Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 4 / 1G / b

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- Solution b

Solution

The discriminant is

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

The discriminant criterion for a quadratic congruence modulo an odd prime therefore says that, for

p \neq = 3

, a solution exists exactly when

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

The quadratic character of minus three gives

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

which equals

1

precisely for

p \equiv 1 (mod 3)

. When

p = 3

, the polynomial is

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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