Number Theory: An Introduction to Mathematics

1 The Law of Quadratic Reciprocity 135

A second proof of the law of quadratic reciprocity will now be given. Letpbe an
odd prime and, for any integeranot divisible byp, with Legendredefine

(a/p)=1or− 1

according asais a quadratic residue or quadratic nonresidue ofp. It follows from
Euler’s criterion (Proposition II.28) that

(ab/p)=(a/p)(b/p)

for any integersa,bnot divisible byp. Also, by Corollary II.29,

(− 1 /p)=(− 1 )(p−^1 )/^2.

Now letqbe an odd prime distinct frompand letK=Fqbe the finite field con-
tainingqelements. Sincep=q, the polynomialtp−1 has no repeated factors inK
and thus haspdistinct roots in some fieldL⊇K.Ifζis any root other than 1, then
the (cyclotomic) polynomial

f(t)=tp−^1 +tp−^2 +···+ 1

has the rootsζk(k= 1 ,...,p− 1 ).
Consider theGauss sum

τ=

∑p−^1

x= 1

(x/p)ζx.

Instead of summing from 1 top−1, we can just as well sum over any set of represen-
tatives ofF×p:

τ=

∑

x≡0modp

(x/p)ζx.

Sinceqis odd,(x/p)q=(x/p)and hence, sinceLhas characteristicq,

τq=

∑

x≡0modp

(x/p)ζxq.

If we puty=xqthen, since

(x/p)=(q^2 x/p)=(qy/p)=(q/p)(y/p),

we obtain

τq=

∑

y≡0modp

(q/p)(y/p)ζy=(q/p)τ.

Furthermore,

τ^2 =

∑

u,v≡0modp

(u/p)(v/p)ζuζv=

∑

u,v≡0modp

(uv/p)ζu+v