Prime Numbers

5.6 Binary quadratic forms 245
is assembled. For each fixed ɛ>0, the running time of this procedure to find
the complete prime factorization of n is O(n 1/3+ɛ ).
For another McKee method of different complexity, see Exercise 5.21.
5.6.3 Composition and the class group
Suppose D is a nonsquare integer, (a1,b,c1), (a2,b,c2) are quadratic forms of
discriminant D, and suppose c1/a2 is an integer. Since the middle coefficients
are equal, we have a1c1 = a2c2, sothatc1/a2 = c2/a1. We claim that the
product of a number represented by the first form and a number represented
by the second form is a number represented by the form (a1a2,b,c1/a2). To
see this assertion, it is sufficient to verify the identity
a1x 2 1 + bx1y1 + c1y 2 1 a2x 2 2 + bx2y2 + c2y 2 2 = a1a2x 2 3 + bx3y3 +(c1/a2)y 2 3,
where
x3 = x1x2 − (c1/a2)y1y2, y3 = a1x1y2 + a2x2y1 + by1y2.
So in some sense, we can combine the two forms (a1,b,c1), (a2,b,c2) of
discriminant D to get a third form (a1a2,b,c1/a2). Note that this third form
is also of discriminant D. This is the start of the definition of composition of
forms.
We say that a binary quadratic form (a, b, c) isprimitive if gcd(a, b, c) =1.
GivenanintegerD that is not a square, but is 0 or 1 (mod 4), let C(D)
denote the set of equivalence classes of primitive binary quadratic forms of
discriminant D; where each class is the set of those forms equivalent to a given
form. We shall use the notation 〈a, b, c〉 for the equivalence class containing
the form (a, b, c).
Lemma 5.6.5. Suppose 〈a1,b,c1〉 = 〈A1,B,C1〉 ∈ C(D), 〈a2,b,c2〉 =
〈A2,B,C2〉 ∈ C(D), and suppose that c1/a2,C1/A2 are integers. Then
〈a1a2,b,c1/a2〉 = 〈A1A2,B,C1/A2〉.
See [Rose 1988], for example.
Lemma 5.6.6. Suppose (a1,b1,c1), (a2,b2,c2) are primitive quadratic forms
of discriminant D. Then there is a form (A1,B,C1) equivalent to (a1,b1,c1)
and a form (A2,B,C2) equivalent to (a2,b2,c2) such that gcd(A1,A2) =1.
Proof. We first show that there are coprime integers x1,y1 such that
a1x 2 1 + b1x1y1 + c1y 2 1 is coprime to a2. Writea2 = m1m2m3, where every
prime that divides m1 also divides a1, but does not divide c1; every prime that
divides m2 also divides c1, but does not divide a1; and every prime that divides
m3 also divides gcd(a1,c1). Find integers u1,v1 such that u1m1+v1m2m3 =1,
and let x1 = u1m1. Find integers u2,v2 such that u2m2 + v2m3x1 =1,and
let y1 = u2m2. Thenx1,y1 have the desired properties.
Make the unimodular change of variables x = x1X −Y, y = y1X +v2m3Y .
This changes (a1,b1,c1) to an equivalent form (A1,B1,C ′ 1), where A1 =