Question about the k-tuples conjecture












2












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On the "Prime Arithmetic Progression" page on Wolfram MathWorld, it states that a special case of the k-tuples conjecture is that there are infinitely long prime arithmetic sequences (arithmetic sequences for which all terms are prime)



But why doesn't this proof disprove that?



Let $a+dn$ be the $n$th term of any arithmetic progression. When $a=n$, the equation becomes $a(1+n)$, which obviously can't be prime because it is the product of two integers. Therefore, all infinite arithmetic progressions have at least one composite number.






proof-verification prime-numbers







share|cite|improve this question









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  • 1




    $begingroup$
    Of course, no arithmetic progression can produce only primes. This is even true for polynomials with integer coefficients. The meaning of the statement is : For every arbitary large $k$, there is an arithmetic progression of length $k$ producing only primes (I think this has been proven, but it is not known whether the statement remains true if the primes have to be consecutive)
    $endgroup$
    – Peter
    Mar 18 '17 at 23:05








  • 1




    $begingroup$
    In the Wolfram article on PAP, if you follow the link in the second paragraph (mathworld.wolfram.com/k-TupleConjecture.html), that article, near the bottom, describes the special case, "prime patterns conjecture", mentioned in the PAP article. It goes on to say that there are arbitrarily long APs of primes. Note arbitrarily long is not the same as infinite. IMO the PAP article, is at best confusing, in talking about the "prime patterns conjecture".
    $endgroup$
    – Χpẘ
    Mar 18 '17 at 23:42






  • 3




    $begingroup$
    Indeed, the Wolfram article uses the exact phrase "infinitely long prime arithmetic progressions", which is not accurate: "arbitrarily long prime arithmetic progressions" is what is meant. So the OP's objection is quite valid!
    $endgroup$
    – Greg Martin
    Mar 18 '17 at 23:43






  • 1




    $begingroup$
    Minor correction: $a(1+n)$ should be $a(1+d)$.
    $endgroup$
    – hardmath
    Mar 19 '17 at 0:14
















2












$begingroup$


On the "Prime Arithmetic Progression" page on Wolfram MathWorld, it states that a special case of the k-tuples conjecture is that there are infinitely long prime arithmetic sequences (arithmetic sequences for which all terms are prime)



But why doesn't this proof disprove that?



Let $a+dn$ be the $n$th term of any arithmetic progression. When $a=n$, the equation becomes $a(1+n)$, which obviously can't be prime because it is the product of two integers. Therefore, all infinite arithmetic progressions have at least one composite number.






proof-verification prime-numbers







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Of course, no arithmetic progression can produce only primes. This is even true for polynomials with integer coefficients. The meaning of the statement is : For every arbitary large $k$, there is an arithmetic progression of length $k$ producing only primes (I think this has been proven, but it is not known whether the statement remains true if the primes have to be consecutive)
    $endgroup$
    – Peter
    Mar 18 '17 at 23:05








  • 1




    $begingroup$
    In the Wolfram article on PAP, if you follow the link in the second paragraph (mathworld.wolfram.com/k-TupleConjecture.html), that article, near the bottom, describes the special case, "prime patterns conjecture", mentioned in the PAP article. It goes on to say that there are arbitrarily long APs of primes. Note arbitrarily long is not the same as infinite. IMO the PAP article, is at best confusing, in talking about the "prime patterns conjecture".
    $endgroup$
    – Χpẘ
    Mar 18 '17 at 23:42






  • 3




    $begingroup$
    Indeed, the Wolfram article uses the exact phrase "infinitely long prime arithmetic progressions", which is not accurate: "arbitrarily long prime arithmetic progressions" is what is meant. So the OP's objection is quite valid!
    $endgroup$
    – Greg Martin
    Mar 18 '17 at 23:43






  • 1




    $begingroup$
    Minor correction: $a(1+n)$ should be $a(1+d)$.
    $endgroup$
    – hardmath
    Mar 19 '17 at 0:14














2












2








2





$begingroup$


On the "Prime Arithmetic Progression" page on Wolfram MathWorld, it states that a special case of the k-tuples conjecture is that there are infinitely long prime arithmetic sequences (arithmetic sequences for which all terms are prime)



But why doesn't this proof disprove that?



