Competition problem — Number theory

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Number theory Difficulty 5.3 AIME, harder Prove it Ukraine

Does there exist a trinomial $f(x) = ax^2 + bx + c$ with integer coefficients such that $a$ isn't divisible by $2022$ and all numbers $f(1), f(2), \ldots, f(2022)$ give different remainders under the division by $2022$ ?

Solution

Consider the following trinomial:
$f(x) = 1011x^2 + 1012x = 1011x(x + 1) + x.$
The first term is always divisible by $2022$ , so $f(x)$ gives remainder $x$ . Proof completed.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.