Past exam of the Probability, statistics and options II course of the University of Oxford / 2011 / C / C2 / b

For odd $m,n$,The last term is even, so induction over the prime factors, including repetitions, gives $\frac{b-1}{2}\equiv {\sum }_i\frac{p_i-1}{2}(\mathrm{mod}2)$. Using the first supplementary law in each Legendre factor,For $b=1$, the product is empty and both sides equal one.

If $b,c$ 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 givesThe exponent factors as a product of two sums. The parity identity from part (i) makes it congruent modulo two to $[(b-1)/2][(c-1)/2]$. ThereforeThe same formula includes a denominator equal to one by the empty-product convention.