For all primes greater than 3, p = 6k±1?

**srulikbd** · May 28th 2006, 03:44 AM

is it true?
what is the proof?
I prefer a hint if it ain't difficult to prove.
tnx

**CaptainBlack** · May 28th 2006, 04:05 AM

Originally Posted by srulikbd

is it true?
what is the proof?
I prefer a hint if it ain't difficult to prove.
tnx

Hint(?):

Let $p>3$ be a prime. Consider the possible values of the remainder
when $p$ is divided by $6$ . As $p>3$ is prime the remainder cannot
be divisible by any factor of $6$ .

RonL

**srulikbd** · May 28th 2006, 05:30 AM

quite easy

**ThePerfectHacker** · May 28th 2006, 09:55 AM

Or to say it another way.
Any number must have form,
$\begin{array}{c}6k\\6k+1\\6k+2\\6k+3\\6k+4\\6k+5$
Note that,
$\begin{array}{c}6k\\6k+2\\6k+4$
are all divisible by 2 and $p>2$ thus not a prime.

And that,
$6k+3$ is divisble by 3 and $p>3$ thus not a prime.
Hence the only possibilites are,
$6k+1$
or,
$6k+5=6k+6-1=6(k+1)-1$
Thus,
$6k\pm 1$ for some integer $k$ .

**srulikbd** · May 29th 2006, 12:30 AM

how do I prove that there are infinite primes that have the form xk+y, y<x, and share no common factors?

**ThePerfectHacker** · May 29th 2006, 10:19 AM

Originally Posted by srulikbd

how do I prove that there are infinite primes that have the form xk+y, y<x, and share no common factors?

Whoa!!!
That is too complicated I believe the proof use's funtional analysis. It was first proven by Dirichelt (my avatar

).

In fact in the book "Introduction to Theory of Numbers" by Hardy and Wright. Which contains elementary,algebraic and analytic number theory; proves all theorems in the book (even the prime number theorem) except this theorem! That is how complicated it is.

**srulikbd** · May 31st 2006, 09:28 AM

coincidence

**ThePerfectHacker** · May 31st 2006, 02:58 PM

I can present a proof to why are there infinitely many primes of form $4k+1 \mbox{ and }4k+3$ if you really wish to know.

**srulikbd** · May 31st 2006, 11:45 PM

i'll be happy!

**ThePerfectHacker** · June 1st 2006, 09:02 AM

Originally Posted by srulikbd

i'll be happy!

Lemma: The product of integers of the form $4k+1$ is again an integer of the form $4k+1$ .

Proof:: If $a=4k+1$ and $b=4j+1$ then, $ab=(4k+1)(4j+1)=4(4kj+k+j)+1$ . Q.E.D.

Theorem: There are infinitely many primes of form $4k+3$ .

Proof: Assume there are finitely many primes of form $4k+3$ call them $\{ p_1,...,p_s} \}$ . Then form an integer,
$n=4p_1p_2\cdot ...\cdot p_s-1$
Prime factorize $n$ as,
$n=q_1q_2\cdot ...\cdot q_r$
Note that $2\not | n$ because it is odd. Thus, $q_i\not =2$ for any $1\leq i\leq r$ .
We cannot have that all the primes factors of $n$ are of the form $4k+1$ for that would imply that $n$ has form $4k+1$ , which is does not because $4k-1=4(k-1)+3$ by the lemma. Thus one of its factors $r_i$ has form $4k+3$ . Because an odd prime has one of two forms $4k+1,4k+3$ . But then $r_i|n$ thus, $r_i|(4p_1p_2\cdot ...\cdot p_s-1)$ thus, $r_i|1$ , because $r_i$ is found among $p_1,...,p_s$ thus an impossibility. Q.E.D.

**srulikbd** · June 2nd 2006, 12:04 AM

I didn't understand the end-what does | means?
and whaat does X means? (2X...) does it means doesn't divide?
tnx

**CaptainBlack** · June 2nd 2006, 04:25 AM

Originally Posted by srulikbd

I didn't understand the end-what does | means?
and whaat does X means? (2X...) does it means doesn't divide?
tnx

$a|b$ is to be read as $a$ divides $b$ .

$a \not| \ b$ is to be read as $a$ does not divide $b$ .

RonL

**srulikbd** · June 2nd 2006, 06:10 AM

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For all primes greater than 3, p = 6k±1?

Ok Understood

another question

cool

ok

tnx both of u now I understand

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