 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ii/paper-2/1g/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

For an odd prime

p

(\frac{a}{p}) = ⎩ ⎨ ⎧ 0, 1, - 1, p ∣ a, a \neq \equiv 0 (mod p) and a is a quadratic residue, a is a quadratic nonresidue .

Gauss lemma says that if

p ∤ a

and

N

of the least positive residues of

a, 2 a, \dots, \frac{p - 1}{2} a

exceed

p /2

, then

(a / p) = (- 1)^{N}

For

a = 2

, none of

2, 4, \dots, p - 1

needs reduction modulo

p

, and the terms exceeding

p /2

are those with

j > p /4

. Hence

N = \frac{p - 1}{2} - ⌊ \frac{p}{4} ⌋ .

Checking

p \equiv 1, 3, 5, 7 (mod 8)

shows that

N

has the same parity as

(p^{2} - 1) /8

. Therefore

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

149 \equiv 1 (mod 4)

(\frac{105}{149}) = (\frac{3}{149}) (\frac{5}{149}) (\frac{7}{149}) = (\frac{2}{3}) (\frac{4}{5}) (\frac{2}{7}) .

The three factors are

- 1, 1, 1

, respectively, so

(\frac{105}{149}) = - 1.

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