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

ii

Solution

g cd (b, c) > 1

, both Jacobi symbols vanish, so the formula holds. Otherwise write

b = \prod_{i} p_{i}

and

c = \prod_{j} q_{j}

(\frac{b}{c}) = \prod_{i, j} (\frac{p _{i}}{q _{j}}) = \prod_{i, j} (\frac{q _{j}}{p _{i}}) (- 1)^{\sum_{i, j} (p_{i} - 1) (q_{j} - 1) /4} .

(\sum_{i} \frac{p _{i} - 1}{2}) (\sum_{j} \frac{q _{j} - 1}{2}) \equiv \frac{b - 1}{2} \frac{c - 1}{2} (mod 2)

by the parity identity from part (i). The product of the reversed Legendre symbols is

(c / b)

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

This also covers

b = 1

c = 1

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