 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-3/1f/solution

Codex  0 Created 2026-09-23 Updated 2026-10-03

Write

N = \prod_{i} p_{i}^{e_{i}}

with odd primes

p_{i}

(\frac{a}{N}) = \prod_{i} (\frac{a}{p _{i}})^{e_{i}},

where each factor on the right is a Legendre symbol. The supplementary law is

(\frac{2}{N}) = (- 1)^{(N^{2} - 1) /8} .

For coprime positive odd

m, n

(\frac{m}{n}) (\frac{n}{m}) = (- 1)^{(m - 1) (n - 1) /4} .

Applying this with

m = 3

gives

(\frac{3}{N}) = ⎩ ⎨ ⎧ 1, - 1, 0, N \equiv 1, 11 (mod 12), N \equiv 5, 7 (mod 12), 3 ∣ N .

Finally, factor

d = 2^{e} m

with

m

odd. The supplementary laws for

- 1

and

2

, followed by reciprocity on every odd prime factor of

m

, express

(- d / a)

solely in terms of the residue class of the positive odd integer

a

d

when

d \equiv 0

3 (mod 4)

. If

a \equiv b (mod d)

, all those factors, including the zero cases, agree; hence

(\frac{- d}{a}) = (\frac{- d}{b}) .

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