is it true?

what is the proof?

I prefer a hint if it ain't difficult to prove.

tnx :)

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- May 28th 2006, 03:44 AMsrulikbdFor all primes greater than 3, p = 6ką1?
is it true?

what is the proof?

I prefer a hint if it ain't difficult to prove.

tnx :) - May 28th 2006, 04:05 AMCaptainBlackQuote:

Originally Posted by**srulikbd**

Let $\displaystyle p>3$ be a prime. Consider the possible values of the remainder

when $\displaystyle p$ is divided by $\displaystyle 6$. As $\displaystyle p>3$ is prime the remainder cannot

be divisible by any factor of $\displaystyle 6$.

RonL - May 28th 2006, 05:30 AMsrulikbdOk Understood
quite easy :)

- May 28th 2006, 09:55 AMThePerfectHacker
Or to say it another way.

Any number must have form,

$\displaystyle \begin{array}{c}6k\\6k+1\\6k+2\\6k+3\\6k+4\\6k+5$

Note that,

$\displaystyle \begin{array}{c}6k\\6k+2\\6k+4$

are all divisible by 2 and $\displaystyle p>2$ thus not a prime.

And that,

$\displaystyle 6k+3$ is divisble by 3 and $\displaystyle p>3$ thus not a prime.

Hence the only possibilites are,

$\displaystyle 6k+1$

or,

$\displaystyle 6k+5=6k+6-1=6(k+1)-1$

Thus,

$\displaystyle 6k\pm 1$ for some integer $\displaystyle k$. - May 29th 2006, 12:30 AMsrulikbdanother question
how do I prove that there are infinite primes that have the form xk+y, y<x, and share no common factors?

- May 29th 2006, 10:19 AMThePerfectHackerQuote:

Originally Posted by**srulikbd**

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. - May 31st 2006, 09:28 AMsrulikbdcool
coincidence :)

- May 31st 2006, 02:58 PMThePerfectHacker
I can present a proof to why are there infinitely many primes of form $\displaystyle 4k+1 \mbox{ and }4k+3$ if you really wish to know.

- May 31st 2006, 11:45 PMsrulikbdok
i'll be happy!

- Jun 1st 2006, 09:02 AMThePerfectHackerQuote:

Originally Posted by**srulikbd**

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

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

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

*Proof:*Assume there are finitely many primes of form $\displaystyle 4k+3$ call them $\displaystyle \{ p_1,...,p_s} \}$. Then form an integer,

$\displaystyle n=4p_1p_2\cdot ...\cdot p_s-1$

Prime factorize $\displaystyle n$ as,

$\displaystyle n=q_1q_2\cdot ...\cdot q_r$

Note that $\displaystyle 2\not | n$ because it is odd. Thus, $\displaystyle q_i\not =2$ for any $\displaystyle 1\leq i\leq r$.

We cannot have that all the primes factors of $\displaystyle n$ are of the form $\displaystyle 4k+1$ for that would imply that $\displaystyle n$ has form $\displaystyle 4k+1$, which is does not because $\displaystyle 4k-1=4(k-1)+3$ by the lemma. Thus one of its factors $\displaystyle r_i$ has form $\displaystyle 4k+3$. Because an odd prime has one of two forms $\displaystyle 4k+1,4k+3$. But then $\displaystyle r_i|n$ thus, $\displaystyle r_i|(4p_1p_2\cdot ...\cdot p_s-1)$ thus, $\displaystyle r_i|1$, because $\displaystyle r_i$ is found among $\displaystyle p_1,...,p_s$ thus an impossibility. Q.E.D. - Jun 2nd 2006, 12:04 AMsrulikbd
I didn't understand the end-what does | means?

and whaat does X means? (2X...) does it means doesn't divide?

tnx :) - Jun 2nd 2006, 04:25 AMCaptainBlackQuote:

Originally Posted by**srulikbd**

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

RonL - Jun 2nd 2006, 06:10 AMsrulikbdtnx both of u now I understand
:)