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Consider a general arithmetic sequence $x_j = y + jn, j \ge 1$. Prove that if $p$ is a prime number such that $p \nmid n$ then there is some element from the sequence $\{x_n\}_{n=1}^\infty$ that is divisible by $p$.

Your proof of this result should give you a method which for a given arithmetic sequence and prime actually allows you to calculate a point in the sequence at which the divisibility property holds. Illustrate your method by presenting an element from this sequence that is divisible by $p$, where $p = 150000001 = 1.5 × 10^8 + 1$, and where $n=14506349$ and $y=2021$.

Does anyone have a clue on how to tackle this?

Bill Dubuque
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math_guru
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    This problem sounds like it was lifted word-for-word off an exam or something. – Mike Jan 22 '20 at 21:42
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    What are your thoughts on this problem? Can you try a couple of examples by hand? – gt6989b Jan 22 '20 at 21:52
  • $!\bmod p!:\ nj+y\equiv 0\iff nj\equiv -y\iff j\equiv -y,n^{-1},,$ and the inverse exists by Bezout and $,p\nmid n., $ See the linked dupe for many ways to compute such inverses (and modular fractions). – Bill Dubuque Jan 22 '20 at 22:09

2 Answers2

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Here's an outline of how to start: There is an algorithm that produces integers x and z such that $$ xp + zn =gcd(p,n) $$ Since p is prime and p does not divide n, gcd(p,n)=1. $$ xp + zn =1 $$ $$ xyp + yzn=y $$ Let j=-yz. You can take it from there.

P. Lawrence
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Hint:

Use congruences: you have to show some $x_j\equiv 0\mod p$ – explicitly $$ y+jn\equiv 0 \mod p.$$ Remember that, if a prime $p$ does not divide $n$, $n$ is invertible modulo $p$ (by Bézout's identity).

Bernard
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