 Solution

ID: past-exam-of-the-probability-statistics-and-options-ii-course-of-the-university-of-oxford/2011/c/c2/b/ii/solution

b, c

have a common prime factor, both Jacobi symbols are zero, and the identity holds. Otherwise write

b = \prod_{i} p_{i}

c = \prod_{j} q_{j}

with multiplicity. All cross-pairs of primes are distinct. Multiplying quadratic reciprocity over these pairs gives

(\frac{b}{c}) = (\frac{c}{b}) (- 1)^{\sum_{i, j} [(p_{i} - 1) /2] [(q_{j} - 1) /2]} .

The exponent factors as a product of two sums. The parity identity from part (i) makes it congruent modulo two to

[(b - 1) /2] [(c - 1) /2]

. Therefore

(\frac{b}{c}) = (\frac{c}{b}) (- 1)^{(b - 1) (c - 1) /4} .

The same formula includes a denominator equal to one by the empty-product convention.

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