Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
Actually, if $k$ is odd then $10^m+1$ is a factor, not $10^k+1$ – Thomas Andrews Nov 23 '12 at 3:08
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. – Will Jagy Nov 23 '12 at 3:09
@ThomasAndrews Thanks for pointing out this. I have edited it. – pipi Nov 23 '12 at 4:35

Many people wonder the same thing you do. Wilfrid Keller keeps track of what they find out. So far: prime for $n=0$ and $n=1$ only; known to be composite for all other $n$, $2\le n\le23$, and many other values of $n$. The first value for which primality status is unknown is $n=24$.

share|cite|improve this answer
@WilfridKeller Thanks for the reference. – pipi Nov 23 '12 at 4:38
pipi, I don't think Wilfrid comes here. – Gerry Myerson Nov 23 '12 at 5:55
Opps... I am so sorry. – pipi Nov 23 '12 at 5:56

If you're interested in quickly determining whether or not $10^{2^n}+1$ is prime (or positive integers in general), I suggest using OpenPFGW. It has (among other things) an efficient implementation of a PRP test.

Using the ABC2 input format, we input this:

    ABC2 10^(2^$a)+1
    a: from 1 to 1000

and it outputs:

PFGW Version [GWNUM 27.8]

CPU Information (From Woltman v26 library code)
Intel(R) Core(TM) i7-2670QM CPU @ 2.20GHz
CPU speed: 2195.32 MHz, 4 hyperthreaded cores
CPU features: RDTSC, CMOV, Prefetch, MMX, SSE, SSE2, SSE4.1, SSE4.2
L1 cache size: 32 KB
L2 cache size: 256 KB, L3 cache size: 6 MB
L1 cache line size: 64 bytes
L2 cache line size: 64 bytes
TLBS: 64

Recognized ABC Sieve file:                                     
ABC2 File
10^(2^0)+1 is trivially prime!: 11                                    
10^(2^1)+1 is trivially prime!: 101                                    
Switching to Exponentiating using GMP                                    
Switching to Exponentiating using Woltman FFT's                                    
10^(2^13)+1 is composite: RES64: [64182BF8406B65C3] (2.4100s+0.0002s)
10^(2^14)+1 is composite: RES64: [C5FF6A4A68324D5A] (12.6942s+0.0003s)
10^(2^15)+1 is composite: RES64: [A874DC2BD3F1B9C8] (58.8378s+0.0003s)
share|cite|improve this answer

Since no one else has mentioned it:

Standard heuristics in number theory suggest that there are only finitely many primes of the form $(2k)^{2^n}+1$ for any integer $k>0.$ The probability that a random number around $(2k)^{2^n}+1$ is prime is roughly $1/(2^n\log(2k))$; if you take into account the congruence conditions for such numbers and treat the chance that such a number is prime as a random variable, then the expectation is $C_k/2^n$ and the sum over these values converges.

If you sum this 'probability' over $n\ge24$ the expected number of primes of this form is less than 0.000001.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.