Number theory, geometry and algebra - Dynamics-approx.jku.at

Such forms satisfy the relation
and as a result
−d = 4ac−b 2 ≥ 3ac
a ≤ 1 3 |d|, c ≤ 1 3 |d|, |b| ≤ 1 3 |d|.
This implies that there are at most finitely many reduced forms with a
given diskriminant d. We write h(d) (the class number of d) for the number
of reduced forms with diskriminant d.
Example For d = −4 we have 3ac ≤ 4 and so a = c = 1 and b = 0. Hence
h(−4) = 1.
Theimportanceofthefunctionhliesinthefactthatircountsthenumber
of equivalent forms. Indeed we have
Proposition 23 If f,g are reduced and d(f) = d(g), then f ∼ g.
Proof. Weshallshowthattwodistinct reducedformf,g arenotequivalent.
We begin with the remark that |x| ≥ |y| implies:
resp.
f(x,y) ≥ |x|(a|x|−|by|)+c|y| 2 ≥ |x| 2 (a−|b|)+c|y| 2 ≥ a−|b|+c.
f(x,y) ≥ a−|b|+c,
if |y| ≥ |x|. Hence the three smallest values of f are a (for (1,0)), c (for
(0,1)) and a−|b|+c (for (1,1) or (1,−1)). Hence a = a ′ , c = c ′ and b = ±b ′ ,
where a ′ ,b ′ ,c ′ are the coefficients of g.
We now show that b = −b ′ implies b = 0. We can assume that −a < b <
a < c. (For −a < −b and a = c implies that b ≥ 0 and −b ≥ 0, i.e. b = 0).
In this case we have
f(x,y) ≥ a−|b|+c > c > a.
If we transform f into g by means of the matrix P above, then a = f(p,r).
Hence p = ±1 and r = 0. The condition ps − qr = 1 then implies that
s = ±1. Further c = f(q,s) and so q = 0. This easily implies that b = 0.
Definition n is represented by f if x,y ∈ Z exists so that
gcd (x,y) = 1, f(x,y) = n.
Proposition 24 n is representable by a form f with discrimant d ⇔ the
equation x 2 = d(mod 4n) has a solution.
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