Let $a+dn$ be the $n$th term of any arithmetic progression. When $a=n$, the equation becomes $a(1+n)$, which obviously can't be prime because it is the product of two integers. Therefore, all infinite arithmetic progressions have at least one composite number.






proof-verification prime-numbers







share|cite|improve this question









$endgroup$




On the "Prime Arithmetic Progression" page on Wolfram MathWorld, it states that a special case of the k-tuples conjecture is that there are infinitely long prime arithmetic sequences (arithmetic sequences for which all terms are prime)



But why doesn't this proof disprove that?



Let $a+dn$ be the $n$th term of any arithmetic progression. When $a=n$, the equation becomes $a(1+n)$, which obviously can't be prime because it is the product of two integers. Therefore, all infinite arithmetic progressions have at least one composite number.






proof-verification prime-numbers




proof-verification prime-numbers






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Mar 18 '17 at 22:57









TreFoxTreFox

1,98411034




1,98411034








  • 1




    $begingroup$
    Of course, no arithmetic progression can produce only primes. This is even true for polynomials with integer coefficients. The meaning of the statement is : For every arbitary large $k$, there is an arithmetic progression of length $k$ producing only primes (I think this has been proven, but it is not known whether the statement remains true if the primes have to be consecutive)
    $endgroup$
    – Peter
    Mar 18 '17 at 23:05








  • 1




    $begingroup$
    In the Wolfram article on PAP, if you follow the link in the second paragraph (mathworld.wolfram.com/k-TupleConjecture.html), that article, near the bottom, describes the special case, "prime patterns conjecture", mentioned in the PAP article. It goes on to say that there are arbitrarily long APs of primes. Note arbitrarily long is not the same as infinite. IMO the PAP article, is at best confusing, in talking about the "prime patterns conjecture".
    $endgroup$
    – Χpẘ
    Mar 18 '17 at 23:42






  • 3




    $begingroup$
    Indeed, the Wolfram article uses the exact phrase "infinitely long prime arithmetic progressions", which is not accurate: "arbitrarily long prime arithmetic progressions" is what is meant. So the OP's objection is quite valid!
    $endgroup$
    – Greg Martin
    Mar 18 '17 at 23:43






  • 1




    $begingroup$
    Minor correction: $a(1+n)$ should be $a(1+d)$.
    $endgroup$
    – hardmath
    Mar 19 '17 at 0:14














  • 1




    $begingroup$
    Of course, no arithmetic progression can produce only primes. This is even true for polynomials with integer coefficients. The meaning of the statement is : For every arbitary large $k$, there is an arithmetic progression of length $k$ producing only primes (I think this has been proven, but it is not known whether the statement remains true if the primes have to be consecutive)
    $endgroup$
    – Peter
    Mar 18 '17 at 23:05








  • 1




    $begingroup$
    In the Wolfram article on PAP, if you follow the link in the second paragraph (mathworld.wolfram.com/k-TupleConjecture.html), that article, near the bottom, describes the special case, "prime patterns conjecture", mentioned in the PAP article. It goes on to say that there are arbitrarily long APs of primes. Note arbitrarily long is not the same as infinite. IMO the PAP article, is at best confusing, in talking about the "prime patterns conjecture".
    $endgroup$
    – Χpẘ
    Mar 18 '17 at 23:42






  • 3




    $begingroup$
    Indeed, the Wolfram article uses the exact phrase "infinitely long prime arithmetic progressions", which is not accurate: "arbitrarily long prime arithmetic progressions" is what is meant. So the OP's objection is quite valid!
    $endgroup$
    – Greg Martin
    Mar 18 '17 at 23:43






