Number Theory: An Introduction to Mathematics

3 The Hasse–Minkowski Theorem 321

aξ^2 +(− 1 )(n−^2 )/^2 aη^2 −ζ^2 ,

is isotropic inQ. Thus it is certainly satisfied ifn ≡0 mod 4. Ifn≡2mod4 it is
satisfied if and only if the quadratic formξ^2 +η^2 −aζ^2 is isotropic. Thus it is satisfied
ifais a sum of two squares. It is not satisfied ifais not a sum of two squares since
then, by Proposition II.39, for some primep≡3 mod 4, the highest power ofpwhich
dividesais odd and

(a,a)p=(a,− 1 )p=(p,− 1 )p=(− 1 )(p−^1 )/^2 =− 1. 

It is worth noting that the last part of this proof shows that if a positive integerais
a sum of two rational squares, then it is also a sum of two squares of integers.
It follows at once from Proposition 42 that, for any positive integern,thereisan
n×n rationalmatrixAsuch thatAtA=nInif and only if eithernis an odd square,
orn≡2 mod 4 andnis a sum of two squares, orn≡0 mod 4 (the Hadamard matrix
case).
In Chapter V we considered not only Hadamard matrices, but also designs. We
now use Proposition 42 to derive the necessary conditions for the existence of square
2-designs which were obtained by Bruck, Ryser and Chowla (1949/50). Letv,k,λbe
integers such that 0<λ<k<vandk(k− 1 )=λ(v− 1 ).Sincek−λ+λv=k^2 ,
it follows from Proposition 42 that there exists av×vrational matrixAsuch that

AtA=(k−λ)Iv+λJv

if and only if,eithervis even and k−λis a square, orvis odd and the quadratic form

(k−λ)ξ^2 +(− 1 )(v−^1 )/^2 λη^2 −ζ^2

is isotropic inQ.
A projective plane of orderdcorresponds to a(d^2 +d+ 1 ,d+ 1 , 1 )(square)
2-design. In this case Proposition 42 tells us that there is no projective plane of order
difdis not a sum of two squares andd≡1 or 2 mod 4. In particular, there is no
projective plane of order 6.
The existence of projective planes of any prime power order follows from the
existence of finite fields of any prime power order. (All known projective planes are of
prime power order, but even ford=9 there are projective planes of the same orderd
which are not isomorphic.) Since there is no projective plane of order 6, the least order
in doubt isd=10. The condition derived from Proposition 42 is obviously satisfied
in this case, since

10 ξ^2 −η^2 −ζ^2 = 0

has the solutionξ=η=1,ζ =3. However, Lam, Thiel and Swiercz (1989) have
announced that, nevertheless, there is no projective plane of order 10. The result was
obtained by a search involving thousands of hours time on a supercomputer and does
not appear to have been independently verified.