Problem: Prove that there are infinitely many natural numbers that can't be written as $6xy+x+y$ where $x,y$ are positive integers.

I noticed that if $A=6xy+x+y$ then $6A+1=(6x+1)(6y+1)$, but I can't proceed from here. A hint would be great!

  • 1
    $\begingroup$ Do natural numbers include zero? $\endgroup$ – Kenny Lau Jul 31 '16 at 12:43
  • $\begingroup$ There are infinitely many primes $\endgroup$ – Kenny Lau Jul 31 '16 at 12:43
  • $\begingroup$ @KennyLau we spontaneously conclude. $\endgroup$ – Zack Ni Jul 31 '16 at 12:45
  • $\begingroup$ Since $\,x,y\ge 1\,$ the factors $\,6x+1,6y+1\,$ are both greater than $1$. So if $\,A\,$ has that form then $\,6A+1\,$ is composite. So if $\,6A+1\,$ is prime then $\,A\,$ does not have that form. But there are infinitely many primes of form $\,6A+1.\, $ QED $\ \ $ $\endgroup$ – Bill Dubuque Jul 31 '16 at 23:41


Prime is infinite of form $6k+1$ where $k$ is an integer and prime cannot be written as two product.

  • $\begingroup$ ...except with 1. $\endgroup$ – Kenny Lau Jul 31 '16 at 12:45
  • $\begingroup$ @kennyLau my typing is so slow so when I go to university, I will train my self lololol $\endgroup$ – Zack Ni Jul 31 '16 at 12:46
  • 4
    $\begingroup$ Sorry...don't you need to show that there are infinitely many primes of the form $6A+1$? That is true, of course, but considerably deeper than the infinitude of primes. $\endgroup$ – lulu Jul 31 '16 at 13:12
  • $\begingroup$ @lulu that's why my word "Hint" there. It only remind the OP of how to carry out the solution rather than give a full solution and give the OP some clue about how to prove infinitely many primes of the form 6A+1. $\endgroup$ – Zack Ni Jul 31 '16 at 13:22
  • 2
    $\begingroup$ @ZackNi I think that "hint" is far too vague. Please give more details about the proof that you have in mind. $\endgroup$ – Bill Dubuque Jul 31 '16 at 15:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.