 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/ia/paper-4/6d/ii/solution

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

x^{2} \equiv a

and

y^{2} \equiv a (mod p)

, then

(x - y) (x + y) \equiv 0

, hence

y \equiv \pm x

; these differ for odd

p

. It fails for composite moduli:

1

8

 New to topics? Read the docs here!