Finding all quads of integers $(a, b, c, p)$ where $p \ge 5$ is prime number such that the remainders of the numbers $am^3 + bm^2 + cm$, $m = 0, 1, . . . , p - 1$, upon division of $p$ are two by two different..
Problem
Source: IFYM - XI International Festival of Young Mathematicians Sozopol 2022, Theme for 11-12 grade, Round 4 p2
Tags: number theory, remainder
12.11.2022 14:12
I have a hunch this problem is basically just a well known theorem for when cubic polynomials $P$ map $\mathbb{F}_p$ to itself bijectively.
note: you don't have to use Cauchy Davenport to finish the problem in that situation, I think there exists a work around by simply using piegonhole principle
12.11.2022 18:14
Another solution: $Lemma$: $\sum_{x=0}^{p-1}{x^d}\equiv 0 \pmod p \Leftrightarrow p-1 \nmid d$. Just a property of primitive roots. Let $P(x)=ax^3+bx^2+cx$. Then $P$ is bijective over $\mathbb{F}_p$. If $a\equiv0 \pmod p$, easy to know $b\equiv 0 \pmod p,c \not \equiv 0 \pmod p$. If $a \not \equiv 0 \pmod p$: Step 1: $p \equiv 2 \pmod 3$.
Step 2: $b^2 \equiv 3ac \pmod p$
We are done.
10.12.2022 20:44
@starchan It probably is indeed well known (though very hard to find somewhere in an elementary language, I guess?) but anyway the point was it to be an exceptionally great NT exercise for the students.