Let $k \in \mathbb{N}$. Prove that there are infinitely many prime numbers ending in $k$ 1's.

I have a couple basic ideas about how to construct a proof of this but I really can't follow any to completion. I thought about trying induction but I can't find a way prove the base case let alone to construct a prime with n + 1 1's. This proof would seem to indicate a way construct arbitrarily large primes so this suggests that induction is out of the question. I'm currently leaning towards a proof by contradiction.

Suppose for some $k \in \mathbb{N}$ we have that there are only finitely many primes ending in $k$ 1's. I want to somehow construct another prime with $k$ 1's that's not on the list. I really don't know how to do this either.

A couple small things that might help:

  • Any number consisting only of a composite number of repeated 1's is composite (the converse is not true)
  • a number ending in $k$ 1's is of the form $\frac{10^k(90a + 1) - 1}{9}$ for some natural number $a$. So equivalently we can try to prove there are infinitly many primes of this form.
  • It is equivalent show that for each $k \in \mathbb{N}$ there is at least 1 prime ending in $k$ 1's, since any prime ending in more than $k$ 1's will also end in $k$ 1's.

What would be best if I could get a hint or some suggestion as to how to attack this problem. A full solution is ok, but a hint is much more appreciated thank you!

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    $\begingroup$ Dirichlet's theorem states that there are infinitely many primes congruent to $a$ mod $m$ whenever $\gcd(a,m) = 1$. Your question is a special case of this where $m=10^k$ and a = $\sum_{i=0}^{k-1}{10^i}$ $\endgroup$ – rikhavshah Aug 14 '16 at 22:24
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    $\begingroup$ Rule of thumb. Don't ever use induction on theorems about prime numbers. There are exceptions but as one can't predict $p_{n+1} $ from $p_n $ they almost never work. $\endgroup$ – fleablood Aug 15 '16 at 0:17

We can just use Dirichlet's theorem, we need to show there's an infinity of primes in the progression $\frac{10^k-1}{9}+d10^k$.

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    $\begingroup$ Genius, thank you so much. We had learned Dirichlet's Theorem earlier but I've never needed it until now. $\endgroup$ – Sean Haight Aug 14 '16 at 22:23
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    $\begingroup$ @SeanHaight do not forget to check you actually can apply D.T. here. It is not hard, but it needs to be done. $\endgroup$ – quid Aug 14 '16 at 22:24
  • $\begingroup$ you are 100% completely welcome $\endgroup$ – Jorge Fernández Hidalgo Aug 14 '16 at 22:24
  • $\begingroup$ I just need to show $10^k$ and $10^k - 1$ are coprime right. $\endgroup$ – Sean Haight Aug 14 '16 at 22:26
  • $\begingroup$ yup ${}{}{}{}{}{}{}{}{}$ $\endgroup$ – Jorge Fernández Hidalgo Aug 14 '16 at 22:27

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