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Math Help - Quadratic Reciprocity

  1. #1
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    Quadratic Reciprocity

    Can anyone point me in the direction of a proof for the generalisation of the Quadratic Reciprocity Law to the Jacobi Symbol?

    Alternatively give me some pointers on how to prove it myself.
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  2. #2
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    Quote Originally Posted by Kiwi_Dave View Post
    Can anyone point me in the direction of a proof for the generalisation of the Quadratic Reciprocity Law to the Jacobi Symbol?

    Alternatively give me some pointers on how to prove it myself.
    If a,b be two odd positive odd numbers >1.

    We want to prove (a/b)(b/a) = (-1)^{(a-1)/2\cdot (b-1)/2}

    Now let a=p_1\cdot .... \cdot p_n and b=q_1\cdot ... \cdot q_m.

    Therefore, (a/b)(b/a) = \prod_{i=1}^n \prod_{j=1}^m (p_i/q_j)(q_j/p_i) = (-1)^{\sum_i \sum_j (q_i-1)/2\cdot(p_j-1)/2}

    It remains to prove,
    \sum_{i} \sum_j (q_i -1)/2\cdot (p_j -1)/2 \equiv (p_1...p_n -1)/2\cdot (q_1...q_m -1)/2 = (a-1)/2\cdot (b-1)/2
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    Quote Originally Posted by ThePerfectHacker View Post
    It remains to prove,
    \sum_{i} \sum_j (q_i -1)/2\cdot (p_j -1)/2 \equiv (p_1...p_n -1)/2\cdot (q_1...q_m -1)/2 = (a-1)/2\cdot (b-1)/2

    Thank's for the help. I can see the truth of the final congruence and will write my argument down. But is there a more susinct argument?

    Each summand on the LHS contributes to the sum (mod 4) if none of the following statements are true:

    q_i\equiv 1 (mod 4)
    p_j\equiv 1 (mod 4)

    That is they contribute iff
    p_j\equiv q_i\equiv 3 (mod 4)

    And the sum is equal to the number of contributions (mod 2). The number of p_j such that p_j\equiv 3 (mod 4) and the number of q_i such that q_i\equiv 3 (mod 4) must both be odd or the sum is zero (mod 2).

    Now

    (p_1...p_n)\equiv 3^r where r is the number of p_j that equal 3 (mod 4). Therefore, (p_1...p_n) \equiv 1 if there are an even number of p_j that equal 3 (mod 4) and (p_1...p_n) \equiv 3 otherwise. In conclusion the sum is only non zero if (p_1...p_n) \equiv 3.

    Similarly the sum is only non zero if (q_1...q_m) \equiv 3.

    Now (p_1...p_n-1)\cdot(p_1...p_n-1)/4\equiv 0 if (p_1...p_n)\equiv 1 or (q_1...q_m)\equiv 1. Conversely (p_1...p_n-1)\cdot(p_1...p_n-1)/4\equiv 1 (mod 2) if (p_1...p_n) \equiv 3 and (q_1...q_m) \equiv 3.
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  4. #4
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    If A,B are odd then (AB-1)/2 \equiv (A-1)/2 + (B-1)/2 (\bmod 2). Proof: (A-1)(B-1)\equiv 0(\bmod 4) thus AB - A - B + 1 \equiv 0 (\bmod 4) and so (AB-1) \equiv (A-1) + (B-1) (\bmod 4). Thus, (AB-1)/2 \equiv (A-1)/2 + (B-1)/2 (\bmod 2).

    Now generalize this by induction for A_1,...,A_s i.e. \sum_i (A_i -1)/2 \equiv (A_1...A_s-1)/2 (\bmod 2).

    Thus, \sum_i \sum_j (p_j-1)/2 \cdot (q_i -1)/2 \equiv (a-1)/2 \cdot \sum_j (p_j-1)/2 \equiv (a-1)/2\cdot (b-1)/2 (\bmod 2).
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