1. Originally Posted by Bacterius
Ok, so the question is : for which $k$ is $10^k + 1$ prime.

Theorem : if $p | 10^k + 1$, then $p | 10^{mk} + 1$ if $m$ is odd.

The proof is trivial, if $10^k \equiv -1 \pmod{p} \ \implies \ 10^{mk} \equiv {(-1)}^m \pmod{p} \ \implies \ 10^{mk} + 1 \equiv 0 \pmod{p}$. This only works for odd $m$ due to the special status of square roots in a congruence ring.

Now, let $k = 1$ and $p = 10^1 + 1 = 11$ (since $11$ is prime). Then $11$ divides all numbers of the form $10^m + 1$ where $m$ is odd. That is, 50% of all concerned numbers.

Then, let $k = 2$ and $p = 10^2 + 1 = 101$ (since $101$ is prime). Then 101 divides all numbers of the form $10^{2m}$ where $m$ is odd (2, 6, 10, ...).

So we're still missing the numbers of the form $10^{2m} + 1$ with $m$ even. Let's look at $10^{2m - 1} + 1$. This one's got to be composite (divisible by $11$) since the exponent is odd.

This is as far as I can go yet. I'm missing numbers of the form $10^{2m} + 1$ for $m$ even
If you look at my post, I showed $10^a+1$ is composite for all $a\neq2^n$ (this is a stronger result). I suspect we can't go much further than this... These resemble Fermat numbers.

2. Originally Posted by chiph588@
If you look at my post, I showed $10^a+1$ is composite for all $a\neq2^n$ (this is a stronger result). I suspect we can't go much further than this... These resemble Fermat numbers.
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 $2m$ with $m$ odd : 2, 6, 10, 14, 18, 22, ...
All exponents of the form $2^n$ : 2, 4, 8, 16, 32, ...

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

Am I right ?

3. Originally Posted by Bacterius
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 $2m$ with $m$ odd : 2, 6, 10, 14, 18, 22, ...
All exponents of the form $2^n$ : 2, 4, 8, 16, 32, ...

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

Am I right ?
I thought that the only candidates were numbers of the form $10^{2^k}+1$. We did not eliminate them.

4. Originally Posted by roninpro
I thought that the only candidates were numbers of the form $10^{2^k}+1$. We did not eliminate them.
Woops, I actually misread that. Sorry. I'll have a think.

5. WOW! I thought that this question would be considerably less difficult to solve...no wonder I was having problems!

6. Originally Posted by yeah:)
WOW! I thought that this question would be considerably less difficult to solve...no wonder I was having problems!
IHMO, if you see "Prime number" in a question, either the question is trivial, either it is excruciatingly difficult to prove.

7. Originally Posted by Bacterius
IHMO, if you see "Prime number" in a question, either the question is trivial, either it is excruciatingly difficult to prove.
Yes, that is true! Any 'brainy' ideas yet, Bacterius (or anyone, for that matter!)?

8. Originally Posted by yeah:)
Yes, that is true! Any 'brainy' ideas yet, Bacterius (or anyone, for that matter!)?
Nah, I'm off to sleep right now. Perhaps I'll get an idea while dreaming

9. Here's a start :

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

So :

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

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

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

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

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

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

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

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

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

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

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

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

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

Multiply by $10$ :

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

$101 \equiv 0 \pmod{p}$

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

Anyone can prove that for any $k > 1$ exists at least one $p$ that satisfies the statement ?

10. 100501 IS a prime number!

11. Originally Posted by Bacterius
Here's a start :

Let $10^{2^k} + 1 \equiv 0 \pmod{n}$ for some prime number $p$, with $p \neq 10^{2^k} + 1$.
I'm having some trouble following you in the first line. Why would you want to assume that $p \neq 10^{2^k} + 1$? What if $10^{2^k}+1$ is prime itself?

Originally Posted by yeah:)
100501 IS a prime number!
I notice that you mentioned 100501 twice. What does this number have to do with the sequence 11, 101, 1001, 10001, ...?

12. You are right - it is completely irrelevant. I apologise about that.

13. 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??

14. So do we have a definitive answer yet?

15. 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

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