Quadratic Forms. Ferdinand Vanmaele. July 5, 2017

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1 Quadratic Forms Ferdinand Vanmaele July 5, Introduction Question. Consider Pythagorean triples (, y, z), that is integers for which 2 + y 2 = z 2. More generally, we wish to find all integers n such that 2 + y 2 = n. The naive way would be to plug in all possible integer values for and y. Doing so results in duplicates (for eample 50 = = ) and does not shed much light on the underlying structures. We will construct a more effective method by using quadratic forms. Definition 1. A quadratic form is a function Q : Z Z Z, (, y) a 2 + by + cy 2 where a, b, c Z. Eample 1. The equation 2 + y 2 = n is represented by a quadratic form with a = c = 1 and b = 0. Remark 1. It is sufficient to find primitive solutions of a quadratic form, i.e. i.e. pairs (, y) where and y are coprime. This follows by dividing with the greatest common divisor d of and y: ( ) ( a 2 + by + cy 2 = d 2 a 2 d 2 + by d 2 + cy2 d 2 = d 2 Q d, y d) Changing signs is negated in the quadratic form, as Q(, y) = Q(, y). The case (1, 0) is not ecluded: Q(1, 0) = a 2. It follows that we can identify solutions (, y) of Q(, y) = n with reduced fractions y on the Farey diagram. 2 The Topograph Definition 2. We construct the dual graph to the Farey diagram, with following properties: Vertices: barycentres of each triangle; Edges: connect vertices such that each edge of the Farey diagram is crossed eactly once; Regions: the areas delimited out by the edges, adjacent to a single verte y assign the value Q(, y) to the region. of the Farey diagram. We As every reduced fraction is represented by a verte on the Farey diagram, every value n of Q(, y) is represented as a region on the dual graph. We call this graph the topograph of Q(, y). (see Figure 1) 1

2 Figure 1: Topograph Eample 2. The quadratic form Q(, y) = 2 + y 2 has the topograph shown in Figure 2. The regions have values Q(1, 1) = 2, Q(1, 2) = 5, and so on. Figure 2: Topograph for 2 + y 2 Remark 2. The topograph has symmetry depending on the chosen quadratic form: 1 Horizontal symmetry, when Q(, y) = Q(, y) = Q(, y); Vertical symmetry, when Q(, y) = Q(y, ). 1 A more complicated form is skew symmetry, described later. 2

3 3 Arithmetic Progression Definition 3. An arithmetic progression is a sequence of numbers such that the difference between consecutive terms is constant. 2 That is: a n = a 0 + nd For our considerations, (a n ) n N Z and d Z. Proposition 1 (Arithmetic progression rule). Let p, q, r and s be the four regions surrounding any given edge in the topograph. Then the three numbers (p, q + r, s) form an arithmetic progression. Eample 3. Let Q(, y) = 2 + y 2, with the edge in the topograph that separates the regions: p := Q(1, 0) = 1, q := Q(1, 1) = 2, r := Q(0, 1) = 1 From the arithmetic progression (p, p + q, s) = (1, = 3, s), it follows that s = 5. Figure 3: Arithmetic progression (regions) Proof. (Arithmetic Progression Rule) Let f : Q Z, y Q(, y) be the evaluation of the topograph on the vertices of the Farey diagram. Let ( ) ( ) f 1 y 1 = q and f 2 y 2 = r be two vertices with corresponding regions q and r. Then by the mediant rule for labeling vertices, the labels on the regions p and s are given by 1 2 y 1 y 2 and 1+2 y 1+y 2 : Figure 4: Mediant rule (These labels are correct even for 1 y 1 = 1 0 and 2 y 2 = 0 1 )

