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I'm wondering if I am correctly interpreting Dirchlet's theorem or not. (that is I can't tell if my statement is trivial or not)

Statement. There are infinitely many primes $x$ such that there are infinitely many primes $6x + 1$.

Proof. Let $x$ be prime. Assemble a sequence $S = 6x+1$, $6x+7$, $6x+13$, $6x+19\ldots$.
Now by Dirichlet Theorem, if $\gcd(a,d)=1$ then there are infinitely many primes within the progression $a$, $a+d$, $a+2d$,$a+3d$, $\ldots a+nd$. But $\gcd(6x+1,6)=1$ for all $x$, therefore there are infinitely many primes in sequence $S$, since assuming $a = 6x+1$ and $d=6$, $S$ is immediately seen in the form $S = a, a+d, a+2d, a+3d\ldots$ QED

Daniele Tampieri
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Jacob Nathaniel
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  • See A. Granville in https://doi.org/10.1016/0022-314X(87)90052-7 – R. J. Mathar Mar 27 '25 at 09:09
  • Yes, that's correct. – lulu Mar 27 '25 at 09:37
  • Worth remarking that there are more elementary proofs of that result, as here – lulu Mar 27 '25 at 09:57
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    To be pedantic, no: "there are infinitely many primes $x$ such that there are infinitely many primes of the form $6x + 1$" is trivially false. Dirichlet's theorem tells you "when $a$ is coprime to $6$, there are infinitely many primes of the form $6n + a$". It follows easily from this that "for each prime $x$, there are infinitely many primes of the form $6x + 6n + 1$", which is what you seem to be talking about in your proof (this remains true if $x$ is not prime, though - no need to assume that!) – Izaak van Dongen Mar 27 '25 at 10:50

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