Every natural number $n$ can be written as $n=s-t$ with $omega(s)=omega(t)$












4












$begingroup$


Can we prove the following statement ?




Every natural number $n$ can be written as $n=s-t$ ($s,t$ positive integers) with $omega(s)=omega(t)$ , in other words , the difference of two positive integers with the same number of distinct prime factors.





  • If $n=1$ , we can choose $ s=3 $ and $ t=2$

  • If $n$ is even , we can choose $ s=2n $ and $ t=n$






number-theory elementary-number-theory prime-factorization







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    What is the question?
    $endgroup$
    – mathreadler
    Jan 11 at 8:30






  • 1




    $begingroup$
    If $n$ is even, then $2n$ and $n$ have the same number of prime factors, so $n = 2n-n$ is a solution
    $endgroup$
    – Charles Madeline
    Jan 11 at 8:39








  • 1




    $begingroup$
    @CharlesMadeline At least we can be sure that a counterexample , if there is actually one, must be huge.
    $endgroup$
    – Peter
    Jan 11 at 9:19






  • 1




    $begingroup$
    We have other possibilities. If we find a number $m$, such that $m$ and $m+1$ are both coprime to $n$ and have the same number of prime factors, we have found a solution as well.
    $endgroup$
    – Peter
    Jan 11 at 9:21








  • 1




    $begingroup$
    @CharlesMadeline Moreover, we only concentrated on solutions of the form $(k+1)n-kn$. This is not required in the question. But I agree that the proof is not yet finished.
    $endgroup$
    – Peter
    Jan 11 at 9:52
















4












$begingroup$


Can we prove the following statement ?




Every natural number $n$ can be written as $n=s-t$ ($s,t$ positive integers) with $omega(s)=omega(t)$ , in other words , the difference of two positive integers with the same number of distinct prime factors.





  • If $n=1$ , we can choose $ s=3 $ and $ t=2$

  • If $n$ is even , we can choose $ s=2n $ and $ t=n$






number-theory elementary-number-theory prime-factorization







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    What is the question?
    $endgroup$
    – mathreadler
    Jan 11 at 8:30






  • 1




    $begingroup$
    If $n$ is even, then $2n$ and $n$ have the same number of prime factors, so $n = 2n-n$ is a solution
    $endgroup$
    – Charles Madeline
    Jan 11 at 8:39








  • 1




    $begingroup$
    @CharlesMadeline At least we can be sure that a counterexample , if there is actually one, must be huge.
    $endgroup$
    – Peter
    Jan 11 at 9:19






  • 1




    $begingroup$
    We have other possibilities. If we find a number $m$, such that $m$ and $m+1$ are both coprime to $n$ and have the same number of prime factors, we have found a solution as well.
    $endgroup$
    – Peter
    Jan 11 at 9:21








  • 1




    $begingroup$
    @CharlesMadeline Moreover, we only concentrated on solutions of the form $(k+1)n-kn$. This is not required in the question. But I agree that the proof is not yet finished.
    $endgroup$
    – Peter
    Jan 11 at 9:52














4












4








4


1



$begingroup$


Can we prove the following statement ?




Every natural number $n$ can be written as $n=s-t$ ($s,t$ positive integers) with $omega(s)=omega(t)$ , in other words , the difference of two positive integers with the same number of distinct prime factors.





  • If $n=1$ , we can choose $ s=3 $ and $ t=2$

  • If $n$ is even , we can choose $ s=2n $ and $ t=n$






number-theory elementary-number-theory prime-factorization







share|cite|improve this question











$endgroup$




Can we prove the following statement ?




Every natural number $n$ can be written as $n=s-t$ ($s,t$ positive integers) with $omega(s)=omega(t)$ , in other words , the difference of two positive integers with the same number of distinct prime factors.





