(\frac{a}{p}) = ⎩ ⎨ ⎧ 0, 1, - 1, p ∣ a, a is a nonzero square modulo p, a is not a square modulo p .

ii

Solution

The Euler criterion states that for

p ∤ a

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

The same modular congruence holds for

p ∣ a

, with both sides zero.

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} .

Again set

ϵ = (a / p) \in {1, - 1}

. The Euler criterion applied to

ϵ

gives

χ (ϵ) = ϵ^{(p - 1) /2}

, as an equality of signs. If

p \equiv 1 (mod 4)

, the exponent is even and the result is

1

; if

p \equiv 3 (mod 4)

, it is odd and the result is

ϵ

. Hence

\frac{( \frac{a}{p} )}{p} = {1, (a / p), p \equiv 1 (mod 4), p \equiv 3 (mod 4) .

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