이차 가우스 합 (편집) (8)

== 이차 상반법칙 ==
아래와 같은 방법으로 이차 상반법칙을 증명할 수 있다.

<math>g(a) = \zeta^{a\cdot0^{2}}+\zeta^{a\cdot1^{2}}+...+\zeta^{a(p-1)^{2}} =\sum_{n=1}^{p-1}\left(\frac{n}{p}\right)\zeta^{an} </math> 이라고 하자. <math>p</math>와 서로소인 홀소수 <math>q</math>에 대해 <math>\{g(1)\}^{q}</math>를 두 가지 방법으로 정리하자.

<math>\textup{(i)}\ \{g(1)\}^{q} =  \{ \sum_{n=1}^{p-1}\left(\frac{n}{p}\right)\zeta^{n} \} ^{q}</math>
* <math> \equiv  \sum_{n=1}^{p-1}\{\left(\frac{n}{p}\right)\zeta^{n}\} ^{q} =  \sum_{n=1}^{p-1}\left(\frac{n}{p}\right) ^{q}\zeta^{nq} =  \sum_{n=1}^{p-1}\left(\frac{n}{p}\right) \zeta^{nq} </math>
* <math>= \sum_{n=1}^{p-1}\left(\frac{q}{p}\right)\left(\frac{q}{p}\right)\left(\frac{n}{p}\right) \zeta^{nq} =  \left(\frac{q}{p}\right)\sum_{n=1}^{p-1}\left(\frac{nq}{p}\right)\zeta^{nq}</math>
* <math>=  \left(\frac{q}{p}\right)\sum_{l=1}^{p-1}\left(\frac{l}{p}\right)\zeta^{l} =  \left(\frac{q}{p}\right)g(1) \;  (mod\; q)</math>

<math>\textup{(ii)}\ \{g(1)\}^{q} =  g(1)\cdot[\{g(1)\}^{2}]^{ \frac{q-1}{2} }</math> 
* <math> =  g(1)\cdot\{(-1)^{\frac{p-1}{2}}p\}^{ \frac{q-1}{2} }</math> 
* <math> =  g(1)\cdot(-1)^{\frac{p-1}{2}\frac{q-1}{2}}p^{ \frac{q-1}{2} }</math> 
* <math> \equiv  g(1)\cdot(-1)^{\frac{p-1}{2}\frac{q-1}{2}}\left(\frac{p}{q}\right)\;  (mod\; q)</math>

<math>\textup{(i), (ii)}</math>에서

<math>\left(\frac{q}{p}\right)g(1)  \equiv  g(1)\cdot(-1)^{\frac{p-1}{2}\frac{q-1}{2}}\left(\frac{p}{q}\right) \;  (mod\; q)  </math>

<math> \left(\frac{q}{p}\right)  \equiv  (-1)^{\frac{p-1}{2}\frac{q-1}{2}}\left(\frac{p}{q}\right) \;  (mod\; q)  </math>

<math> \left(\frac{p}{q}\right)\left(\frac{q}{p}\right)  \equiv  (-1)^{\frac{p-1}{2}\frac{q-1}{2}} \;  (mod\; q)  </math>

* <math> \therefore \left(\frac{p}{q}\right)\left(\frac{q}{p}\right) = (-1)^{\frac{p-1}{2}\frac{q-1}{2}}\;  \;  (\because  \left(\frac{p}{q}\right), \left(\frac{q}{p}\right), (-1)^{\frac{p-1}{2}\frac{q-1}{2}} \in \{1, -1\})</math>

따라서 이차 상반법칙이 증명된다.

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