x$ .) For the quadratic case, take $f(x)=cx^2+dx+2020$ . Then $f(a)=2021$ and $f(b)=2022$ give, respectively,
$ca^2+da=a(ca+d)=1$ and
$cb^2+db=b(cb+d)=2.$ The first equation tells us that $a=\pm 1,$ from which we get
$a=1\implies c+d=1$ and
$a=-1\implies -c+d=-1.$ We can do similar analysis for the second equation:
$b=2\implies 2c+d=1,$ $b=1\implies c+d=2,$ $b=-1\implies -c+d=-2,$ and
$b=-2\implies -2c+d=-1.$
It is easy to see that no pair of $a$ and $b$ produce a nonzero, integral value for $c$ (nonzero since we are looking for quadratics), and hence no such quadratics exist.