There exists a prime congruent to $m$ mod $p$











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Q: Is every element of $mathbb Z/pmathbb Z$ represented by a prime number? More generally, let $m, n in mathbb Z$ be coprime. Is there a prime number congruent to $m$ modulo $n$?




the affirmative answer is given by Dirichlet's theorem.




A: Yes, in fact there are infinitely many such primes.




I don't want to prove Dirichlet's theorem. Is it obvious that there always exists at least one such prime?










share|cite|improve this question






















  • no, it is not obvious. True, though.
    – Will Jagy
    Dec 2 '15 at 5:35










  • @WillJagy Is there a faster way to prove this, or is it considered a corollary of Dirichlet's theorem?
    – Ben
    Dec 2 '15 at 5:38








  • 1




    I find this an interesting question. It would seem that proving there is only one should be easier that proving that there are an infinite number.
    – marty cohen
    Dec 2 '15 at 5:41






  • 1




    The great majority of number theory constructions that I have ever seen require one such prime. The great American number theorist, Leonard Eugene Dickson, hated Dirichlet's theorem and all those methods. He did what he could to never use it in his 1939 book, Modern Elementary Theory of Numbers. In the end, though, he needed it. He did ask someone else to write the appendix with the proof, he couldn't stand the idea.
    – Will Jagy
    Dec 2 '15 at 5:45










  • There we go, the appendix, pages 269-305 so pretty long, was written by W. T. Reid.
    – Will Jagy
    Dec 2 '15 at 5:49















up vote
4
down vote

favorite
1













Q: Is every element of $mathbb Z/pmathbb Z$ represented by a prime number? More generally, let $m, n in mathbb Z$ be coprime. Is there a prime number congruent to $m$ modulo $n$?




the affirmative answer is given by Dirichlet's theorem.




A: Yes, in fact there are infinitely many such primes.




I don't want to prove Dirichlet's theorem. Is it obvious that there always exists at least one such prime?










share|cite|improve this question






















  • no, it is not obvious. True, though.
    – Will Jagy
    Dec 2 '15 at 5:35










  • @WillJagy Is there a faster way to prove this, or is it considered a corollary of Dirichlet's theorem?
    – Ben
    Dec 2 '15 at 5:38








  • 1




    I find this an interesting question. It would seem that proving there is only one should be easier that proving that there are an infinite number.
    – marty cohen
    Dec 2 '15 at 5:41






  • 1




    The great majority of number theory constructions that I have ever seen require one such prime. The great American number theorist, Leonard Eugene Dickson, hated Dirichlet's theorem and all those methods. He did what he could to never use it in his 1939 book, Modern Elementary Theory of Numbers. In the end, though, he needed it. He did ask someone else to write the appendix with the proof, he couldn't stand the idea.
    – Will Jagy
    Dec 2 '15 at 5:45










  • There we go, the appendix, pages 269-305 so pretty long, was written by W. T. Reid.
    – Will Jagy
    Dec 2 '15 at 5:49













up vote
4
down vote

favorite
1









up vote
4
down vote

favorite
1






1






Q: Is every element of $mathbb Z/pmathbb Z$ represented by a prime number? More generally, let $m, n in mathbb Z$ be coprime. Is there a prime number congruent to $m$ modulo $n$?




the affirmative answer is given by Dirichlet's theorem.




A: Yes, in fact there are infinitely many such primes.




I don't want to prove Dirichlet's theorem. Is it obvious that there always exists at least one such prime?










share|cite|improve this question














Q: Is every element of $mathbb Z/pmathbb Z$ represented by a prime number? More generally, let $m, n in mathbb Z$ be coprime. Is there a prime number congruent to $m$ modulo $n$?




the affirmative answer is given by Dirichlet's theorem.




A: Yes, in fact there are infinitely many such primes.




