infinite primes $p\equiv1\pmod n$ without cyclotomic polynomial

Question

Without cyclotomic polynomial, is there an elementary proof of the following: for each integer $n>1$, there are infinitely many primes $p$ such that $p\equiv1\pmod n$ ?

please don't refer to Dirichlet's theorem on arithmetic progressions, or analytic number theory. Thanks

For $n$ prime this can be done by taking a prime divisor $p$ of $2^n-1$. It follows that $n\mid p-1$. I'm not sure if this can be extended for composite $n$... I suppose you don't want to use Zsigmondy's theorem? — Bart Michels, Apr 13 '14 at 10:53
See Qiaochu's answer here.You can prove a special case of the theorem using his answer. — rah4927, Apr 19 '14 at 13:22

score 2 · Answer 1 · answered Apr 12 '14 at 04:22

2

An elementary proof of this result is outlined in problem 36 on p. 108 of An Introduction to the Theory of Numbers by Niven, Zuckerman and Montgomery. That problem references the article by I. Niven and B. Powell "Primes in certain arithmetic progressions," Amer. Math. Monthly, 83 (1976), 467-469, as simplified by R.W. Johnson.

answered Apr 12 '14 at 04:22

Pete Klimek

516

For the Niven book, I dont think you are referencing the right page. I am looking at the 4th edition right now and it is not there. However, I cannot find it. Maybe you mistyped? – Sandeep Silwal Apr 12 '14 at 05:26
@SandeepSilwal fifth edition – ziang chen Apr 12 '14 at 05:57
I was citing the fifth edition. The problem is from section 2.8 (Primitive Roots and Power Residues). – Pete Klimek Apr 12 '14 at 06:01

infinite primes $p\equiv1\pmod n$ without cyclotomic polynomial

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