Past exam of the mathematics course of the University of Cambridge / 2001 / ia / Paper 4 / 5E / b

Table of contents
- Solution b

Solution

Write

r = (p - 1) /2

,

O = 1 \cdot 3 \dots (p - 2)

and

E = 2 \cdot 4 \dots (p - 1)

p - 2, p - 4, \dots, p - 2 r

are exactly the odd factors of

O

, in reverse order. Hence

O \equiv (- 1)^{r} E (mod p), OE = (p - 1)! \equiv - 1 (mod p)

by Wilson's theorem. It follows that the odd-residue squared product modulo a prime is

1^{2} \cdot 3^{2} \dots (p - 2)^{2} = O^{2} \equiv (- 1)^{r + 1} = (- 1)^{(p + 1) /2} (mod p) .

Thus it is

- 1

for

p \equiv 1 (mod 4)

and

1

for

p \equiv 3 (mod 4)

.

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