  • If $n=1$ , we can choose $ s=3 $ and $ t=2$

  • If $n$ is even , we can choose $ s=2n $ and $ t=n$






number-theory elementary-number-theory prime-factorization




number-theory elementary-number-theory prime-factorization






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 11 at 9:56









Peter

47.2k1039127




47.2k1039127










asked Jan 11 at 8:26









rafarafa

598212




598212








  • 2




    $begingroup$
    What is the question?
    $endgroup$
    – mathreadler
    Jan 11 at 8:30






  • 1




    $begingroup$
    If $n$ is even, then $2n$ and $n$ have the same number of prime factors, so $n = 2n-n$ is a solution
    $endgroup$
    – Charles Madeline
    Jan 11 at 8:39








  • 1




    $begingroup$
    @CharlesMadeline At least we can be sure that a counterexample , if there is actually one, must be huge.
    $endgroup$
    – Peter
    Jan 11 at 9:19






  • 1




    $begingroup$
    We have other possibilities. If we find a number $m$, such that $m$ and $m+1$ are both coprime to $n$ and have the same number of prime factors, we have found a solution as well.
    $endgroup$
    – Peter
    Jan 11 at 9:21








  • 1




    $begingroup$
    @CharlesMadeline Moreover, we only concentrated on solutions of the form $(k+1)n-kn$. This is not required in the question. But I agree that the proof is not yet finished.
    $endgroup$
    – Peter
    Jan 11 at 9:52














  • 2




    $begingroup$
    What is the question?
    $endgroup$
    – mathreadler
    Jan 11 at 8:30






  • 1




    $begingroup$
    If $n$ is even, then $2n$ and $n$ have the same number of prime factors, so $n = 2n-n$ is a solution
    $endgroup$
    – Charles Madeline
    Jan 11 at 8:39








  • 1




    $begingroup$
    @CharlesMadeline At least we can be sure that a counterexample , if there is actually one, must be huge.
    $endgroup$
    – Peter
    Jan 11 at 9:19






  • 1




    $begingroup$
    We have other possibilities. If we find a number $m$, such that $m$ and $m+1$ are both coprime to $n$ and have the same number of prime factors, we have found a solution as well.
    $endgroup$
    – Peter
    Jan 11 at 9:21








  • 1




    $begingroup$
    @CharlesMadeline Moreover, we only concentrated on solutions of the form $(k+1)n-kn$. This is not required in the question. But I agree that the proof is not yet finished.
    $endgroup$
    – Peter
    Jan 11 at 9:52








2




2




$begingroup$
What is the question?
$endgroup$
– mathreadler
Jan 11 at 8:30




$begingroup$
What is the question?
$endgroup$
– mathreadler
Jan 11 at 8:30




1




1




$begingroup$
If $n$ is even, then $2n$ and $n$ have the same number of prime factors, so $n = 2n-n$ is a solution
$endgroup$
– Charles Madeline
Jan 11 at 8:39






$begingroup$
If $n$ is even, then $2n$ and $n$ have the same number of prime factors, so $n = 2n-n$ is a solution
$endgroup$
– Charles Madeline
Jan 11 at 8:39






1




1




$begingroup$
@CharlesMadeline At least we can be sure that a counterexample , if there is actually one, must be huge.
$endgroup$
– Peter
Jan 11 at 9:19




$begingroup$
@CharlesMadeline At least we can be sure that a counterexample , if there is actually one, must be huge.
$endgroup$
– Peter
Jan 11 at 9:19




1




1




$begingroup$
We have other possibilities. If we find a number $m$, such that $m$ and $m+1$ are both coprime to $n$ and have the same number of prime factors, we have found a solution as well.
$endgroup$
– Peter
Jan 11 at 9:21






$begingroup$
We have other possibilities. If we find a number $m$, such that $m$ and $m+1$ are both coprime to $n$ and have the same number of prime factors, we have found a solution as well.
$endgroup$
– Peter
Jan 11 at 9:21






1




1




$begingroup$
@CharlesMadeline Moreover, we only concentrated on solutions of the form $(k+1)n-kn$. This is not required in the question. But I agree that the proof is not yet finished.
$endgroup$
– Peter
Jan 11 at 9:52




$begingroup$
@CharlesMadeline Moreover, we only concentrated on solutions of the form $(k+1)n-kn$. This is not required in the question. But I agree that the proof is not yet finished.
$endgroup$
– Peter
Jan 11 at 9:52










2 Answers
2






active

oldest

votes


















3












$begingroup$

There is a way to prove this directly though, and it is pretty simple:



Case 1: Both 2 and 3 divide $n$, or neither 2 nor 3 divide $n$: Then let $s=3n$ and $t=2n$.



