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Question. Show $(-2/p)$ equals $1$ when $p\equiv 1,3\bmod 8$ and $-1$ when $p\equiv 5,7\bmod 8$.

So using the multiplicativity of the symbol; we have $$\Big(\frac{-1}{p}\Big)\Big(\frac{2}{p}\Big),$$ and I know the rules for $(2/p)$ and $(-1/p)$. But then what? Usually I use Chinese remainder theorem to try and obtain a bigger modulus but I can't here since $8$ and $4$ aren't co-prime.

Edit. So to combine the congruences we have $$\Big(\frac{-2}{p}\Big)=1,$$ when $p\equiv 1\bmod 8$ and $p\equiv 3\bmod 8$ since we can't have $p$ being congruent to both $7$ and $5$ modulo $8$ at the same time.

Similarly, we have $$\Big(\frac{-2}{p}\Big)=-1,$$ when $p\equiv 7\bmod 8$ or $5\bmod 8$ because this is where the overlaps are when comparing $-1$ and $1$ from the results gained from $(2/p)$ and $(-1/p)$.

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The proof that

  • $\Big(\frac{-1}{p}\Big) = \begin{cases} 1 & \text{if } p \equiv 1 \pmod 4 \\ -1 & \text{if } p \equiv 3 \pmod 4 \end{cases}$ and
  • $\Big(\frac{2}{p}\Big) = \begin{cases} 1 & \text{if } p \equiv \pm1 \pmod 8 \\ -1 & \text{if } p \equiv \pm3 \pmod 8 \end{cases}$

are classic. The first bullet point becomes $\Big(\frac{-1}{p}\Big) = \begin{cases} 1 & \text{if } p \equiv 1,5 \pmod 8 \\ -1 & \text{if } p \equiv 3,7 \pmod 8 \end{cases}$.

Multiplying the two Legendre symbols gives

$$\Big(\frac{-2}{p}\Big) = \begin{cases} 1 & \text{if } p \equiv 1,3 \pmod 8 \\ -1 & \text{if } p \equiv 5,7 \pmod 8 \end{cases}.$$

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  • $\begingroup$ I had quite a bit of trouble doing the multiplication of the symbols; so if you could check my edit and see whether I've done it right I'd be very grateful. $\endgroup$ – thesmallprint Apr 26 '18 at 17:11
  • $\begingroup$ @thesmallprint Sorry, English isn't my first language and I don't get what's overlapping. I suggest using begin{cases} ... \end{cases} to make things clear. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Apr 26 '18 at 17:43
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    $\begingroup$ I think I understand the answer now. Thank you very much :) $\endgroup$ – thesmallprint Apr 26 '18 at 18:41

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