3
$\begingroup$

I have seen the past threads but I think I have another proof, though am not entirely convinced.

Suppose there are only finitely many primes $p_1, ..., p_n$ of the form $6k-1$ and then consider the number

$N = (p_1.p_2. ... .p_n)^2 - 1$

If all of the primes dividing N are of the form $6k+1$ then $N$ leaves remainder $1$ upon division by $6$. However, the RHS has remainder $1-1 = 0$ or $-1-1 = -2$ upon division by $6$, depending on whether $n$ is even or odd.

Therefore there must be at least one prime of the form $6k-1$ dividing $N$, and thus some $p_j$ divides N, and so divides $(p_1.p_2. ... .p_n)^2 - N$ which is $1$, a contradiction.

Apologies for this question if it appears unnecessary I'm just beginning number theory and trying to get to grips with it.

$\endgroup$
3
$\begingroup$

It's potentially possible that $N$ is divisible only by $2$ and $3$.

For example, if $5$ was the only prime of this form, then $5^2-1=24$ is not divisible by a prime of the form $6k-1$. Also $(5\cdot 11)^2-1 = 3024=2^4\cdot 3^3\cdot 7$. So your argument is wrong even when there is another prime divisor not equal to $2,3$.

Instead, try: $N=6p_1p_2\cdots p_n-1$.

$\endgroup$
  • $\begingroup$ Nice spot I feel very stupid now. However, I just had a thought - if we note that there are at least two primes of this form, that is 5 and 11 are of the form 6k-1 for some integers k and are prime, then does the proof not then work? Your comment is very insightful however, how did you come up with such a number if you don't mind me asking? $\endgroup$ – apple Jun 9 '15 at 14:20
  • $\begingroup$ No, as you can see, if $5,11$ were all the primes, then $55^2-1$ doesn't give you a new prime - the only prime divisors are $2,3,7$. The problem is that your $N$ is always divisible by $6$, so you can't actually say anything about the other prime factors known that $N\equiv 0\pmod 6$. @apple $\endgroup$ – Thomas Andrews Jun 9 '15 at 14:21
  • $\begingroup$ Well that's that - haha - got me again! Apologies for the stupid comments I think I need some coffee ... cheers. $\endgroup$ – apple Jun 9 '15 at 14:23
3
$\begingroup$

Suppose, that there are finite amount of primes with the form $6k-1$. Then try $N = 6(p_1 . . . p_n) − 1$, it has the form $6K+5$. $2$ and $3$ can't divide that number, so the divisors have the form of $6k+1$ or $6k+5$. Suppose, that all of them have the form $6k+1$. In that case, $N$ has that form too, which is contradiction. Therefore, $N$ has a divisor with form $6k+5$, but that number can't be in our list, we found a new prime, contradiction $\rightarrow$ the set is infinite. (Or just simply use Dirichlet's theorem :) )

$\endgroup$
  • 2
    $\begingroup$ The whole point of this kind of problem is to prove special cases of Dirichlet using elementary means, of course. Basically, it is a start on the road to Dirichlet. $\endgroup$ – Thomas Andrews Jun 9 '15 at 14:22
  • $\begingroup$ I proved it in an elemantary way aswell, just thought it could be interesting. :) $\endgroup$ – Atvin Jun 9 '15 at 14:23

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.