3
$\begingroup$

Using the fact that $$\left ( \frac{2}{p} \right )=(-1)^{\frac{p^2-1}{8}}$$ for each prime $p>2$,prove that there infinitely many primes of the form $8k-1$.

I thought that we could I assume that there is a finite number of primes of the form $8k-1$: $p_1,p_2 \dots ,p_k$

Could we maybe set $N=8p_1p_2 \cdots p_k-1 >1$

Then $N$ has a prime divisor $p$.$p$ can be of the form $8n+1,8n+3,8n+5 \text{ or } 8n+7$..

How could I continue?? Also...how can I use this: $\left ( \frac{2}{p} \right )=(-1)^{\frac{p^2-1}{8}}$ ?

$\endgroup$
7
$\begingroup$

Let $p_1,p_2, \ldots, p_k$ be the list of ALL primes of the form $8s+7$. Let $$N=(p_1p_2 \dotsb p_k)^2-2.$$ Note that $N \equiv 7 \pmod{8}$ and is odd. If $p$ is a prime that divides $N$, then $$(p_1p_2 \dotsb p_k)^2 \equiv 2 \pmod{p}.$$ Thus $$\left(\frac{2}{p}\right)=1.$$ Thus $p \equiv \pm 1 \pmod{8}$.

So all primes that divide $N$ must be of the from $8s+1$ or $8s+7$. But not all of them can be of the form $8s+1$ (ask why???)

So there must be one of the form $q=8s+7$. Now see if you can proceed from here.

$\endgroup$
  • $\begingroup$ Anurag A: Why $N \equiv 7 \pmod 8$? We know that $p_1, \dots , p_k \equiv 7 \pmod 8$,but why also the difference $(p_1 \cdots p_k)^2-2 \equiv 7 \pmod 8$ ?? $$$$ If $N$ has a prime divisor $p$ of the form $8s-1$,$p$ must be of of $p_1, \dots, p_k$,so $p \mid p_1 \cdot p_2 \cdots p_k \Rightarrow p \mid (p_1 \cdots p_k)^2 \text{ and as } p \mid N \Rightarrow p \mid 2 \text{ that is a contradiction.}$ $\endgroup$ – evinda Jun 22 '14 at 23:06
  • $\begingroup$ I understood it now...thank you very much!!!! $\endgroup$ – evinda Jun 22 '14 at 23:27
  • $\begingroup$ @Anurag , Please fill in the why? part so that answer is complete $\endgroup$ – Breaking Benjamin Jun 16 at 10:30

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.