 Solution

ID: past-exam-of-the-options-i-course-of-the-university-of-oxford/2010/c/c1/iii/solution

Write

χ (a) = (a / p)

. If

p ∣ a

, the claimed identity has both sides zero. Otherwise let

ϵ = χ (a) \in {1, - 1}

. Applying the Euler criterion gives

χ (ϵ a) \equiv (ϵ a)^{(p - 1) /2} \equiv ϵ^{(p - 1) /2} χ (a) = ϵ^{(p + 1) /2} (mod p) .

Both sides are in

{1, - 1}

, which are distinct residue classes because

p

is odd. Therefore the modular congruence is an equality:

\frac{( \frac{a}{p} ) a}{p} = (\frac{a}{p})^{(p + 1) /2} .

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