 Solution

ID: past-exam-of-the-options-ii-course-of-the-university-of-oxford/2011/c/c2/a/iv/solution

p

divides

a

b

, both sides of the claimed multiplicativity identity are zero. Otherwise the Euler criterion gives

(\frac{ab}{p}) \equiv (ab)^{(p - 1) /2} \equiv a^{(p - 1) /2} b^{(p - 1) /2} \equiv (\frac{a}{p}) (\frac{b}{p}) (mod p) .

Each side is

1

- 1

. Since

p

is odd, these two integers represent different residue classes. Hence the modular congruence gives equality:

(\frac{ab}{p}) = (\frac{a}{p}) (\frac{b}{p}) .

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