 Solution

ID: past-exam-of-the-options-ii-course-of-the-university-of-oxford/2012/c/c2/a/iii/solution

(\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) .

If either factor is divisible by

p

, both asserted values are zero. Otherwise the two values lie in

{1, - 1}

, whose elements are distinct modulo an odd prime. The modular congruence is therefore equality:

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

Finally,

a \equiv b (mod p)

means they give the same residue class and thus the same set of square-congruence solutions. Divisibility by

p

is also identical. Hence

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

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