4 For a quadratic form Q(, y) = a 2 + by + cy 2 we then have: s = Q( 1 + 2, y 1 + y 2 ) = a( ) 2 + b( )(y 1 + y 2 ) + c(y 1 + y 2 ) 2 Similarly we have: = a( ) + b( 1 y y y y 2 ) + c(y y 1 y 2 + y 2 2) = a b 1 y 1 + cy1 2 + a b 2 y 2 + cy2 2 + (2a b( 1 y y 1 ) + 2cy 1 y 2 ) }{{}}{{}}{{} Q( 1,y 1)=q Q( 2,y 2)=r =:R p = Q( 1 2, y 1 y 2 ) = a( 1 2 ) 2 + b( 1 2 )(y 1 y 2 ) + c(y 1 y 2 ) 2 = a( ) + b( 1 y 1 1 y 2 2 y y 2 ) + c(y 2 1 2y 1 y 2 + hy 2 2) = a b 1 y 1 + cy1 2 + a b 2 y 2 + cy2 2 (2a b( 1 y y 1 ) + 2cy 1 y 2 ) }{{}}{{} Q( 1,y 1)=q Q( 2,y 2)=r Computing p + s results in the canceling of the R term (which is the same for both p and s), leaving: p + s = 2(q + r) This equation can be rewritten as (q + r) p = s (q + r), which just says that (p, q + r, s) forms an arithmetic progression. Remark 3. For any given verte in the topograph, compute the three surrounding regions by inserting the corresponding fractions on the Farey diagram. Using the proposition, we can then calculate all other regions (and with that, all other values of Q(, y)) from these three regions only. For Q(, y) = a 2 + by + cy 2, an easy place to start is Q(1, 0) = a, Q(0, 1) = c and Q(1, 1) = a + b + c. Conversely, specify any three values a, b, c around any verte of the topograph. Consider the quadratic form: Q(, y) = a 2 + (b a c)y + cy 2 We then have: Q(1, 0) = a, Q(0, 1) = c, Q(1, 1) = a + (b a c) + c = b Eample 4. Consider the quadratic form with both positive and negative values Q(, y) = 2 2y 2. As we will see later, this is an eample of a hyperbolic form. Let p = Q(1, 0) = 1, q = Q(1, 1) = 1, r = Q(0, 1) = 2. What is s? (p, q + r, s) = (1, 3, s) s = 7 Proposition 1 p = 1, q = 2, r = 7 (p, q + r, s) = ( 1, 9, s) s = 17 p = Q(0, 1) = 2, q = Q(1, 1) = 1, r = Q(1, 0) = 1. (p, q + r, s) = ( 2, 0, s) s = 2. We can arbitrarily continue this for other regions of the topograph. (see Figure 5) 4

5 Figure 5: Topograph for 2 2y 2 4 Application to continued fractions 4.1 Periodic separator line Definition 4. Let Q(, y) be a quadratic form that takes on both positive and negative values. We get the separator line if we straighten out the zigzag path of edges in the topograph that separate negative values to a line. Figure 6: Separator line for 2 2y 2 Eample 5. We again consider Q(, y) = 2 2y 2. We will construct the separator line using the arithmetic progression rule. a) p = Q(0, 1) = 2, q = Q(1, 1) = 1, r = Q(1, 0) = 1. The horizontal line segments separates positive from negative values. b) As in the eample for the arithmetic progression rule, we get (p, q + r, s) = ( 2, 0, s) s = 2. Positive, place above the separator line. 5

6 Figure 7: Construction of the separator line c) p = 1, q = 1, r = 2 (p, q + r, s) = (1, 1, s) s = 1. Positive, place above the separator line. d) p = 2, q = 1, r = 1 (p, q + r, s) = (2, 0, s) s = 2. Negative, place below the separator line. e) p = 1, q = 1, r = 2 (p, q + r, s) = ( 1, 1, s) s = 1. Negative, place below the separator line. As we have now returned to a), further repetitions produce a periodically repeating pattern as we move to the right. The arithmetic progression rule implies that it also repeats perodically to the left, so it is periodic in both directions. We have found a periodic separator line for Q(, y) = 2 2y 2. Remark 4. As we move upward from the separator line, the values of Q(, y) become larger and larger, approaching + monotonically. As we move downward, the values approach monotonically. The reason for this will become clear when we discuss the Monotonicity Property in the net talk. Eample 6. Let Q(, y) = 2 y 2. This is an eample of a 0-hyperbolic form, i.e. a form which takes on positive and negative values as well as 0. In particular, Q = 2 dy 2 with d a square (1 = 1 2 ). This form does not have a periodic (but monotonic) separator line: Figure 8: Separator line for 2 y 2 This follows from the second arithmetic progression rule, where we label the boundaries of the topograph regions with the difference of terms in an arithmetic progression (see net talk): Proposition 2. Q 1 (, y) = 2 2y 2 has the same positive and negative values. Q 2 (, y) = 2 3y 2 has different positive and negative values. Proof. Construct the periodic separator line for Q 2 (, y) in a similar way as we have done for Q 1 (, y). Below the separator line, we have the periodic pattern ( 2, 3,...) and above it (1, 1,...) (see Figure 10). Continued application of the arithmetic progression rule results in the regions ( 2, 3, 11, 23, 26...), repeated horizontally. These sequences decrease monotonically and go towards, by the Monotonicity Property. 6

