If you look at my post, I showed $\displaystyle 10^a+1 $ is composite for all $\displaystyle a\neq2^n $ (this is a stronger result). I suspect we can't go much further than this... These resemble Fermat numbers.

- Jun 5th 2010, 09:03 PMchiph588@
- Jun 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 $\displaystyle 2m$ with $\displaystyle m$ odd : 2, 6, 10, 14, 18, 22, ...

All exponents of the form $\displaystyle 2^n$ : 2, 4, 8, 16, 32, ...

We're still missing a small - yet infinite - subset of numbers, which can be expressed as $\displaystyle m = 12k$, $\displaystyle k \in \mathbb{N}$, that is, 12, 24, 36, ...

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

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

Let $\displaystyle 10^{2^k} + 1 \equiv 0 \pmod{n}$ for some prime number $\displaystyle p$, with $\displaystyle p \neq 10^{2^k} + 1$.

So :

$\displaystyle 10^{2^k} + 1 \equiv 0 \pmod{p}$

$\displaystyle 10^{2^k} + 10 - 10 + 1 \equiv 0 \pmod{p}$

$\displaystyle 10^{2^k} + 10 - 9 \equiv 0 \pmod{p}$

$\displaystyle 10^{2^k} + 10 \equiv 9 \pmod{p}$

$\displaystyle 10(10^{2^k - 1} + 1) \equiv 9 \pmod{p}$

$\displaystyle 10^{2^k - 1} + 1 \equiv 9 \times 10^{-1} \pmod{p}$

$\displaystyle 10^{2^k - 1} \equiv 9 \times 10^{-1} - 1 \pmod{p}$

Therefore, if $\displaystyle 10^{2^k - 1} - 9 \times 10^{-1} + 1$ divides some $\displaystyle p$ in $\displaystyle \mathbb{Z}/p\mathbb{Z}$, then $\displaystyle 10^{2^k}$ is divisible by $\displaystyle p$, thus is composite.

So we basically have to show that such a $\displaystyle p$ exists for all $\displaystyle k > 1$. I'll give the proof that it doesn't exist for $\displaystyle k = 1$.

Let $\displaystyle k = 1$. Then, hypothetically, for some prime $\displaystyle p$ :

$\displaystyle 10^{2^1 - 1} - 9 \times 10^{-1} + 1 \equiv 0 \pmod{p}$

$\displaystyle 10 - 9 \times 10^{-1} + 1 \equiv 0 \pmod{p}$

$\displaystyle 11 - 9 \times 10^{-1} \equiv 0 \pmod{p}$

Multiply by $\displaystyle 10$ :

$\displaystyle 110 - 9 \equiv 0 \pmod{p}$

$\displaystyle 101 \equiv 0 \pmod{p}$

Which is only true for the prime number $\displaystyle 101$. However, $\displaystyle 101 = 10^{2^1} + 1$ and we assumed that $\displaystyle p \neq 10^{2^k} + 1$ at the beginning, so we have a contradiction that leads to the conclusion that such a $\displaystyle p$ does not exist for $\displaystyle k = 1$.

Anyone can prove that for any $\displaystyle k > 1$ exists at least one $\displaystyle p$ that satisfies the statement ? - Jun 6th 2010, 07:06 AMyeah:)
100501 IS a prime number!

- Jun 6th 2010, 07:31 AMroninpro
I'm having some trouble following you in the first line. Why would you want to assume that $\displaystyle p \neq 10^{2^k} + 1$? What if $\displaystyle 10^{2^k}+1$ is prime itself?

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

- Jun 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?? - Jun 6th 2010, 10:37 AMyeah:)
So do we have a definitive answer yet?

- Jun 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