AoPS - Problem

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Problem

Source: Mathematics Regional Olympiad of Mexico Southeast 2021 P2

Tags: number theory, prime numbers, arithmetic sequence

AlanLG

22.10.2021 21:47

Let $n\geq 2021$ . Let $a_1<a_2<\cdots<a_n$ an arithmetic sequence such that $a_1>2021$ and $a_i$ is a prime number for all $1\leq i\leq n$ . Prove that for all $p$ prime with $p<2021, p$ divides the diference of the arithmetic sequence.

DottedCaculator

22.10.2021 22:03

Suppose that the difference is

$d$ , and

$p\nmid d$ for some

$p$ . Then,

$a_1$ ,

$a_2$ ,

$\ldots$ ,

$a_p$ are distinct mod

$p$ , so at least one of them is a multiple of

$p$ , so it is not prime since

$a_i>2021>p$ . Therefore,

$p\mid d$ for all

$p<2021$ .

sanyalarnab

21.03.2022 19:00

Let there exist a prime $p< M$ such that $p$ does not divide $d=a_2-a_1$ . $a_i \equiv a_1+(i-1)d \mod p$ Also as $n \ge M > p$ , by PHP we get a complete residue system of $i$ in modulo $p$ . So there exists $N \le n$ such that $N \equiv (1-a_1x) \mod p$ , where $x$ is inverse of $d$ in modulo $p$ . Simply it follows that $a_N \equiv 0 \mod p$ $\implies a_N = p$ But $a_N > a_1 > M > p$ which is a contradiction. So the desired statement follows.