3 The Hasse–Minkowski Theorem 315We are going to show that there exists an integercsuch thatc^2 ≡amodb.
Since±bis a product of distinct primes, it is enough to show that the congruence
x^2 ≡amodpis soluble for each primepwhich dividesb(by Corollary II.38). Since
this is obvious ifa≡0or1modp, we may assume thatpis odd andanot divisible
byp. Then, since fis isotropic inQp,(a,b)p=1. Henceais a square modpby
Proposition 25.
Consequently there exist integersc,dsuch thata=c^2 −bd. Moreover, by adding
toca suitable multiple ofbwe may assume that|c|≤|b|/2. Then
|d|=|c^2 −a|/|b|≤|b|/ 4 + 1 <|b|andd=0, sinceais square-free anda=1. We have
bd(aξ^2 +bη^2 −ζ^2 )=aX^2 +dY^2 −Z^2 ,where
X=cξ+ζ, Y=bη, Z=aξ+cζ.Moreover the linear transformationξ,η,ζ→X,Y,Zis invertible in any field of zero
characteristic, sincec^2 −a=0. Hence, sincefis isotropic inQvfor allv,soalsois
g=aξ^2 +dη^2 −ζ^2 .Sincefis not isotropic inQ, by hypothesis, neither isg.But
this contradicts the original choice off,since|ad|<|ab|.
It may be noted that forn=3 it need only be assumed thatfis isotropic inQpfor
all primesp. For the preceding proof used the fact thatfis isotropic inQ∞only to
exclude from consideration the quadratic form−ξ^2 −η^2 −ζ^2 and this quadratic form
is anisotropic also inQ 2 , by Proposition 27. In fact forn=3itneedonlybeassumed
thatfis isotropic inQvfor allvwith at most one exception since, by Proposition 35,
the number of exceptions must be even.
(iii)n≥4: We have
f=a 1 ξ 12 +···+anξn^2 ,wherea 1 ,...,anare square-free integers. We writef=g−h,where
g=a 1 ξ 12 +a 2 ξ 22 ,h=−a 3 ξ 32 −···−anξn^2.LetSbe the finite set consisting of∞and all primespwhich divide 2a 1 ···an.By
Proposition 7, for eachv∈Sthere existscv∈Q×vwhich is represented inQvby both
gandh. We will show that we can takecvto be the same nonzero integercfor every
v∈S.
Letv=pbe a prime inS. By multiplying by a square inQ×pwe may assume that
cp=pεpc′p,whereεp=0or1and|c′p|p=1. Ifpis odd and ifbpis an integer
such that|cp−bp|p≤p−εp−^1 ,then|bp|p=|cp|pand|bpc−p^1 − 1 |p≤p−^1. Hence
bpc−p^1 is a square inQ×p, by Proposition VI.16, and we can replacecpbybp. Similarly
ifp=2andifb 2 is an integer such that|c 2 −b 2 | 2 ≤ 2 −ε^2 −^3 ,then|b 2 | 2 =|c 2 | 2 and
|b 2 c− 21 − 1 | 2 ≤ 2 −^3. Henceb 2 c− 21 is a square inQ× 2 and we can replacec 2 byb 2.