7 Figure 9: Second arithmetic progression rule Using a similar approach above the separator line, we get have sequences (1, 6, 13, 22,...) that go towards +. This implies that Q 2 (, y) has at least two different positive and negative values (for eample, 2 which does not appear in the sequence 1, 6, 13, 22,...). Figure 10: Separator line for 2 3y 2 Now consider the separator line of Q 1 (, y). There is a skew symmetry that moves the negative values on the separator line to the positive values (see Figure 6). By the arithmetic progression rule, it follows that Q 1 (, y) has the same positive and negative values. 4.2 Continued fractions Question. Let Q(, y) = 2 dy 2, where d is a positive integer that is not a square. Can we compute the infinite continued fraction for d using the topograph of Q(, y)? Notation ((Continued fractions)). [m 0 : m 1, m 2, m 3,...] := m 0 + If m 1, m 2, m 3 repeat periodically, we write [m 0 : m 1, m 2, m 3 ]. 1 m m 2+ 1 m 3 + Remark 5. The topograph of the form 2 dy 2 always has a periodic separator line whenever d is a positive integer that is not a square. We will prove this in the net talk. Since the form takes the positive value 1 on 1 0 and the negative value d on 0 1, this separator line always includes the edge separating these fractions. 7

8 Eample 7. We again consider the form Q(, y) = 2 2y 2. Superimpose the triangles of the Farey diagram corresponding to this part of the topograph to obtain an infinite strip of triangles. Figure 11: Strip of triangles for [1 : 2] This strip corresponds to the infinite fraction [1 : 2] (the triangles correspond to the cutting sequence L 1 R 2 L 2...) 3. We will use the quadratic form Q(, y) = 2 2y 2 to calculate the value of the infinite fraction. The values n on the separator line for 2 2y 2 are either ±1 or ±2. We rewrite this equation as: 4 Moving right of the diagram, we get for, y : ( ) 2 = 2 + n y y 2 lim (2 + ny ),y 2 = lim 2 = 2 n {±1, ±2},y lim,y y = 2 Thus [1 : 2] = 2. Similarly, we obtain [1 : 1, 2] = 3 using the quadratic form 2 3y 2 (see Figure 12). Figure 12: Strip of triangles for [1 : 1, 2] Eample 8. Taking a closer look at the figure, we notice that it is not necessary to superimpose the above triangles to compute the period of a continued fraction. It suffices to count the downward and upward edges of the topograph (instead of triangles), starting from the edge separating 1 0 and 0 1. This can be illustrated with the quadratic form Q(, y) = 2 7y 2 with continued fraction 7 = [ 2 : 1, 1, 1, 4 ] (see Figure 13). 3 See Proposition 1 in the talk Continued Fractions And Cutting Sequences. 4 Again noting that 1 is a valid fraction in the Farey diagram. 0 8

9 Figure 13: Separator line for 2 7y 2 Remark 6. The periodicity implies (with similar reasoning as for 2 2y 2 ) that the continued fraction for d has the form: d = [a0 : a 1, a 2,..., a n ] Conversely, every infinite, periodic continued fraction has a corresponding hyperbolic form (but not necessarily 2 dy 2 ), as proved in the net talk. Remark 7. The separator line has horizontal symmetry: Q(, y) = 2 dy 2 = Q(, y) Furthermore, for a continued fraction [a 0 : a 1, a 2,..., a n ], a n = 2a 0 and the intermediate terms a 1, a 2,..., a n 1 form a palindrome. Quadratic forms where d has an uneven period (such as 2 13y 2 ) also have a skew symmetry or an additional glide-reflection (that is, reflection plus translation) along the strip that interchanges the positive and negative values of the form. 5 Figure 14: Separator line for 2 13y 2 Remark 8. Squares from fractions p q can be computed similarly to d using the quadratic form q 2 py 2. As with 2 dy 2, these forms always possess a periodic separator line, assuming that p and q are not both squares. The palindrome property and the relation a n = 2a 0 still hold for the continued fractions of p q, assuming that p a 0 > 0, i.e. q > 1. 5 Doubling the period corresponds to ignoring the glide-reflection, and just considering the translational periodicity. 9