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Given an integer $n>2$, prove that there are infinitely many primes $p$ such that $p\not\equiv1\text{ mod }n$.

I read about the question here when $n=5$, where they used that $n$ itself is a prime. However, how should I prove it for any natural $n$? I got the hint "consider the polynomial $nx-1$" and tried to prove by contradiction, buy I couldn't go any further.

Thank you!

Bill Dubuque
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Jeremy
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    It's the special case of the generalization of Euclid's proof in the linked dupe, where $,S = 1 + n\Bbb Z,\ c(k) = nk-1\ \ $ – Bill Dubuque May 20 '21 at 07:09

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Assume that there were only finitely many primes not congruent to 1 modulo $n.$ Consider $nx - 1$ for $x$ a positive integer. If $p \mid x$ is a prime factor of $x,$ then obviously it cannot be a prime factor of $nx-1,$ since $p \mid nx$ but $p \nmid -1.$ But now you're done, for $nx - 1$ has to have at least one factor not congruent to 1 modulo $n$ (if all of its factors were $1$ modulo $n,$ then $nx-1$ would be congruent to $+1$ modulo $n,$ which is impossible unless $n=2,$ and you can show there are infinitely many odd primes on your own).

Thus if $p_1, ..., p_k$ are the finitely many primes not congruent to 1 mod $n$, take $x = p_1\cdots p_k$ to get a contradiction.

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    This seems convoluted in phrasing. Is the last line supposed to be " are the finitely many not congruent to 1 mod n primes"? (If so, I believe you have a correct approach.) – Calvin Lin May 20 '21 at 06:38
  • @CalvinLin You're right, I changed it. –  May 20 '21 at 06:46