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I have been assigned this homework problem and I'm having trouble figuring out how to prove these statements.

If $n_1,\dotsc,n_k\in${$6z+1\mid z\in\mathbb{Z}$}, show that $n_1n_2\cdot\cdot\cdot n_k\in${$6z+1\mid z\in\mathbb{Z}$}.

Next, show that {$6z+5\mid z\in\mathbb{Z}$} contains infinitely many primes. (Without using Dirichlet's Theorem).

Thank you in advance!

Alex M.
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  • What have you tried? Have you noticed any prime >3 is either of the form $6z+1$ or $6z+5$?. – Jack Yoon Sep 25 '15 at 07:57
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    Have you checked some previous posts on the same problem? For example: http://math.stackexchange.com/questions/825917/proof-that-there-are-infinitely-many-primes-of-the-form-6k1-proof-verificati, http://math.stackexchange.com/questions/1318537/proving-infintely-many-primes-of-the-form-6k-1 or http://math.stackexchange.com/questions/507160/finding-infinetely-many-primes-in-three-different-forms-of-prime-numbers-below, http://math.stackexchange.com/questions/407259/how-do-you-prove-that-there-are-infinitely-many-primes-of-the-form-5-6n – Martin Sleziak Sep 25 '15 at 10:58

1 Answers1

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Just mimic Euclid's proof of the infinitude of primes.

Assume that the set of primes of the form $6k-1$ is finite, say given by $\{p_1,\ldots,p_n\}$, and consider:

$$ N = 6p_1 p_2\cdot\ldots\cdot p_n-1. \tag{1}$$ We obviously have $N\equiv -1\pmod{6}$, hence any prime number $p\mid N$ is $\equiv \pm 1\pmod{6}$ and there is at least one prime $p\mid N$ such that $p\equiv -1\pmod{6}$. However, $p\neq p_i$ for every $i\in [1,n]$, since $\gcd(N,p_1)=1$. That leads to a contradiction.

Jack D'Aurizio
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