I don't want to prove Dirichlet's theorem. Is it obvious that there always exists at least one such prime?







elementary-number-theory






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share|cite|improve this question










asked Dec 2 '15 at 5:31









Ben

1,971616




1,971616












  • no, it is not obvious. True, though.
    – Will Jagy
    Dec 2 '15 at 5:35










  • @WillJagy Is there a faster way to prove this, or is it considered a corollary of Dirichlet's theorem?
    – Ben
    Dec 2 '15 at 5:38








  • 1




    I find this an interesting question. It would seem that proving there is only one should be easier that proving that there are an infinite number.
    – marty cohen
    Dec 2 '15 at 5:41






  • 1




    The great majority of number theory constructions that I have ever seen require one such prime. The great American number theorist, Leonard Eugene Dickson, hated Dirichlet's theorem and all those methods. He did what he could to never use it in his 1939 book, Modern Elementary Theory of Numbers. In the end, though, he needed it. He did ask someone else to write the appendix with the proof, he couldn't stand the idea.
    – Will Jagy
    Dec 2 '15 at 5:45










  • There we go, the appendix, pages 269-305 so pretty long, was written by W. T. Reid.
    – Will Jagy
    Dec 2 '15 at 5:49


















  • no, it is not obvious. True, though.
    – Will Jagy
    Dec 2 '15 at 5:35










  • @WillJagy Is there a faster way to prove this, or is it considered a corollary of Dirichlet's theorem?
    – Ben
    Dec 2 '15 at 5:38








  • 1




    I find this an interesting question. It would seem that proving there is only one should be easier that proving that there are an infinite number.
    – marty cohen
    Dec 2 '15 at 5:41






  • 1




    The great majority of number theory constructions that I have ever seen require one such prime. The great American number theorist, Leonard Eugene Dickson, hated Dirichlet's theorem and all those methods. He did what he could to never use it in his 1939 book, Modern Elementary Theory of Numbers. In the end, though, he needed it. He did ask someone else to write the appendix with the proof, he couldn't stand the idea.
    – Will Jagy
    Dec 2 '15 at 5:45










  • There we go, the appendix, pages 269-305 so pretty long, was written by W. T. Reid.
    – Will Jagy
    Dec 2 '15 at 5:49
















no, it is not obvious. True, though.
– Will Jagy
Dec 2 '15 at 5:35




no, it is not obvious. True, though.
– Will Jagy
Dec 2 '15 at 5:35












@WillJagy Is there a faster way to prove this, or is it considered a corollary of Dirichlet's theorem?
– Ben
Dec 2 '15 at 5:38






@WillJagy Is there a faster way to prove this, or is it considered a corollary of Dirichlet's theorem?
– Ben
Dec 2 '15 at 5:38






1




1




I find this an interesting question. It would seem that proving there is only one should be easier that proving that there are an infinite number.
– marty cohen
Dec 2 '15 at 5:41




I find this an interesting question. It would seem that proving there is only one should be easier that proving that there are an infinite number.
– marty cohen
Dec 2 '15 at 5:41




1




1




The great majority of number theory constructions that I have ever seen require one such prime. The great American number theorist, Leonard Eugene Dickson, hated Dirichlet's theorem and all those methods. He did what he could to never use it in his 1939 book, Modern Elementary Theory of Numbers. In the end, though, he needed it. He did ask someone else to write the appendix with the proof, he couldn't stand the idea.
– Will Jagy
Dec 2 '15 at 5:45




The great majority of number theory constructions that I have ever seen require one such prime. The great American number theorist, Leonard Eugene Dickson, hated Dirichlet's theorem and all those methods. He did what he could to never use it in his 1939 book, Modern Elementary Theory of Numbers. In the end, though, he needed it. He did ask someone else to write the appendix with the proof, he couldn't stand the idea.
– Will Jagy
Dec 2 '15 at 5:45












There we go, the appendix, pages 269-305 so pretty long, was written by W. T. Reid.
– Will Jagy
Dec 2 '15 at 5:49




There we go, the appendix, pages 269-305 so pretty long, was written by W. T. Reid.
– Will Jagy
Dec 2 '15 at 5:49










3 Answers
3






active

oldest

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up vote
1
down vote



accepted











I don't want to prove Dirichlet's theorem. Is it obvious that there always exists at least one such prime?




It is not obvious, and in fact the claim "if $gcd(m, n) = 1$ then there is at least one prime congruent to $m bmod n$" is equivalent to the claim "if $gcd(m, n) = 1$ then there are infinitely many primes congruent to $m bmod n$."



