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Equivalence and Reduction of Binary Quadratic Forms
| Content Provider | Semantic Scholar |
|---|---|
| Abstract | We have already defined the discriminant D of f (x, y) = ax 2 + bxy + cy 2 as the quantity D = ±d = b 2 − 4ac where d is always > 0. D is congruent to 0 or 1 (mod 4) and has the same parity as b. So, b is even only when D ≡ 0 (mod 4). By completing the square of f (x, y) we get 4af (x, y) = (2ax + by) 2 − Dy 2. If D > 0 then f (1, 0) = a and f (−b, 2a) = −aD have opposite signs unless a = 0. Similarly, f (0, 1) = c and f (2c, −b) = −cD have opposite signs unless c = 0. If a = c = 0, then f (x, y) = bxy with b = 0 since D = 0. In this case, f (1, 1) and f (−1, 1) have opposite signs. Therefore, if D > 0 then f (x, y) takes on both positive and negative values and is said to be indefinite. If D < 0, then 4af (x, y) > 0 whenever (x, y) = (0, 0). So, a and f (x, y) have the same sign. The form thus takes on positive values when a > 0 and negative values when a < 0. Hence, to simplify matters, we only consider positive definite forms, multiplying f through by −1 if necessary. Finally, note c > 0 since ac = (b 2 − D)/4 > 0. Let f (x, y) = x 2 + y 2 and F (X, Y) = X 2 + 2XY + 2Y 2. What can we say about the integers represented by f (x, y) and F (X, Y)? Let's try some computations. and so on. As you can see, the same numbers are represented by both forms. The y-values remain the same while the x-values change. If n = x 2 +y 2 then n = X 2 +2XY +2Y 2 where X = x − y and Y = y. Also, if n = X 2 + 2XY + 2Y 2 then n = x 2 + y 2 where x = X + Y and y = Y. So, we have: x y = 1 1 0 1 X Y and X Y = 1 −1 0 1 x y. Definition. Two … |
| File Format | PDF HTM / HTML |
| Alternate Webpage(s) | http://www.dms.umontreal.ca/~andrew/PDFpre/FITCh2.pdf |
| Language | English |
| Access Restriction | Open |
| Content Type | Text |
| Resource Type | Article |
