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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: $\left ( \frac{2}{p} \right )=(-1)^{\frac{p^2-1}{8}}$ ?

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1 Answer 1

up vote 5 down vote accepted

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.

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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 $p \mid p_1 \cdot p_2 \cdots p_k \Rightarrow p \mid (p_1 \cdots p_k)^2 \text{ and as } p \mid N \Rightarrow p \mid 2 \text{ that is a contradiction.}$ –  evinda Jun 22 at 23:06
    
I understood it now...thank you very much!!!! –  evinda Jun 22 at 23:27

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