If you look at my post, I showed is composite for all (this is a stronger result). I suspect we can't go much further than this... These resemble Fermat numbers.

- June 5th 2010, 09:03 PMchiph588@
- June 5th 2010, 09:09 PMBacterius
Nice one, you just wiped some more numbers.

So let's resume. We eliminated :

All odd exponents : 1, 3, 5, 7, 9, ...

All exponents of the form with odd : 2, 6, 10, 14, 18, 22, ...

All exponents of the form : 2, 4, 8, 16, 32, ...

We're still missing a small - yet infinite - subset of numbers, which can be expressed as , , that is, 12, 24, 36, ...

Am I right ? - June 5th 2010, 09:33 PMroninpro
- June 5th 2010, 09:34 PMBacterius
- June 6th 2010, 01:51 AMyeah:)
WOW! I thought that this question would be considerably less difficult to solve...no wonder I was having problems!

- June 6th 2010, 02:40 AMBacterius
- June 6th 2010, 04:44 AMyeah:)
- June 6th 2010, 05:20 AMBacterius
- June 6th 2010, 06:12 AMBacterius
**Here's a start :**

Let for some prime number , with .

So :

Therefore, if divides some in , then is divisible by , thus is composite.

So we basically have to show that such a exists for all . I'll give the proof that it doesn't exist for .

Let . Then, hypothetically, for some prime :

Multiply by :

Which is only true for the prime number . However, and we assumed that at the beginning, so we have a contradiction that leads to the conclusion that such a does not exist for .

Anyone can prove that for any exists at least one that satisfies the statement ? - June 6th 2010, 07:06 AMyeah:)
100501 IS a prime number!

- June 6th 2010, 07:31 AMroninpro
I'm having some trouble following you in the first line. Why would you want to assume that ? What if is prime itself?

I notice that you mentioned 100501 twice. What does this number have to do with the sequence 11, 101, 1001, 10001, ...? - June 6th 2010, 07:52 AMyeah:)
You are right - it is completely irrelevant. I apologise about that.

- June 6th 2010, 08:57 AMhmmmm
ok so if we have that 10^m+1 is coposite for all odd m, then we assume m is even and so

10^m + 1 = 10^2k +1 = 100^k + 1 and so if k is odd then we have the number is composite, if it is not then we just repeat the process until we have the exponent as an odd number which we will get as m must be a finite, and so there are no more primes of this form, is this helpful?? - June 6th 2010, 10:37 AMyeah:)
So do we have a definitive answer yet?

- June 6th 2010, 11:07 AMgmatt
Best bet at figuring this out is probably modifying Pepin's test and running an optimized computer program:

Pépin's test - Wikipedia, the free encyclopedia