The implication $Leftarrow$ is obvious. To prove the implication $Rightarrow$, suppose we want to prove that there are infinitely many primes congruent to $m bmod n$. We can do this by using the existence of




  • at least one prime congruent to $m bmod n$

  • at least one prime congruent to $m + n bmod 2n$

  • at least one prime congruent to $m + 2n bmod 3n$


etc. This produces a sequence $p_k$ of primes congruent to $m + kn bmod (k+1)n$, which gives a sequence of primes congruent to $m bmod n$ where $p_k to infty$ and so the sequence $p_k$ contains infinitely many primes.






share|cite|improve this answer























  • Thanks Qiaochu!
    – Ben
    Nov 19 at 5:30










  • A slight problem is $m+n$ and $2n$ aren’t coprime if $m$ and $n$ are odd. If one uses $m+n$ mod $n^2$ (and then $m+n^2$ mod $n^3$ and so on) I think this issue goes away. In other words, $m+n^i$ mod $n^k$ for $i = 1,ldots,k-1$ should give $k-1$ distinct (even mod $n^k$) primes congruent to $m$ mod $n$, for any $k$.
    – Ben
    Nov 19 at 6:23


















up vote
0
down vote













It is possible to prove some cases without Diriclet's Theorem for $ax+b$ with $a$ and $b$ coprime. Without going into too much deatail for the proofs, they are:



any integer $a$ and $b=1, -1$



$a=2, 3, 4, 6, 8, 12, 24$ and $b$ is coprime to $a$



If you are given forms $ax+b$ and $ax+c$, say, it is also possible to prove that there are infintely many primes of the form $ax+b$ OR $ax+c$ (but does not prove the form individually). This happens exactly when $b^4=1pmod a$ and $b^3=cpmod a$.






share|cite|improve this answer




























    up vote
    0
    down vote













    This answer is inspired by Qiaochu's idea, with a small correction mentioned in the comments there. Call the statement below ${bf P}(q)$:




    If $(m,n) = 1$ then there exist $q$ distinct primes congruent to $m text{mod } n$.




    The goal is to show that ${bf P}(1) implies {bf P}(infty)$ by repeatedly applying ${bf P}(1)$ to higher and higher multiples of $n$, with different lifts of $m$.





    To makes this work, one must find numbers of the form $m + ell$ for which $n, |, ell$ and which are coprime to some large multiple of $n$. This will be done using the following (obvious) fact:




    Let $m$ and $n$ be coprime, let $ell geq 1$, and let $N$ be a multiple of $n$ which is also coprime to $m$. Then $m+ell$ is coprime to $N$ if $text{rad}(N), |, ell$.




    In particular, the numbers $m_i := m + n^i$ are coprime to $n^{k+1}$ for each $i in [1,k]$. Applying ${bf P}(1)$ to each $m_i$, one finds $k$ distinct primes congruent to $m$ mod $n$, deducing ${bf P}(k)$.



    ${bf P}(1) implies {bf P}(k)$ for any $k in mathbb N$, so ${bf P}(infty)$ follows.






    share|cite|improve this answer





















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      3 Answers
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      3 Answers
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      active

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      up vote
      1
      down vote



      accepted











      I don't want to prove Dirichlet's theorem. Is it obvious that there always exists at least one such prime?




      It is not obvious, and in fact the claim "if $gcd(m, n) = 1$ then there is at least one prime congruent to $m bmod n$" is equivalent to the claim "if $gcd(m, n) = 1$ then there are infinitely many primes congruent to $m bmod n$."



      The implication $Leftarrow$ is obvious. To prove the implication $Rightarrow$, suppose we want to prove that there are infinitely many primes congruent to $m bmod n$. We can do this by using the existence of




      • at least one prime congruent to $m bmod n$

      • at least one prime congruent to $m + n bmod 2n$

      • at least one prime congruent to $m + 2n bmod 3n$


      etc. This produces a sequence $p_k$ of primes congruent to $m + kn bmod (k+1)n$, which gives a sequence of primes congruent to $m bmod n$ where $p_k to infty$ and so the sequence $p_k$ contains infinitely many primes.






      share|cite|improve this answer























      • Thanks Qiaochu!
        – Ben
        Nov 19 at 5:30










      • A slight problem is $m+n$ and $2n$ aren’t coprime if $m$ and $n$ are odd. If one uses $m+n$ mod $n^2$ (and then $m+n^2$ mod $n^3$ and so on) I think this issue goes away. In other words, $m+n^i$ mod $n^k$ for $i = 1,ldots,k-1$ should give $k-1$ distinct (even mod $n^k$) primes congruent to $m$ mod $n$, for any $k$.
        – Ben
        Nov 19 at 6:23















      up vote
      1
      down vote



      accepted











      I don't want to prove Dirichlet's theorem. Is it obvious that there always exists at least one such prime?




      It is not obvious, and in fact the claim "if $gcd(m, n) = 1$ then there is at least one prime congruent to $m bmod n$" is equivalent to the claim "if $gcd(m, n) = 1$ then there are infinitely many primes congruent to $m bmod n$."



