2 The Hilbert Symbol 305Proposition 25Let p be an odd prime and a,b∈Qpwith|a|p=|b|p= 1 .Then
(i)(a,b)p= 1 ,
(ii)(a,pb)p= 1 if and only if a=c^2 for some c∈Qp.
In particular, for any integers a,b not divisible by p,(a,b)p= 1 and(a,pb)p=
(a/p),where(a/p)is the Legendre symbol.
Proof LetS⊆Zpbe a set of representatives, with 0∈S, of the finite residue field
Fp=Zp/pZp. There exist non-zeroa 0 ,b 0 ∈Ssuch that
|a−a 0 |p< 1 ,|b−b 0 |p< 1.But Lemma 11 implies that there existx 0 ,y 0 ∈Ssuch that
|a 0 x^20 +b 0 y 02 − 1 |p< 1.Since|x 0 |p≤1,|y 0 |p≤1, it follows that
|ax 02 +by 02 − 1 |p< 1.Hence, by Proposition VI.16,ax 02 +by 02 =z^2 for somez∈Qp.Sincez=0, this
implies(a,b)p=1. This proves (i).
Ifa=c^2 for somec∈Qp,then(a,pb)p=1, by Proposition 23. Conversely,
suppose there existx,y∈Qpsuch thatax^2 +pby^2 =1. Then|ax^2 |p=|pby^2 |p,since
|a|p=|b|p=1. It follows that|x|p =1,|y|p≤1. Thus|ax^2 − 1 |p <1and
henceax^2 =z^2 for somez∈Q×p. This proves (ii).
The special case whereaandbare integers now follows from Corollary VI.17.
Corollary 26If p is an odd prime and if a,b,c ∈Qpare p-adic units, then the
quadratic form aξ^2 +bη^2 +cζ^2 is isotropic.
Proof In fact, the quadratic form −c−^1 aξ^2 −c−^1 bη^2 −ζ^2 is isotropic, since
(−c−^1 a,−c−^1 b)p=1, by Proposition 25.
Proposition 27Let a,b∈Q 2 with|a| 2 =|b| 2 = 1 .Then
(i)(a,b) 2 = 1 if and only if at least one of a,b,a− 4 ,b− 4 is a square inQ 2 ;
(ii)(a, 2 b) 2 = 1 if and only if either a or a+ 2 b is a square inQ 2.
In particular, for any odd integers a,b,(a,b) 2 = 1 if and only if a ≡ 1 or
b≡1mod4, and(a, 2 b) 2 = 1 if and only if a≡ 1 or a+ 2 b≡1mod8.
Proof Suppose there existx,y∈Q 2 such thatax^2 +by^2 =1 and assume, for exam-
ple, that|x| 2 ≥|y| 2 .Then|x| 2 ≥1and|x| 2 = 2 α,whereα≥0. By Corollary VI.14,
x= 2 α(x 0 + 4 x′), y= 2 α(y 0 + 4 y′),wherex 0 ∈{ 1 , 3 },y 0 ∈{ 0 , 1 , 2 , 3 }andx′,y′∈Z 2 .Ifaandbare not squares inQ 2
then, by Proposition VI.16,|a− 1 | 2 > 2 −^3 and|b− 1 | 2 > 2 −^3. Thus
a=a 0 + 8 a′, b=b 0 + 8 b′,