  • 1




    $begingroup$
    Minor correction: $a(1+n)$ should be $a(1+d)$.
    $endgroup$
    – hardmath
    Mar 19 '17 at 0:14








1




1




$begingroup$
Of course, no arithmetic progression can produce only primes. This is even true for polynomials with integer coefficients. The meaning of the statement is : For every arbitary large $k$, there is an arithmetic progression of length $k$ producing only primes (I think this has been proven, but it is not known whether the statement remains true if the primes have to be consecutive)
$endgroup$
– Peter
Mar 18 '17 at 23:05






$begingroup$
Of course, no arithmetic progression can produce only primes. This is even true for polynomials with integer coefficients. The meaning of the statement is : For every arbitary large $k$, there is an arithmetic progression of length $k$ producing only primes (I think this has been proven, but it is not known whether the statement remains true if the primes have to be consecutive)
$endgroup$
– Peter
Mar 18 '17 at 23:05






1




1




$begingroup$
In the Wolfram article on PAP, if you follow the link in the second paragraph (mathworld.wolfram.com/k-TupleConjecture.html), that article, near the bottom, describes the special case, "prime patterns conjecture", mentioned in the PAP article. It goes on to say that there are arbitrarily long APs of primes. Note arbitrarily long is not the same as infinite. IMO the PAP article, is at best confusing, in talking about the "prime patterns conjecture".
$endgroup$
– Χpẘ
Mar 18 '17 at 23:42




$begingroup$
In the Wolfram article on PAP, if you follow the link in the second paragraph (mathworld.wolfram.com/k-TupleConjecture.html), that article, near the bottom, describes the special case, "prime patterns conjecture", mentioned in the PAP article. It goes on to say that there are arbitrarily long APs of primes. Note arbitrarily long is not the same as infinite. IMO the PAP article, is at best confusing, in talking about the "prime patterns conjecture".
$endgroup$
– Χpẘ
Mar 18 '17 at 23:42




3




3




$begingroup$
Indeed, the Wolfram article uses the exact phrase "infinitely long prime arithmetic progressions", which is not accurate: "arbitrarily long prime arithmetic progressions" is what is meant. So the OP's objection is quite valid!
$endgroup$
– Greg Martin
Mar 18 '17 at 23:43




$begingroup$
Indeed, the Wolfram article uses the exact phrase "infinitely long prime arithmetic progressions", which is not accurate: "arbitrarily long prime arithmetic progressions" is what is meant. So the OP's objection is quite valid!
$endgroup$
– Greg Martin
Mar 18 '17 at 23:43




1




1




$begingroup$
Minor correction: $a(1+n)$ should be $a(1+d)$.
$endgroup$
– hardmath
Mar 19 '17 at 0:14




$begingroup$
Minor correction: $a(1+n)$ should be $a(1+d)$.
$endgroup$
– hardmath
Mar 19 '17 at 0:14










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As others have pointed out in the comment section, there can be no arithmetic progression that produces only primes (see here for a quick explanation). So, as you rightly say, all infinite arithmetic progressions have at least one composite number.






share|cite|improve this answer









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    $begingroup$

    As others have pointed out in the comment section, there can be no arithmetic progression that produces only primes (see here for a quick explanation). So, as you rightly say, all infinite arithmetic progressions have at least one composite number.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      As others have pointed out in the comment section, there can be no arithmetic progression that produces only primes (see here for a quick explanation). So, as you rightly say, all infinite arithmetic progressions have at least one composite number.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        As others have pointed out in the comment section, there can be no arithmetic progression that produces only primes (see here for a quick explanation). So, as you rightly say, all infinite arithmetic progressions have at least one composite number.






        share|cite|improve this answer









        $endgroup$



        As others have pointed out in the comment section, there can be no arithmetic progression that produces only primes (see here for a quick explanation). So, as you rightly say, all infinite arithmetic progressions have at least one composite number.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 20 '18 at 11:30









        KlangenKlangen

        1,75811334




        1,75811334






























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