 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ii/paper-4/11g/b/solution

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

Because

I

and

A

(I + A)^{p} = \sum_{j = 0}^{p} (j p) A^{j} .

For prime

p

, every intermediate coefficient

(j p)

is divisible by

p

(I + A)^{p} \equiv I + A^{p} (mod p) .

For

A = (01 - 1 0), A^{2} = - I,

one has

(I + A)^{2} = 2 A

and

(I + A)^{4} = - 4 I

. If

p = 4 k + 1

, then

(I + A)^{p} = (- 4)^{k} (I + A) \equiv I + A,

(- 4)^{k} \equiv 1 (mod p)

. If

p = 4 k - 1

, then

A^{p} = - A

and

(I + A)^{p} = (- 4)^{k} (I + A)^{- 1} = \frac{( - 4 ) ^{k}}{2} (I - A) \equiv I - A,

(- 4)^{k} \equiv 2 (mod p)

(\frac{2}{p}) \equiv 2^{(p - 1) /2} (mod p)

. In either case the preceding congruence gives

(\frac{2}{p}) = (- 1)^{k}

. Since

(p^{2} - 1) /8

has the same parity as

k

for

p = 4 k \pm 1

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

 New to topics? Read the docs here!