 Solution

ID: past-exam-of-the-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}

is even. Induction over the prime factors of

b

, repeated according to multiplicity, therefore gives

\frac{b - 1}{2} \equiv \sum_{j = 1}^{r} \frac{p _{j} - 1}{2} (mod 2) .

Using the supplementary law for each Legendre symbol in the definition of the Jacobi symbol,

(\frac{- 1}{b}) = \prod_{j} (- 1)^{(p_{j} - 1) /2} = (- 1)^{(b - 1) /2} .

For

b = 1

, the empty product and the displayed expression are both

1

 New to topics? Read the docs here!