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dirichlet-thm.6neg-case - A special case of Dirichlets...

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A special case of Dirichlet’s theorem Jonathan L.F. King [email protected] 28 October, 2009 (at 15:16 ) Entrance. An arithmetic progression ( A.P. ) means a set T + M Z of integers, 1 where the gap ( or modulus ) M is a posint and translation ( or target ) T is an integer. I’ll also use comb for “arithmetic progression”. An A.P. C := T + M Z is coprime if T M . 1 : Dirichlet’s Theorem. Each coprime arithmetic pro- gression contains infinitely many prime numbers. While this is difficult to prove in general, there are three easy special 2 cases, the combs 1 + 3 Z 1 + 4 Z and 1 + 6 Z . We will establish this last case. Henceforth, let mean 6 and let congruent mean “mod-6 congruent”. 2a : Lemma. Suppose a product q 1 · q 2 · . . . · q · . . . · q L of integers 3 is coprime to 6 . Then each multiplicand q is coprime to 6 . Proof. Exercise ; prove the contrapositive. Where does your argument use that each q is is an integer ? Does the lemma generalize to “6” being replaced by N , an arbitrary posint? Henceforth, symbols r and q , with or without ap- pendages, range over the integers. 2b : Corollary. Suppose product r := q 1 · q 2 · . . . · q L is congruent to 1 . For oddly-many indices in [ 1 .. L ] , then, q 1 .
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