Prime Numbers

196 Chapter 4 PRIMALITY PROVING
The next result begins to show a possible relevance of Gauss sums to
primality testing. It may be viewed as an analogue to Fermat’s little theorem.
Lemma 4.4.2. Suppose p, q, n are primes with p|q − 1 and gcd(pq, n) =1.
Then
G(p, q) np−1 −1 ≡ χp,q(n) (modn).
Proof. Let χ = χp,q. Sincen is prime, the multinomial theorem implies that
G(p, q) np−1
=
q−1
m=1
χ(m)ζ m q
n p−1
≡
q−1
m=1
χ(m) np−1
ζ mnp−1
q (mod n).
By Fermat’s little theorem, np−1 ≡ 1(modp), so that χ(m) np−1 = χ(m).
Letting n−1 denote a multiplicative inverse of n modulo q, wehave
q−1
m=1
χ(m) np−1
ζ mnp−1
q−1
q = χ(m)ζ
m=1
mnp−1
q =
= χ(n)
q−1
m=1
q−1
m=1
χ(mn p−1 )ζ mnp−1
q
χ(n −(p−1) )χ(mn p−1 )ζ mnp−1
q
= χ(n)G(p, q),
where the next to last equality uses that χ(n p )=χ(n) p =1andthelast
equality follows from the fact that mn p−1 traverses a reduced residue system
(mod q) asm does this. Thus,
G(p, q) np−1
≡ χ(n)G(p, q) (modn).
Let q −1 be a multiplicative inverse of q modulo n and multiply this last display
by q −1 G(p, q). Lemma 4.4.1 then gives the desired result. ✷
The next lemma allows one to replace a congruence with an equality, in
some cases.
Lemma 4.4.3. If m, n are natural numbers with m not divisible by n and
ζ j m ≡ ζ k m (mod n), then ζ j m = ζ k m.
Proof. By multiplying the congruence by ζ −k
m , we may assume the given
congruence is ζ j m ≡ 1(modn). Note that m−1
l=1 (x − ζl m)=(x m − 1)/(x − 1),
so that m−1
l=1 (1−ζl m)=m. Thus no factor in this last product is zero modulo
n, which proves the result. ✷
Definition 4.4.4. Suppose p, q are distinct primes. If α ∈ Z[ζp,ζq] \{0},
where α = p−2 q−2 i=0 k=0 aikζ i pζ k q , denote by c(α) the greatest common divisor
of the coefficients aik. Further, let c(0) = 0.
We are now ready to describe the deterministic Gauss sums primality test.
Algorithm 4.4.5 (Gauss sums primality test). We are given an integer n>
1. This deterministic algorithm decides whether n is prime or composite, returning
“n is prime” or “n is composite” in the appropriate case.