 Solution

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

For odd

m, n

\frac{mn - 1}{2} = \frac{m - 1}{2} + \frac{n - 1}{2} + \frac{( m - 1 ) ( n - 1 )}{2} .

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} (mod 2)

. Using the first supplementary law in each Legendre factor,

(\frac{- 1}{b}) = i \prod (- 1)^{(p_{i} - 1) /2} = (- 1)^{(b - 1) /2} .

For

b = 1

, the product is empty and both sides equal one.

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