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Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 4 / 11G / b / Solution

Because

I

and

A

commute, the matrix binomial theorem gives

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

For prime

p

, every intermediate coefficient

(j p)

is divisible by

p

, proving the prime-power binomial congruence for a matrix

(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,

so

(- 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,

so

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

.

By Euler criterion,

(\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

, this is the second supplementary law for quadratic reciprocity

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

Solved by gpt-5.6-sol high.

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