Proof
(related to Theorem: First Supplementary Law to the Quadratic Reciprocity Law)
- By hypothesis, $p > 2$ is an odd prime number.
- By the Euler's criterion for quadratic residues we have the congruence $\left(\frac {-1}p\right)(p)\equiv (-1)^{\frac{p-1}{2}}(p).$
- Now, the difference $\left(\frac {-1}p\right)-(-1)^{\frac{p-1}{2}}$ equals one of the integers $0$, $-2$ or $2.$
- Because $p$ is odd, only $0$ is possible.
- It remains to be shown that the identity $\left(\frac {-1}p\right)=(-1)^{\frac{p-1}{2}}$ is equivalent to
$$\left(\frac {-1}p\right)=\begin{cases}1&\text{if }p\equiv 1\mod 4,\\-1&\text{if }p\equiv -1\mod 4.\end{cases}$$
- If $p(4)\equiv 1(4)$ then $p=4k+1$ for some $k\in\mathbb Z.$
* Then $\frac{p-1}2=2k$ is even and $(-1)^{2k}=1.$
* Thus, in this case $\left(\frac {-1}p\right)=1,$ as required.
- If $p(4)\equiv -1(4)$ then $p=4k-1$ for some $k\in\mathbb Z.$
* Then $\frac{p-1}2=2k-1$ is odd and $(-1)^{2k-1}=-1.$
* Thus, in this case $\left(\frac {-1}p\right)=-1,$ as required.
- Since $p$ is odd, no other congruence classes modulo $4$ except $\pm 1$ exist.
∎
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References
Bibliography
- Landau, Edmund: "Vorlesungen über Zahlentheorie, Aus der Elementaren Zahlentheorie", S. Hirzel, Leipzig, 1927
