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