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I'm trying to prove the first supplement of the quadratic reciprocity law in its particular form:

$$-\bar{1} \text{is a square in} \mathbb{Z}/p\mathbb{Z} \iff p \equiv 1 \pmod 4$$

For the forward part, I'm thinking about applying Euler's criterion, since we know $-\bar{1}$ is a square. For the backward part, my starting point is that I have proved that $x^{p-1}=\bar{1}$ and that implies that:

$$\Bigl(x^{\frac{p-1}{2}} - \bar{1} \Bigr)\Bigl(x^{\frac{p-1}{2}} + \bar{1} \Bigr) = \bar{0}$$

RFTexas
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1 Answers1

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If $-1$ is a square mod $p$, say $-1 \equiv a^2 \pmod{p}$, then $\overline{a}$ has order $4$ in $\Bbb F_p^\times$. By Lagrange we get that $4 \mid (p-1)=|\Bbb F_p^\times|$, so $p \equiv 1\pmod{4}$

Lukas Heger
  • 20,801
  • Oh ok! I get it for the forward part! I'm new to number theory. Could you elaborate on why $\bar{a}$ has order 4? Thanks. – RFTexas Dec 03 '20 at 17:47