 Solution

ID: past-exam-of-the-short-options-course-of-the-university-of-oxford/2021/1/c/solution

Write

P = \prod_{i} p_{i}

Q = \prod_{j} q_{j}

, counting repetitions. Since

P, Q

are coprime, every pair

p_{i}, q_{j}

(\frac{p _{i}}{q _{j}}) = (- 1)^{((p_{i} - 1) /2) ((q_{j} - 1) /2)} (\frac{q _{j}}{p _{i}}) .

Multiplying over all

i, j

(\sum_{i} \frac{p _{i} - 1}{2}) (\sum_{j} \frac{q _{j} - 1}{2}) \equiv \frac{P - 1}{2} \frac{Q - 1}{2} (mod 2),

by the odd-product parity identity proved in (b)(i). Thus the Jacobi reciprocity law is

(\frac{P}{Q}) = (- 1)^{(P - 1) (Q - 1) /4} (\frac{Q}{P}) .

If either integer is

1

, its factorization is empty and both sides equal

1

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