      The implication $Leftarrow$ is obvious. To prove the implication $Rightarrow$, suppose we want to prove that there are infinitely many primes congruent to $m bmod n$. We can do this by using the existence of




      • at least one prime congruent to $m bmod n$

      • at least one prime congruent to $m + n bmod 2n$

      • at least one prime congruent to $m + 2n bmod 3n$


      etc. This produces a sequence $p_k$ of primes congruent to $m + kn bmod (k+1)n$, which gives a sequence of primes congruent to $m bmod n$ where $p_k to infty$ and so the sequence $p_k$ contains infinitely many primes.






      share|cite|improve this answer























      • Thanks Qiaochu!
        – Ben
        Nov 19 at 5:30










      • A slight problem is $m+n$ and $2n$ aren’t coprime if $m$ and $n$ are odd. If one uses $m+n$ mod $n^2$ (and then $m+n^2$ mod $n^3$ and so on) I think this issue goes away. In other words, $m+n^i$ mod $n^k$ for $i = 1,ldots,k-1$ should give $k-1$ distinct (even mod $n^k$) primes congruent to $m$ mod $n$, for any $k$.
        – Ben
        Nov 19 at 6:23













      up vote
      1
      down vote



      accepted







      up vote
      1
      down vote



      accepted







      I don't want to prove Dirichlet's theorem. Is it obvious that there always exists at least one such prime?




      It is not obvious, and in fact the claim "if $gcd(m, n) = 1$ then there is at least one prime congruent to $m bmod n$" is equivalent to the claim "if $gcd(m, n) = 1$ then there are infinitely many primes congruent to $m bmod n$."



      The implication $Leftarrow$ is obvious. To prove the implication $Rightarrow$, suppose we want to prove that there are infinitely many primes congruent to $m bmod n$. We can do this by using the existence of




      • at least one prime congruent to $m bmod n$

      • at least one prime congruent to $m + n bmod 2n$

      • at least one prime congruent to $m + 2n bmod 3n$


      etc. This produces a sequence $p_k$ of primes congruent to $m + kn bmod (k+1)n$, which gives a sequence of primes congruent to $m bmod n$ where $p_k to infty$ and so the sequence $p_k$ contains infinitely many primes.






      share|cite|improve this answer















      I don't want to prove Dirichlet's theorem. Is it obvious that there always exists at least one such prime?




      It is not obvious, and in fact the claim "if $gcd(m, n) = 1$ then there is at least one prime congruent to $m bmod n$" is equivalent to the claim "if $gcd(m, n) = 1$ then there are infinitely many primes congruent to $m bmod n$."



      The implication $Leftarrow$ is obvious. To prove the implication $Rightarrow$, suppose we want to prove that there are infinitely many primes congruent to $m bmod n$. We can do this by using the existence of




      • at least one prime congruent to $m bmod n$

      • at least one prime congruent to $m + n bmod 2n$

      • at least one prime congruent to $m + 2n bmod 3n$


      etc. This produces a sequence $p_k$ of primes congruent to $m + kn bmod (k+1)n$, which gives a sequence of primes congruent to $m bmod n$ where $p_k to infty$ and so the sequence $p_k$ contains infinitely many primes.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Nov 18 at 5:47

























      answered Nov 18 at 4:56









      Qiaochu Yuan

      274k32578914




      274k32578914












      • Thanks Qiaochu!
        – Ben
        Nov 19 at 5:30










      • A slight problem is $m+n$ and $2n$ aren’t coprime if $m$ and $n$ are odd. If one uses $m+n$ mod $n^2$ (and then $m+n^2$ mod $n^3$ and so on) I think this issue goes away. In other words, $m+n^i$ mod $n^k$ for $i = 1,ldots,k-1$ should give $k-1$ distinct (even mod $n^k$) primes congruent to $m$ mod $n$, for any $k$.
        – Ben
        Nov 19 at 6:23


















      • Thanks Qiaochu!
        – Ben
        Nov 19 at 5:30










      • A slight problem is $m+n$ and $2n$ aren’t coprime if $m$ and $n$ are odd. If one uses $m+n$ mod $n^2$ (and then $m+n^2$ mod $n^3$ and so on) I think this issue goes away. In other words, $m+n^i$ mod $n^k$ for $i = 1,ldots,k-1$ should give $k-1$ distinct (even mod $n^k$) primes congruent to $m$ mod $n$, for any $k$.
        – Ben
        Nov 19 at 6:23
















