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I do not know if this what I am going to ask is immediate consequence of something known but if not it may have an easy answer which I do not see, so any help would be great. Let us define sequence that depends on three numbers: $n$, $k$ and $l$, as $a(n,k,l)=n^k+l$, let us now choose some $l=l_0\in\mathbb N$ and keep it fixed from now on. So we have sequence $a(n,k)=n^k+l_0$. And my question is: If we denote by $S$ the image set of $a(n,k)$, $k>1$ (the set of values of sequence $a$ when $(n,k)$ runs through the $\mathbb N\times ({{\mathbb N}\setminus\{1\}}$) does $S$ have an infinity of primes?

I do know that it is not known whether polynomials of degree greater than one have in their image set an infinity of primes but this problem is much weaker, it asks to show that for fixed $l=l_0$ the image set of family of polynomials contains infinity of primes, can someone prove this?

EDIT: Because I can see that there are some misunderstandings with the formulation of the problem I will try here to clarify it. Suppose we choose $l=l_0=1$. We now have family of sequences:

$2\to a_2(n)=n^2+1$

$3\to a_3(n)=n^3+1$

....

$k\to a_k(n)=n^k+1$

...

The question is not is there any sequence of those above listed sequences that have an infinity of primes but the question is do the sequence of sequences ($k\to a_k(n)$, $k>1$, $l=l_0$) contains infinity of primes or, stated differently, do all above listed sequences taken together have an infinity of primes (note that it is enough that for every $k$ there exists ONE prime in $a_k(n)$ because then there will be infinity of them in the sequence $k\to a_k(n)$)?

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I fixed the title to make it clear what's being asked, but I'd encourage doing the same with the body of the question itself - the three-parameter sequence is just a diversion, and so is the notion of 'image sets'; you could quite simply ask 'is it known if there are infinitely many primes of the specified form?'. –  Steven Stadnicki Feb 13 '13 at 18:50
    
No, you didn`t do a good thing, do you see what I ask? –  A.P. Feb 13 '13 at 18:51
    
I have added something to make it more clear. Now it looks good. –  A.P. Feb 13 '13 at 18:57

1 Answer 1

Let $l_0=-1$. Then the problem becomes whether there are infinitely many Mersenne primes. So the conjecture is not all that weak.

Edit: The post has changed, $l_0$ now must be positive.

For $l_0=1$, we end up with the question of whether there are infinitely many primes of the Fermat family, that is, of the shape $a^{2^k}+1$, where $a$ is even. I believe the answer to that is not known.

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(To clarify: this is because no number of the form $n^k-1$ can be prime if $n\gt 2$, since it's divisible by $n-1$.) –  Steven Stadnicki Feb 13 '13 at 18:51
    
I know that, thank you. –  A.P. Feb 13 '13 at 18:53
    
Yes Andre, i wasn`t aware of that when posting, I know that but I didn`t think about that at the moment, now the question is different because I restricted that $l=l_0\in\mathbb N$, can you say something now that will clarify the things? –  A.P. Feb 13 '13 at 18:55
    
And I do not ask whether one particular type of sequence have an infinity of primes, my question is for the whole family of polynomials, maybe you didn`t exactly get what I am asking? –  A.P. Feb 13 '13 at 18:59
1  
From your post, it seems that $l_0$ is to be considered fixed. If we choose $l_0=1$, and $a^q+1$ is prime, then except in trivial cases we must have that $q$ has no odd prime factors. So $q$ must be a power of $2$, which leads to the Fermat-like family. –  André Nicolas Feb 13 '13 at 19:03

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