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A famous theorem of Dirichlet says that infinitely many primes are of the form:$\alpha n+\beta$, but are there infinitely many of the form: $\alpha ^n+\beta$, where $\beta$ is even and $\alpha$ is prime to $\beta$? or of the form $\alpha!+\gamma$, where $\gamma$ is odd?
Out of mere curiosity has this question come, thus any help is greatly appreciated.

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i think that you can use this method.because b is prime and a is prime to b,then it means that a is odd number,so represent b=2*k and a=2*s+1,if use power notation ,you get +1 at the end of polynomial,so even number plus 1 is sometimes only odd number,sometimes odd and prime together – dato datuashvili Jun 16 '12 at 7:16
and about second form does ! means factorial?then sure no,because a! for any a>1 is even,so you get even+even=even – dato datuashvili Jun 16 '12 at 7:18
@dato: You mean to reduce the problem to polynomials having primes as values? Well, since little do I know results in this direction, I can only conjecture so. Notice however that for $n!+1$, we found that, if n=4,5,6, $n!+1$ is not a prime. This invokes some critical doubts about the validity of the conjecture... – awllower Jun 16 '12 at 7:21
If $\gamma\gt1$, then $\gamma$ divides $\alpha!+\gamma$ for all $\alpha\ge\gamma$, so only finitely many primes there. – Gerry Myerson Jun 16 '12 at 7:43
The cases $\alpha=2$, $\beta=\pm 1$ are rather famous open problems. (That is, whether there are infinitely many Mersenne or Fermat primes). – Henning Makholm Jun 16 '12 at 16:17
up vote 10 down vote accepted

There are relatively prime non-trivial $\alpha$ and $\beta$, with $\beta$ even, such that $\alpha^n +\beta$ is not prime if $n \ge 1$. Easy, let $\beta$ have decimal expansion that ends in $4$, and let $\alpha>1$ have decimal expansion that ends in $1$.

A more subtle class of examples is illustrated by $625^n+4$. For this one we use the algebraic identity $$x^4+4=(x^2-2x+2)(x^2+2x+2)$$ to prove compositeness.

For the factorial question, a necessary condition for primality if $\alpha \gt 1$ is $\gamma=\pm 1$. Unfortunately it is not known whether there are infinitely many primes of the form $n!\pm 1$.

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Then in general are there $\alpha$ and $\beta$ such that the first form is always prime? Also thanks for the attention. – awllower Jun 16 '12 at 7:25
yes sure a=1,b=2 – dato datuashvili Jun 16 '12 at 7:30
I am pretty sure that there are no known $\alpha\gt 1$, $\beta$ such that $\alpha^n+\beta$ is prime for infinitely many $n$. Doesn't mean there aren't any, it is just that such questions are very difficult. the question in your comment may be easier to settle (in the negative). – André Nicolas Jun 16 '12 at 7:30
Ah! So one could show the negative answer to the statement concerning $n!+1$? Might I ask for a description or a reference? Thanks. Also the comment of Dato is really, well..., funny!! – awllower Jun 16 '12 at 7:32
I did not make any comment about the factorial. It is not known whether there are infinitely many primes of the form $n!+1$, and it is not for lack of trying. – André Nicolas Jun 16 '12 at 7:37

Numbers $n$ such that $n! - 1$ is prime is The list begins, 3, 4, 6, 7, 12, 14, 30, 32, 33, 38, 94, 166, 324, 379, 469, 546, 974, 1963, 3507, 3610, 6917, 21480, 34790, 94550, 103040. Presumably the list is infinite, but it appears that no one has proved it.

Numbers $n$ such that $n! + 1$ is prime is The list begins, 0, 1, 2, 3, 11, 27, 37, 41, 73, 77, 116, 154, 320, 340, 399, 427, 872, 1477, 6380, 26951, 110059, 150209. As before, presumably the list is infinite, but it appears that no one has proved it.

Many references are given at those two webpages.

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Thanks for the reference! – awllower Jun 18 '12 at 17:56

if you are interested if there is infinite prime number of the form $n!+1$ for infinite many n,then first use some example $n=2$ then $n!+1=3$ is prime,for n=3,$n!+1=7$ but comes question who can calculate $n!$ for n=50 for example,so it is difficult to say if there is infinity number of prime of this form

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Thanks. Maybe this is a misunderstanding for @André Nicolas expressed in his comment that it is easy to answer the question in my comment, I thought he meant that about $n!+1$. – awllower Jun 16 '12 at 7:43
may be,he has added last comment that it is difficult to say something about n!+1. you are welcome – dato datuashvili Jun 16 '12 at 7:47
I don't know who can calculate $n!$ for $n=50$, but I do know that someone managed to prove that $150209!+1$ is prime. There are some very clever people out there, and there is more than one way to prove that a number is prime. See – Gerry Myerson Jun 18 '12 at 7:17

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