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It is clear to see that 11 and 101 are primes which sum of digit is 2. I wonder are there more or infinte many of such prime.

At first, I was think of the number $10^n+1$. Soon, I knew that $n\neq km$ for odd $k>1$, otherwise $10^m+1$ is a factor.

So, here is my question:

Are there infinite many integer $n\ge 0$ such that $10^{2^n}+1$ prime numbers?

After a few minutes: I found that if $n=2$, $10^{2^n}+1=10001=73\times137$, not a prime; if $n=3$, $10^{2^n}+1=17\times5882353$, not a prime; $n=4$, $10^{2^n}+1=353\times449\times641\times1409\times69857$, not a prime.

Now I wonder if 11 and 101 are the only two primes with this property.

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    Actually, if $k$ is odd then $10^m+1$ is a factor, not $10^k+1$2012-11-23
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    Nobody knows. An affirmative answer would confirm the conjecture that there are infinitely many primes of the form $w^2 + 1.$ A negative answer would not settle things. Note that nobody knows whether there are infinitely many Fermat primes either. http://en.wikipedia.org/wiki/Fermat_number2012-11-23
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    @ThomasAndrews Thanks for pointing out this. I have edited it.2012-11-23

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