 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-4/1g/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

1 1^{2} - 4 \cdot 31 = - 3.

p \neq = 3

, a solution exists exactly when

(\frac{- 3}{p}) = 1.

(\frac{- 3}{p}) = (\frac{p}{3}),

which equals

1

precisely for

p \equiv 1 (mod 3)

. When

p = 3

X^{2} + 2 X + 1 = (X + 1)^{2},

so it also has a solution. Hence

p \equiv 0 or 1 (mod 3) .

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