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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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    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.2011-11-01

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