      Thanks Qiaochu!
      – Ben
      Nov 19 at 5:30




      Thanks Qiaochu!
      – Ben
      Nov 19 at 5:30












      A slight problem is $m+n$ and $2n$ aren’t coprime if $m$ and $n$ are odd. If one uses $m+n$ mod $n^2$ (and then $m+n^2$ mod $n^3$ and so on) I think this issue goes away. In other words, $m+n^i$ mod $n^k$ for $i = 1,ldots,k-1$ should give $k-1$ distinct (even mod $n^k$) primes congruent to $m$ mod $n$, for any $k$.
      – Ben
      Nov 19 at 6:23




      A slight problem is $m+n$ and $2n$ aren’t coprime if $m$ and $n$ are odd. If one uses $m+n$ mod $n^2$ (and then $m+n^2$ mod $n^3$ and so on) I think this issue goes away. In other words, $m+n^i$ mod $n^k$ for $i = 1,ldots,k-1$ should give $k-1$ distinct (even mod $n^k$) primes congruent to $m$ mod $n$, for any $k$.
      – Ben
      Nov 19 at 6:23










      up vote
      0
      down vote













      It is possible to prove some cases without Diriclet's Theorem for $ax+b$ with $a$ and $b$ coprime. Without going into too much deatail for the proofs, they are:



      any integer $a$ and $b=1, -1$



      $a=2, 3, 4, 6, 8, 12, 24$ and $b$ is coprime to $a$



      If you are given forms $ax+b$ and $ax+c$, say, it is also possible to prove that there are infintely many primes of the form $ax+b$ OR $ax+c$ (but does not prove the form individually). This happens exactly when $b^4=1pmod a$ and $b^3=cpmod a$.






      share|cite|improve this answer

























        up vote
        0
        down vote













        It is possible to prove some cases without Diriclet's Theorem for $ax+b$ with $a$ and $b$ coprime. Without going into too much deatail for the proofs, they are:



        any integer $a$ and $b=1, -1$



        $a=2, 3, 4, 6, 8, 12, 24$ and $b$ is coprime to $a$



        If you are given forms $ax+b$ and $ax+c$, say, it is also possible to prove that there are infintely many primes of the form $ax+b$ OR $ax+c$ (but does not prove the form individually). This happens exactly when $b^4=1pmod a$ and $b^3=cpmod a$.






        share|cite|improve this answer























          up vote
          0
          down vote










          up vote
          0
          down vote









          It is possible to prove some cases without Diriclet's Theorem for $ax+b$ with $a$ and $b$ coprime. Without going into too much deatail for the proofs, they are:



          any integer $a$ and $b=1, -1$



          $a=2, 3, 4, 6, 8, 12, 24$ and $b$ is coprime to $a$



          If you are given forms $ax+b$ and $ax+c$, say, it is also possible to prove that there are infintely many primes of the form $ax+b$ OR $ax+c$ (but does not prove the form individually). This happens exactly when $b^4=1pmod a$ and $b^3=cpmod a$.






          share|cite|improve this answer












          It is possible to prove some cases without Diriclet's Theorem for $ax+b$ with $a$ and $b$ coprime. Without going into too much deatail for the proofs, they are:



          any integer $a$ and $b=1, -1$



          $a=2, 3, 4, 6, 8, 12, 24$ and $b$ is coprime to $a$



          If you are given forms $ax+b$ and $ax+c$, say, it is also possible to prove that there are infintely many primes of the form $ax+b$ OR $ax+c$ (but does not prove the form individually). This happens exactly when $b^4=1pmod a$ and $b^3=cpmod a$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 18 at 4:45









          J. Linne

          829315




          829315






















              up vote
              0
              down vote













              This answer is inspired by Qiaochu's idea, with a small correction mentioned in the comments there. Call the statement below ${bf P}(q)$:




              If $(m,n) = 1$ then there exist $q$ distinct primes congruent to $m text{mod } n$.




              The goal is to show that ${bf P}(1) implies {bf P}(infty)$ by repeatedly applying ${bf P}(1)$ to higher and higher multiples of $n$, with different lifts of $m$.





              To makes this work, one must find numbers of the form $m + ell$ for which $n, |, ell$ and which are coprime to some large multiple of $n$. This will be done using the following (obvious) fact:




              Let $m$ and $n$ be coprime, let $ell geq 1$, and let $N$ be a multiple of $n$ which is also coprime to $m$. Then $m+ell$ is coprime to $N$ if $text{rad}(N), |, ell$.




