MathLabs

Problem 2

Let a1,a2,a3,…a_1, a_2, a_3, \ldots be an infinite increasing sequence of positive integers. Prove that for every p≥1p \ge 1 there are infinitely many ama_m which can be written in the form am=xap+yaqa_m = xa_p + ya_q with x,yx, y positive integers and q>pq > p.
Step 2 of 4: Pick the infinite subsequence sharing that remainder
ai1<ai2<ai3<⋯ ,aik≡r(modap)a_{i_1}<a_{i_2}<a_{i_3}<\cdots,\quad a_{i_k}\equiv r \pmod{a_p}
Detailed analysis

Let ai1<ai2<ai3<⋯a_{i_1} < a_{i_2} < a_{i_3} < \cdots be the (infinite) subsequence of all terms with aik≡r(modap)a_{i_k} \equiv r \pmod{a_p}. Because the original sequence is infinite, we may discard finitely many terms and assume every index iki_k satisfies ik>pi_k > p; set q=i1q = i_1, so q>pq > p. Every term aika_{i_k} with k≥2k \ge 2 now satisfies aik≡aq≡r(modap)a_{i_k} \equiv a_q \equiv r \pmod{a_p}, so ap∣(aik−aq)a_p \mid (a_{i_k} - a_q).