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First thing that to my mind is 101. I thought to prove that there are infinite primes of form

$1000.......00001$

It is of the form $10^n+1$ .So it is suffice to prove that there are infinite primes of form $10^n+1$ .Any leads??

J. W. Tanner
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Mathematical Curiosity
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  • It is not known whether or not there are infinitely many primes of the form $10^n+1$. – lulu Dec 25 '19 at 13:09
  • Do you care to share your motivation? – Praphulla Koushik Dec 25 '19 at 13:09
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    Dirichlet's theorem on Arithmetic progressions should suffice, though. For instance, show there are infinitely many primes which end in $1001$ and so on. – lulu Dec 25 '19 at 13:11
  • @lulu How can we say "It is not known whether or not there are infinitely many primes of the form $10^n+1$"? – Praphulla Koushik Dec 25 '19 at 13:11
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    @PraphullaKoushik I don't understand. We don't currently have the techniques needed to solve questions of that form. More broadly, we don't know if there are infinitely many primes of the form $2^n+1$ nor even something seemingly simpler like $n^2+1$. What are you asking? – lulu Dec 25 '19 at 13:12
  • @lulu May be I did not ask clearly. How do we know there are no techniques available? Is it mentioned in some referred publication that as of today or any other day, there is no technique available for this kind of problem? Is it ok now? – Praphulla Koushik Dec 25 '19 at 13:16
  • Still not following. To the best of my knowledge, there are no proofs known regarding such problems (in either direction). Is it possible that current methods suffice to prove it but nobody has yet sorted out the details? Well, sure. But my guess is that we'll need a deep new idea or two. – lulu Dec 25 '19 at 13:20
  • Remarkably the answer to the $10^n+1$ question is still unsolved. Can the question in the title be answered though? – aman Dec 25 '19 at 13:39

3 Answers3

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The desired result follows quickly from Dirichlet's Theorem on primes in arithmetic progressions.

Indeed, we remark that $\gcd(10^n+1,10^m)=1$ so there are infinitely many primes congruent to $10^n+1\pmod {10^m}$. Thus, for instance, there are infinitely many primes which end in $10000001$, say, since there are infinitely many primes congruent to $10000001\pmod {10^8}$.

Note: it is also the case that there are infinitely many primes which begin with any given string of digits, see this for instance. But I think Dirichlet's theorem is easier to work with.

lulu
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  • Dirichlet's Theorem is deep and difficult. I would be surprised to see a simple proof that ${k10^{n+1}+1: k\in \Bbb N}$ contains infinitely many primes...+1 – DanielWainfleet Dec 25 '19 at 15:19
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We can show that there are infinitely many solutions to the congruence $p\equiv10^n+1\mod{10^{n+1}}$ as such a prime would end in $10.....01$

As $10^{n+1}, 10^n+1$ are coprime, the Dirichlet theorem applies, and so there are infinitely many solutions for any $n$

aman
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There is another way of looking at this apart from Dirichlet's theorem. The density of primes $\le 10^n$ approximates $1/(n\ln 10)$, but the density of integers $\le 10^n$ with no zero digits approximates $\frac98(9/10)^n$. And $\frac98(9/10)^n$ is less than $1/(n\ln 10)$ for large enough $n$.

TonyK
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