# Thread: For all primes greater than 3, p = 6ką1?

1. ## For 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

2. 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

3. ## Ok Understood

quite easy

4. 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 $k$.

5. ## another question

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

6. 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.

7. ## cool

coincidence

8. 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.

9. ## ok

i'll be happy!

10. 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.

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

12. 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