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!
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1Do natural numbers include zero? – Kenny Lau Jul 31 '16 at 12:43
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There are infinitely many primes – Kenny Lau Jul 31 '16 at 12:43
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@KennyLau we spontaneously conclude. – Zau Jul 31 '16 at 12:45
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1Since $,x,y\ge 1,$ 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 $\ \ $ – Bill Dubuque Jul 31 '16 at 23:41
2 Answers
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Hint:
Prime is infinite of form $6k+1$ where $k$ is an integer and prime cannot be written as two product.
Zau
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@kennyLau my typing is so slow so when I go to university, I will train my self lololol – Zau Jul 31 '16 at 12:46
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4Sorry...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. – lulu Jul 31 '16 at 13:12
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@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. – Zau Jul 31 '16 at 13:22
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2@ZackNi I think that "hint" is far too vague. Please give more details about the proof that you have in mind. – Bill Dubuque Jul 31 '16 at 15:02
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@BillDubuque uh, if I write, it will duplicate this post:http://math.stackexchange.com/questions/671820/proving-an-infinite-number-of-primes-of-the-form-6n1 – Zau Jul 31 '16 at 22:59
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Preamble: Zoo's hint has a gap of requiring that there are infinitely many primes of the form $6k+1$, which takes quite a bit of work to establish. I'm presenting a solution which doesn't require that.
- OP showed that $ A = 6xy+x+y \Leftrightarrow 6A+1 = (6x+1)(6y+1)$, where $x, y$ are positive integers.
- There are infinitely many primes of the form $6k\pm 1$.
- If there are infinitely many primes of the form $p = 6k+1$, consider $p$. The only way that $p$ factorizes into a form of $(6x+1)(6y+1)$ is via $ 1 \times p$, which leads to $xy = 0$. Hence, $A = \frac{ p-1} { 6}$ cannot be written in the given form.
- Else, there are infinitely many primes of the form $p = 6k+5$, consider $p^2$. The only way that $p^2$ factorizes into a form of $(6x+1)(6y+1)$ is via $ 1 \times p^2$, which leads to $xy = 0 $. Hence, $ A = \frac{ p^2 - 1 } { 6 } $ cannot be written in the given form.
Calvin Lin
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i.e. there are infinitely many Hilbert primes in $,1!+!6:!\Bbb N,,$ viz. $,\color{#c00}{p},$ or $,\color{#0af}{p^2},,$ for prime $,p\color{#c00}{\equiv 1},$ or $,\color{#0af}{-1}\pmod{!6}.,$ Likely a duplicate but I have no time to search. $\ \ $ – Bill Dubuque Jan 25 '25 at 23:12
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@BillDubuque I'm not sure what your point is. Are you saying that this problem should be closed as a duplicate of Hilbert primes in 6N+1, assuming that such a post exists? If so, why don't you just directly close it as a duplicate of "infinitely many primes of the form 6N+1", per the above solution? – Calvin Lin Jan 25 '25 at 23:19
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My point is primarily to emphasize that this is more conceptually expressed as primality (irreducibility) in said Hilbert numbers. I'm sure your argument appears elsewhere here, either in your language or Hilbert form. – Bill Dubuque Jan 25 '25 at 23:23