5
$\begingroup$

Theorem. there are infinitely many primes of the form $6k+1$.

I've just proved that there are infinitely many primes of the form $6k+1$.

Could you please check my proof?

At first, I proved that

$$ for \ \ p:prime, \ p \ge 5 \\\ \\ \left(\frac{-3}{p}\right)= \begin{cases} 1,& \ p \equiv 1 \pmod 6 \\ -1, & \ p \equiv 5 \pmod 6 \end{cases} $$ (I will use this lemma for proving Theorem.)

$$$$ NOW assume that there are finite primes of the form $6k+1$.

then we can say $ \ p_1, \ p_2, \cdots, p_k \ $ : all the primes of the form $6k+1$.

Let

$$ n=(p_1\cdot p_2\cdots\ p_k)^2 +3, $$

then (by Fundamental Theorem of Arithmetic) there is prime factor $p$ of $n$.

$$ $$ Id est, $$(p_1\cdot p_2\cdots\ p_k)^2 \equiv -3 \pmod p $$

So, $$p \equiv 1 \pmod 6$$

Thus $$p=p_i \ for \ some \ i=1, \cdots , k$$

This time

$$p=p_i \ \ \ divides \ \ (p_1\cdot p_2\cdots\ p_k)^2 \\ p=p_i \ \ \ can't \ \ divide \ \ 3 \\ $$

So, $$p=p_i \ \ can't \ \ divide \ \ n. \\$$

It is contradiction with "$p$ is prime factor of $n$"

$\therefore \ $ There are infinitely many primes of the form $6k+1$.

$Q.E.D.$

$$ $$ What about my proof?

After proving, I saw someone's proof,

BUT he set $$ n=(2p_1\cdot p_2\cdots\ p_k)^2 +3 $$

I don't know why he set $ n=(2p_1\cdot p_2\cdots\ p_k)^2 +3 $, instead of $ n=(p_1\cdot p_2\cdots\ p_k)^2 +3 $.

Is my proof wrong?

$$ $$ Please give me some hand. Thanks in advance.

$\endgroup$
  • 1
    $\begingroup$ Your $n$ is congruent to $4$ mod $6$. $\endgroup$ – Ted Shifrin Jun 8 '14 at 15:12
  • $\begingroup$ @TedShifrin, Ah.. You mean if $n \equiv 4 \pmod 6$, then 2 is prime factor of n... Thus this is wrong, right?? $\endgroup$ – user143993 Jun 8 '14 at 15:19
  • $\begingroup$ I wouldn't say wrong, but incomplete. Note that you can also solve the issue by noting that $x^2\equiv1\pmod8$ for odd $x$, which means your $n$ cannot be a power of $2$, hence has an odd prime divisor. $\endgroup$ – barto Jun 8 '14 at 15:24
  • $\begingroup$ @barto, Ummm $x^2\equiv1\pmod8$ iff $x \equiv1, 3, 5, 7\pmod8$ iff $x \equiv1 \pmod2$.. How does it go "It means your n cannot be a power of 2" Could you explain more , please? $\endgroup$ – user143993 Jun 8 '14 at 15:36
3
$\begingroup$

There is a minor issue. Not every prime divisor of your $n$ is necessarily of the form $6k+1$. (Because, as you noted, the theorem with the Légendre-symbol is only valid for $p\geqslant5$.) You would have to argue that $2$ and $3$ do not divide $n$.
This can be fixed by letting $n=(2p_1,\cdots,p_k)^2+3$ as you noted, or alternatively by observing that $x^2\equiv1\pmod8$ for odd $x$, which means your $n$ cannot be a power of $2$, hence has an odd prime divisor.

$\endgroup$
  • 1
    $\begingroup$ THANKS a lot for answering. I agree "argue that 2 do not divide n", BUT why I argue that '3 do not divide n'? $n=(p_1\cdot p_2\cdots\ p_k)^2 +3$ is the form 6k+1+3=6k+4 .. isn't it? Do I get wrong thing? Give me some advice, please:-) $\endgroup$ – user143993 Jun 8 '14 at 15:26
  • $\begingroup$ It is indeed obvious that $3$ does not divide $n$, but it's worth noting because it's still a crucial step in your proof. $\endgroup$ – barto Jun 8 '14 at 15:28
  • $\begingroup$ THANK you sincerely. :-) AND Just to be sure...I'm sorry if it offended you. $\endgroup$ – user143993 Jun 8 '14 at 15:33
  • $\begingroup$ Not at all ;-), you're welcome. $\endgroup$ – barto Jun 8 '14 at 15:34
  • 1
    $\begingroup$ Well done! Apologies for not answering your questions immediately, I wasn't at home last evening. $\endgroup$ – barto Jun 9 '14 at 7:03

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.