# Infinitely many primes of the $\sum_{i=0}^{a} n^{i}+m\cdot\sum_{i=0}^{b} n^{i}$ form?

How to show that there is infinitely many prime numbers of the form:

$p=\sum_{i=0}^{a} n^{i}+m\cdot\sum_{i=0}^{b} n^{i}$

where: $m\in \mathbb{Z}^{*}$ , $a,b,n\in \mathbb{N}$ , $\gcd(a+1,b+1)=1$

For example:

$\sum_{i=0}^{6} 10^{335\cdot i}+8648\cdot \sum_{i=0}^{4} 10^{335\cdot i}$ is prime number.

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Can you tell us the source of this problem? – Srivatsan Nov 1 '11 at 8:14
@Srivatsan,there is no source. this form is result of my research – pedja Nov 1 '11 at 8:18
Do you know this to be true or is it a conjecture? Also do you have any heuristic evidence? – Srivatsan Nov 1 '11 at 8:19
While it is ok to post a conjecture and ask for approaches, I feel you should clearly state that it is your conjecture in the question. Your phrasing is misleading in that it seems to suggest that you already know it to be true (that would be the case with, for instance, a textbook exercise). People will be interested to know the source of the question. – Srivatsan Nov 1 '11 at 8:29
As I wrote, I feel it is unlikely that the answer is "no". Of course, the prime numbers have never cared much about what I feel. – Gerry Myerson Nov 1 '11 at 21:45

I think that I have figured out answer on this question so here it is :

If we apply formula for the sum of the numbers in a geometric progression we may write :

$\sum_{i=0}^{a} n^{i}=\frac{n^{a+1}-1}{n-1}$ and $\sum_{i=0}^{b} n^{i}=\frac{n^{b+1}-1}{n-1}$ . Therefore we have to show that sequence :

$p_m=\frac{n^{a+1}-1}{n-1}+m\cdot \frac{n^{b+1}-1}{n-1}$ , contains infinitely many primes for any $a$ and $b$ .

According to Dirichlet's theorem there are infinitely many prime numbers in sequence : $a+n\cdot d$ , for any $(a,d)$ pair if $\gcd(a,d)=1$ , so in order to prove that sequence $p_m$ contains infinitely many primes we have to show that :

$$\gcd(\frac{n^{a+1}-1}{n-1},\frac{n^{b+1}-1}{n-1})=1$$

There is a known theorem that states :

$$\gcd(a^n-1,a^m-1)=a^{\gcd(n,m)}-1$$

so we may write following :

$$\gcd(n^{a+1}-1,n^{b+1}-1)=n^{\gcd(a+1,b+1)}-1=n-1 \Rightarrow$$

$$\Rightarrow \gcd(\frac{n^{a+1}-1}{n-1},\frac{n^{b+1}-1}{n-1})=1$$

The last equality implies that sequence :

$p_m=\sum_{i=0}^{a} n^{i}+m\cdot\sum_{i=0}^{b} n^{i}$ ,contains infinitely many prime numbers for any fixed $(a,b,n)$ triple,so there are infinitely many primes of the form : $\sum_{i=0}^{a} n^{i}+m\cdot\sum_{i=0}^{b} n^{i}$

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