Binary Quadratic Forms as Arnold’s Perfect Forms
By Titus Piezas III
Abstract: We discuss two explicit identities that can establish certain binary quadratic forms as perfect forms as defined by Vladimir Arnold. The latter one is connected to fundamental discriminants d, positive or negative, with class number h(d) = 3m. A conjecture will also be given as well as a possible extension to diagonal quaternary quadratic forms. (An update is given at the end.)
Contents:
I. Introduction
II. Derivation
III. Negative discriminants d with class number 3m
IV. Positive discriminants d with class number 3m
V. Conclusion: Ramanujan’s quaternary forms
It has been known since Brahmagupta (598-670 AD), that the product of two sums of two squares is itself the sum of two squares. This can be illustrated by the nice identity,
(u2+v2)(x2+y2) = (ux-vy)2 + (uy+vx)2
which is just a special case of Bramagupta’s Lemma, important to solutions of Pell equations. This is also known as the Fibonacci Two-Square Identity since Leonardo Fibonacci (1170-1250) included it in his influential book Liber Abaci (a book which, among other things, introduced Hindu-Arabic numerals to the West as well as, of course, the Fibonacci numbers1). There are generalizations to this identity, namely the Lagrange Four-Square and the Degen-Graves Eight-Square (and no more)2, important to division algebras. However, another kind of generalization is possible as was considered by V. Arnold3 (of Hilbert’s 13th). The form x2+y2 is just a special case of the binary quadratic form,
F(x,y) := ax2+bxy+cy2
abbreviated as (a,b,c). If we explicitly specify (from this point onwards) that a,b,c, as well as the x,y, are to be in the integers, then the more general question to ask would be:
“For what a,b,c such that (au2+buv+cv2)(ax2+bxy+cy2) = az12+bz1z2+cz22 can one always find integral zi for all integral u,v,x,y?”
In other words, what binary quadratic forms, with all variables integral, are such that the product of two forms is of like form? Arnold called these as perfect forms (or simply, perfect) and as possessing the semi-group property. The Brahmagupta-Fibonacci identity is obviously the case (1,0,1). Are there other cases? Before answering that, if we consider the product of three forms then all possess the trigroup property (Arnold, Aicardi). This can easily be shown by the identity,
(ap2+bpq+cq2)(ax2+bxy+cy2)(az2+bzw+cw2) = az12+bz1z2+cz22
where z1 = j1p+j2q, z2 = j3p+j4q, and,
j1 = a(xz)-c(wy)
j2 = c(yz+wx)+b(xz)
j3 = a(wx+yz)+b(wy)
j4 = c(wy)-a(xz)
Not all possess the semi-group property and it is still an unsolved problem to characterize all (a,b,c) which possesses this (Aicardi, 2004). A broad class was found by Arnold as the case F(x,y)1 = 1 or if ax2+bxy+cy2 = 1 for some x1, y1, then F(x,y) is perfect. There is a second broad class which we will explicitly give here with has a different condition F(x,y)1 = ac or if ax2+bxy+cy2 = ac for some x1, y1, then F(x,y) is also perfect. For the latter, a small particular example can be given by (2,1,3),
(2u2+uv+3v2)(2x2+xy+3y2) = 2z12+z1z2+3z22
where,
z1 = -u(x-y)+v(x+2y)
z2 = v(x-y)+u(x+y)
Note that (2,1,3) never attains the value 1 (Aicardi, Timorin). This form has discriminant d = -23 and has class number h(d) = 3. By the second broad class of (a,b,c), all fundamental d with h(d) = 3 has a non-monic reduced binary quadratic form representation that is perfect. In fact, this second class has three conditions and seem to operate on the reduced forms with d that has class number h(d) a multiple of three! Whether this is true for all such d still remains to be seen. However, both these classes can be derived from the same assumption and that is what we’ll do in the next section.
