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Dirichlet's Theorem on arithmetic progressions states that there are infinitely many primes of the form

$$an+b$$

Being $a$ and $b$ coprimes.

  1. Very elemental question: Can we say that this Theorem:

$$an-b$$

is equivalent to Dirichlet's one? I supose it is, as it would be the same as $a(n-1)+(n-b)$, but I would like to get sure, as I am not used to work with this kind of formulae

  1. For which functions $f(n)$ can we say that there are infinitely many primes of the form

$$a[f(n)]+b$$ ? For example, $a2^n+b$ or $a(5n)+b$

Thank you!

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  • $\begingroup$ Just a comment but your "proof" of writing $an-b$ as $a(n-1)+(n-b)$ does not work since the constant should be independent of $n$. $\endgroup$ – Asvin Nov 19 '16 at 0:27
  • $\begingroup$ Well, $a(5n)+b$ where $a,b$ are merely required to be coprime won't work ... we could have $b$ divisible by $5$. $\endgroup$ – GEdgar Nov 19 '16 at 0:46
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  1. Yes, Dirichlet's theorem is equivalent to the statement that there are infinitely primes of the form $an - b$ for any relatively prime $a$ and $b$.

  2. The short answer is that we don't know. Other than Dirichlet's theorem, there isn't another known family of functions with similar properties. There are many conjectures. For instance, it is thought that $x^2 + 1$ should be prime infinitely often. More generally, it is thought that any "reasonable" quadratic polynomial should be prime infinitely often (where "reasonable" is analogous to $a,b$ being relatively prime in the linear case). One such grand conjecture is Schinzel's Hypothesis.

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Question 1 answer

Primes of the form an+b are primes that are $\;\equiv \;$b (mod a)

Therefore there are infinitely many primes $\;\equiv \;$b (mod a)

If b is negative, find positive $\; b_1 \;\equiv \;$-b (mod a)

Then$\;\;$ an-b $\;$ = $\;$ a$n_1+b_1\;\;$and will yield infinitely many primes.

Question 2 may not have a finite answer.

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