              In particular, the numbers $m_i := m + n^i$ are coprime to $n^{k+1}$ for each $i in [1,k]$. Applying ${bf P}(1)$ to each $m_i$, one finds $k$ distinct primes congruent to $m$ mod $n$, deducing ${bf P}(k)$.



              ${bf P}(1) implies {bf P}(k)$ for any $k in mathbb N$, so ${bf P}(infty)$ follows.






              share|cite|improve this answer

























                up vote
                0
                down vote













                This answer is inspired by Qiaochu's idea, with a small correction mentioned in the comments there. Call the statement below ${bf P}(q)$:




                If $(m,n) = 1$ then there exist $q$ distinct primes congruent to $m text{mod } n$.




                The goal is to show that ${bf P}(1) implies {bf P}(infty)$ by repeatedly applying ${bf P}(1)$ to higher and higher multiples of $n$, with different lifts of $m$.





                To makes this work, one must find numbers of the form $m + ell$ for which $n, |, ell$ and which are coprime to some large multiple of $n$. This will be done using the following (obvious) fact:




                Let $m$ and $n$ be coprime, let $ell geq 1$, and let $N$ be a multiple of $n$ which is also coprime to $m$. Then $m+ell$ is coprime to $N$ if $text{rad}(N), |, ell$.




                In particular, the numbers $m_i := m + n^i$ are coprime to $n^{k+1}$ for each $i in [1,k]$. Applying ${bf P}(1)$ to each $m_i$, one finds $k$ distinct primes congruent to $m$ mod $n$, deducing ${bf P}(k)$.



                ${bf P}(1) implies {bf P}(k)$ for any $k in mathbb N$, so ${bf P}(infty)$ follows.






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                  This answer is inspired by Qiaochu's idea, with a small correction mentioned in the comments there. Call the statement below ${bf P}(q)$:




                  If $(m,n) = 1$ then there exist $q$ distinct primes congruent to $m text{mod } n$.




                  The goal is to show that ${bf P}(1) implies {bf P}(infty)$ by repeatedly applying ${bf P}(1)$ to higher and higher multiples of $n$, with different lifts of $m$.





                  To makes this work, one must find numbers of the form $m + ell$ for which $n, |, ell$ and which are coprime to some large multiple of $n$. This will be done using the following (obvious) fact:




                  Let $m$ and $n$ be coprime, let $ell geq 1$, and let $N$ be a multiple of $n$ which is also coprime to $m$. Then $m+ell$ is coprime to $N$ if $text{rad}(N), |, ell$.




                  In particular, the numbers $m_i := m + n^i$ are coprime to $n^{k+1}$ for each $i in [1,k]$. Applying ${bf P}(1)$ to each $m_i$, one finds $k$ distinct primes congruent to $m$ mod $n$, deducing ${bf P}(k)$.



                  ${bf P}(1) implies {bf P}(k)$ for any $k in mathbb N$, so ${bf P}(infty)$ follows.






                  share|cite|improve this answer












                  This answer is inspired by Qiaochu's idea, with a small correction mentioned in the comments there. Call the statement below ${bf P}(q)$:




                  If $(m,n) = 1$ then there exist $q$ distinct primes congruent to $m text{mod } n$.




                  The goal is to show that ${bf P}(1) implies {bf P}(infty)$ by repeatedly applying ${bf P}(1)$ to higher and higher multiples of $n$, with different lifts of $m$.





                  To makes this work, one must find numbers of the form $m + ell$ for which $n, |, ell$ and which are coprime to some large multiple of $n$. This will be done using the following (obvious) fact:




                  Let $m$ and $n$ be coprime, let $ell geq 1$, and let $N$ be a multiple of $n$ which is also coprime to $m$. Then $m+ell$ is coprime to $N$ if $text{rad}(N), |, ell$.




                  In particular, the numbers $m_i := m + n^i$ are coprime to $n^{k+1}$ for each $i in [1,k]$. Applying ${bf P}(1)$ to each $m_i$, one finds $k$ distinct primes congruent to $m$ mod $n$, deducing ${bf P}(k)$.



                  ${bf P}(1) implies {bf P}(k)$ for any $k in mathbb N$, so ${bf P}(infty)$ follows.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 19 at 8:07









                  Ben

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