Past exam of the mathematics course of the University of Cambridge / 2001 / ia / Paper 4 / 7E / c

Solution

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

with

p

an odd prime, then

x^{4} \equiv 1

but

x^{2} \neq \equiv 1

. The multiplicative order consequently divides four and is neither one nor two, so it is four. The preceding divisibility result makes

4 ∣ p - 1

necessary.

For sufficiency, suppose

p \equiv 1 (mod 4)

and put

r = (p - 1) /2

, an even integer. Pair opposite factors in the factorial:

(p - 1)! = \prod_{j = 1}^{r} j (p - j) \equiv (- 1)^{r} (r!)^{2} = (r!)^{2} (mod p) .

(r!)^{2} \equiv - 1 (mod p)

For odd p, x^{2} \equiv - 1 (mod p) is soluble ⟺ p \equiv 1 (mod 4) .

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