Maths Olympiad Prep

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, 2021

Number theory Difficulty 5.7 AIME, harder Prove it Taiwan

Given a positive integer kk, show that there exists a prime pp and such that one can choose distinct integers a1,a2,,ak+3{1,2,,p1}a_1, a_2, \dots, a_{k+3} \in \{1, 2, \dots, p-1\} such that pp divides aiai+1ai+2ai+3ia_i a_{i+1} a_{i+2} a_{i+3} - i for all i=1,2,,ki = 1, 2, \dots, k.

Solution

First, construct distinct positive rational numbers r1,r2,,rk+3r_1, r_2, \dots, r_{k+3} such that riri+1ri+2ri+3=ir_i r_{i+1} r_{i+2} r_{i+3} = i for all i=1,2,,ki = 1, 2, \dots, k. Since ri+4=i+1irir_{i+4} = \frac{i+1}{i} r_i, {r4n+r}n\{r_{4n+r}\}_n are strictly increasing sequences for r=1,2,3,4r = 1, 2, 3, 4. Choose r1,r2,r3r_1, r_2, r_3 to be distinct primes larger than kk and r4=1r1r2r3r_4 = \frac{1}{r_1 r_2 r_3}, it is easy to see that {ri}\{r_i\} are distinct rational numbers. Write ri=uivir_i = \frac{u_i}{v_i} as irreducible fractions. It is easy to see that ui(vi)1=uj(vj)1u_i (v_i)^{-1} = u_j (v_j)^{-1} in Zp\mathbb{Z}_p only if uivjujvi(modp)u_i v_j \equiv u_j v_i \pmod{p}. Choose prime p>(maxi{ui,vi})2p > (\max_i\{u_i, v_i\})^2, then ai=ui(vi)1a_i = u_i (v_i)^{-1} in Zp\mathbb{Z}_p satisfies the desired properties.

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from zh; metadata (topic, difficulty) added by this project.