Case 2: 3 does not divide $n$ but $2$ does: Then let $s=2n$ and $t=n$.



Case 3: 3 divides $n$ but 2 does not: Then let $p$ be the smallest odd prime that does not divide $n$. Then let $s=pn$ and let $t=(p-1)n$. [Note that $p-1$ is the product of smaller odd primes, all of which divide $n$ [by def'n of $p$], and 2, which does not. So $omega(pn) = omega(n) +1$ [because $p$ doesn't divide $n$], while $omega((p-1)n) = omega(n) +1$ as well, because the one prime factor of $p-1$ that does not divide $n$ is 2.]






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Nice solution !
    $endgroup$
    – Kolja
    Jan 12 at 0:22



















1












$begingroup$

The Bunyakovsky-conjecture implies that we always find a solution.



To show this, assume that there are infinite many positive integers $p$, such that $$2p+1$$ $$2p+3$$ $$2p^2+4p+1$$ are simultaneously prime. Then, we can choose $p>n$ satisfying this property. Then, $$(2p+1)(2p+3)n-2(2p^2+4p+1)n=n$$ is a solution, whenever $n$ is odd (the even case has already been solved).



There are plenty of other possibilities to choose the expressions, so the given statement is in fact much weaker than Bunyakovsky's conjecture.






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    There is a way to prove this directly though, and it is pretty simple:



    Case 1: Both 2 and 3 divide $n$, or neither 2 nor 3 divide $n$: Then let $s=3n$ and $t=2n$.



    Case 2: 3 does not divide $n$ but $2$ does: Then let $s=2n$ and $t=n$.



    Case 3: 3 divides $n$ but 2 does not: Then let $p$ be the smallest odd prime that does not divide $n$. Then let $s=pn$ and let $t=(p-1)n$. [Note that $p-1$ is the product of smaller odd primes, all of which divide $n$ [by def'n of $p$], and 2, which does not. So $omega(pn) = omega(n) +1$ [because $p$ doesn't divide $n$], while $omega((p-1)n) = omega(n) +1$ as well, because the one prime factor of $p-1$ that does not divide $n$ is 2.]






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Nice solution !
      $endgroup$
      – Kolja
      Jan 12 at 0:22
















    3












    $begingroup$

    There is a way to prove this directly though, and it is pretty simple:



    Case 1: Both 2 and 3 divide $n$, or neither 2 nor 3 divide $n$: Then let $s=3n$ and $t=2n$.



    Case 2: 3 does not divide $n$ but $2$ does: Then let $s=2n$ and $t=n$.



    Case 3: 3 divides $n$ but 2 does not: Then let $p$ be the smallest odd prime that does not divide $n$. Then let $s=pn$ and let $t=(p-1)n$. [Note that $p-1$ is the product of smaller odd primes, all of which divide $n$ [by def'n of $p$], and 2, which does not. So $omega(pn) = omega(n) +1$ [because $p$ doesn't divide $n$], while $omega((p-1)n) = omega(n) +1$ as well, because the one prime factor of $p-1$ that does not divide $n$ is 2.]






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Nice solution !
      $endgroup$
      – Kolja
      Jan 12 at 0:22














    3












    3








    3





    $begingroup$

    There is a way to prove this directly though, and it is pretty simple:



    Case 1: Both 2 and 3 divide $n$, or neither 2 nor 3 divide $n$: Then let $s=3n$ and $t=2n$.



    Case 2: 3 does not divide $n$ but $2$ does: Then let $s=2n$ and $t=n$.



