Let p be a prime number. Prove that ($\frac{-2}{p}$)= 1 if and only if p is of the form $8k + 1$ or $8k + 3$, and then from there conclude that there are infinitely many primes of the form $8k + 3$ (see numbers $N = (p_1p_2 ··· p_n)^2 + 2$).

I've seen this proof Prove that there are infinitely many primes of the form $8k + 3$ but still dont know what to do with the first part of the assigment. I've tried this

$(\frac{-2}p)=1 \Rightarrow$ $(\frac{-2}p)=(\frac{-1}p)(\frac{2}p)=(-1)^{\frac{p-1}2}(-1)^{\lfloor\frac{p^2+1}8\rfloor}=1 \Rightarrow$ $\frac{p-1}2+\lfloor\frac{p^2+1}8\rfloor=2k$
but this is probably not the way to prove that. Can anyone please help me?

  • $\begingroup$ Is that bracket floor function?? As if k=100 then it will become 0 $\endgroup$ Jan 10 '16 at 15:10
  • $\begingroup$ Nope @ArchisWelankar $\endgroup$
    – mod
    Jan 10 '16 at 15:11
  • $\begingroup$ Where do you get $\frac{p+1}{6}$ from? $\frac{p^2-1}{8}$ is relevant. $\endgroup$ Jan 10 '16 at 15:13

This follows from the fact that $$\left (\frac{2}{p} \right )=(-1)^{\frac{p^2-1}{8}}$$ . We also know that $$\left (\frac{-1}{p} \right )=(-1)^{\frac{p-1}{2}}$$ so using them together :

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

Now $\left (\frac{-2}{p} \right )=1$ when $\frac{p^2-1}{8}+\frac{p-1}{2}$ is even .

This is equivalent with :

$$\frac{(p-1)(p+5)}{8}$$ being even .

Both factors are even but at most one is divisible with $4$ because their difference is $6$ . This means that one is divisible with $8$ which leads to the two possible solutions :

$$p \equiv 1 \pmod{8}$$ and $$p \equiv 3 \pmod{8}$$


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