Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 125 / 1 / b

Solution

For

x = 0, 1, 2

, the value of

x^{3} - x - 1

is the nonsquare

2

F_{3}

# E (F_{3}) = 1, a_{3} = 3 + 1 - 1 = 3.

α, β = \frac{3 \pm i 3}{2} = 3 e^{\pm iπ /6}

T^{2} - 3 T + 3

# E (F_{3^{r}}) = 3^{r} + 1 - α^{r} - β^{r} = 3^{r} + 1 - 2 \cdot 3^{r /2} cos (r π /6) .

The last term vanishes exactly when

r π /6 \equiv π /2 (mod π)

. Hence

# E (F_{3^{r}}) = 3^{r} + 1 ⟺ r \equiv 3 (mod 6) .

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