    Case 3: 3 divides $n$ but 2 does not: Then let $p$ be the smallest odd prime that does not divide $n$. Then let $s=pn$ and let $t=(p-1)n$. [Note that $p-1$ is the product of smaller odd primes, all of which divide $n$ [by def'n of $p$], and 2, which does not. So $omega(pn) = omega(n) +1$ [because $p$ doesn't divide $n$], while $omega((p-1)n) = omega(n) +1$ as well, because the one prime factor of $p-1$ that does not divide $n$ is 2.]






    share|cite|improve this answer











    $endgroup$



    There is a way to prove this directly though, and it is pretty simple:



    Case 1: Both 2 and 3 divide $n$, or neither 2 nor 3 divide $n$: Then let $s=3n$ and $t=2n$.



    Case 2: 3 does not divide $n$ but $2$ does: Then let $s=2n$ and $t=n$.



    Case 3: 3 divides $n$ but 2 does not: Then let $p$ be the smallest odd prime that does not divide $n$. Then let $s=pn$ and let $t=(p-1)n$. [Note that $p-1$ is the product of smaller odd primes, all of which divide $n$ [by def'n of $p$], and 2, which does not. So $omega(pn) = omega(n) +1$ [because $p$ doesn't divide $n$], while $omega((p-1)n) = omega(n) +1$ as well, because the one prime factor of $p-1$ that does not divide $n$ is 2.]







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 11 at 23:56

























    answered Jan 11 at 23:06









    MikeMike

    3,542411




    3,542411








    • 1




      $begingroup$
      Nice solution !
      $endgroup$
      – Kolja
      Jan 12 at 0:22














    • 1




      $begingroup$
      Nice solution !
      $endgroup$
      – Kolja
      Jan 12 at 0:22








    1




    1




    $begingroup$
    Nice solution !
    $endgroup$
    – Kolja
    Jan 12 at 0:22




    $begingroup$
    Nice solution !
    $endgroup$
    – Kolja
    Jan 12 at 0:22











    1












    $begingroup$

    The Bunyakovsky-conjecture implies that we always find a solution.



    To show this, assume that there are infinite many positive integers $p$, such that $$2p+1$$ $$2p+3$$ $$2p^2+4p+1$$ are simultaneously prime. Then, we can choose $p>n$ satisfying this property. Then, $$(2p+1)(2p+3)n-2(2p^2+4p+1)n=n$$ is a solution, whenever $n$ is odd (the even case has already been solved).



    There are plenty of other possibilities to choose the expressions, so the given statement is in fact much weaker than Bunyakovsky's conjecture.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      The Bunyakovsky-conjecture implies that we always find a solution.



      To show this, assume that there are infinite many positive integers $p$, such that $$2p+1$$ $$2p+3$$ $$2p^2+4p+1$$ are simultaneously prime. Then, we can choose $p>n$ satisfying this property. Then, $$(2p+1)(2p+3)n-2(2p^2+4p+1)n=n$$ is a solution, whenever $n$ is odd (the even case has already been solved).



      There are plenty of other possibilities to choose the expressions, so the given statement is in fact much weaker than Bunyakovsky's conjecture.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        The Bunyakovsky-conjecture implies that we always find a solution.



        To show this, assume that there are infinite many positive integers $p$, such that $$2p+1$$ $$2p+3$$ $$2p^2+4p+1$$ are simultaneously prime. Then, we can choose $p>n$ satisfying this property. Then, $$(2p+1)(2p+3)n-2(2p^2+4p+1)n=n$$ is a solution, whenever $n$ is odd (the even case has already been solved).



        There are plenty of other possibilities to choose the expressions, so the given statement is in fact much weaker than Bunyakovsky's conjecture.






        share|cite|improve this answer









        $endgroup$



        The Bunyakovsky-conjecture implies that we always find a solution.



        To show this, assume that there are infinite many positive integers $p$, such that $$2p+1$$ $$2p+3$$ $$2p^2+4p+1$$ are simultaneously prime. Then, we can choose $p>n$ satisfying this property. Then, $$(2p+1)(2p+3)n-2(2p^2+4p+1)n=n$$ is a solution, whenever $n$ is odd (the even case has already been solved).



        There are plenty of other possibilities to choose the expressions, so the given statement is in fact much weaker than Bunyakovsky's conjecture.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 11 at 9:45









        PeterPeter

        47.2k1039127




        47.2k1039127






























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