In the conclusion of Aicardi and Timorin’s paper, “On binary quadratic forms with semigroup property”, they mentioned that (2,1,3), (2,1,4), (3,1,5) with discriminants d = -23,-31,-59, respectively, were perfect foorms that do not integrally attain 1 and hence were distinct from Arnold’s class. (Incidentally, these are also the first three d with class number h(d)=3.) This author thought it might be interesting to find the corresponding identity for these d analogous to the one for the sum of two squares. The one for d = -23 was found by brute force, looking at patterns in small solutions by keeping u,v,x constant while varying y, then varying x, and so on. Once found, it was apparent that it had the form,
(au2+buv+cv2)(ax2+bxy+cy2)-(az12+bz1z2+cz22) = 0 (eq.1)
where,
z1 = u(r1x+r2y)+v(r3x+r4y)
z2 = u(r5x+r6y)+v(r7x+r8y)
for some unknown constants ri. We can solve eq.1 using the same approach in “Ramanujan and the cubic equation 33+43+53 = 63”, namely by using the “internal structure” of the equation to generate a system of equations in the unknowns. This can be done by expanding eq.1, then collecting powers of u,v,x,y, to get,
(p1x2+ p2xy+ p3y2)u2+( p4x2+ p5xy+ p6y2)uv+( p7x2+ p8xy+ p9y2)v2 = 0 (eq.2)
with,
p1 = a2-ar12-br1r5-cr52
p2 = ab-2ar1r2-br2r5-br1r6-2cr5r6
p3 = ac-ar22-br2r6-cr62
p4 = ab-2ar1r3-br3r5-br1r7-2cr5r7
p5 = b2-2a(r1r4+r2r3)-b(r1r8+r2r7+r3r6+r4r5)-2c(r5r8-r6r7)
p6 = bc-2ar2r4-br4r6-br2r8-2cr6r8
p7 = ac-ar32-br3r7-cr72
p8 = bc-2ar3r4-br4r7-br3r8-2cr7r8
p9 = c2-ar42-br4r8-cr82
To satisfy eq.2, all we have to do is to set all pi = 0. While it looks horrendous to solve, with modern computer algebra systems it is quite feasible, though one eventually has to use a lot of tedious simplication. It only has eight ri with nine pi but fortunately this system is still solvable and in fact has at least two distinct non-trivial solutions based on the two possible quadratic equations on the ri, namely p1 and p3 (p7 and p9 are just equivalent after changes in variables). These yield the two broad classes mentioned earlier. Note that by letting r1 = am and r5 = an, then p1 = 0 becomes,
am2+bmn+cn2 = 1
precisely the condition specified by Arnold, while p3 = 0 (or p7) is equivalent to the condition,
ap2+bpq+cq2 = ac
for some integer p,q. To solve this system, one can start with the linear equations given by the pi with even subscripts. Solving for r1, r2, r3 in p2, p4, p6, and using those expressions on p8 = 0, we get,
(b2-4ac)(-r6+r7)(cr5-ar8) = 0
and giving the first constraint on the ri. With r7 = r6 (the third factor seems to yield the same results, with changes in the variables) one can then solve for the rest of the unknown ri using the pi with odd subscripts. After much algebraic manipulation and simplification, we find two distinct solutions:
Theorem 1. (Arnold’s)
“Given the binary quadratic form F(x,y) = ax2+bxy+cy2 where a,b,c,x,y are integers, if there are integers n1, n2, such that,
an12+bn1n2+cn22 = 1
then F(x,y) is a perfect form.”
Proof:
This is simply given by the identity (and variations thereof),
(au2+buv+cv2)(ax2+bxy+cy2)-(az12+bz1z2+cz22) =
-(an12+bn1n2+cn22-1)(au2+buv+cv2)(ax2+bxy+cy2)
where,
z1 = u(n3x+cn2y)+cv(n2x-n1y)
z2 = v(an1x+n4y)-au(n2x-n1y)
and,
n3 = an1+bn2
n4 = bn1+cn2
Note: For monic forms when a = 1 automatically it is a perfect form since a solution to n12+bn1n2+cn22 = 1 can be given as n1 = 1, n2 = 0. (Or equivalently, when c = 1, after exchanging variables.)
Theorem 2.
“Given the binary quadratic form F(x,y) = ax2+bxy+cy2 where a,b,c,x,y are integers, if there are integers r1, r2, r3, r4 such that,
ar12+br1r2+cr22 = ac
and,
r3 = (ar1+br2)/c
r4 = (br1+cr2)/a
then F(x,y) is a perfect form.”
Proof:
Given by the identity:
(au2+buv+cv2)(ax2+bxy+cy2)-(az12+bz1z2+cz22) =
-(ar12+br1r2+cr22-ac)(au2+buv+cv2)(ax2+bxy+cy2)/(ac)
where,
z1 = -u(r2x-r1y)+v(r1x+r4y)
z2 = v(r2x-r1y)+u(r3x+r2y)
and,
r3 = (ar1+br2)/c
r4 = (br1+cr2)/a
Note: The subscripts of the ri have been changed from the original.
To find (a,b,c) which satisfy theorem 2, one can look at class numbers. To recall, the class number h(d) of an order of a quadratic field with d < 0 is equal to the number of reduced binary quadratic forms with discriminant d (Weisstein). For example, d = -23 only has three, (1,1,6), (2,1,3), (2,-1,3), hence has class number h(d) = 3. By theorems 1 and 2, if (a,b,c) is perfect then so is (a,-b,c) so we will just be focusing on the first. The non-monic (a,b,c) for all 16 d with h(d) = 3 turns out to be all perfect.
A. Class number 3 (16)
|
d |
a |
b |
c |
r1 |
r2 |
r3 |
r4 |
|
-23 |
2 |
1 |
3 |
1 |
1 |
1 |
2 |
|
-31 |
2 |
1 |
4 |
2 |
0 |
1 |
1 |
|
-59 |
3 |
1 |
5 |
-2 |
1 |
-1 |
1 |
|
-83 |
3 |
1 |
7 |
2 |
1 |
1 |
3 |
|
-107 |
3 |
1 |
9 |
3 |
0 |
1 |
1 |
|
-139 |
5 |
1 |
7 |
1 |
2 |
1 |
3 |
|
-211 |
5 |
3 |
11 |
1 |
2 |
1 |
5 |
|
-283 |
7 |
5 |
11 |
-3 |
2 |
-1 |
1 |
|
-307 |
7 |
1 |
11 |
3 |
1 |
2 |
2 |
|
-331 |
5 |
3 |
17 |
-4 |
1 |
-1 |
1 |
|
-379 |
5 |
1 |
19 |
-4 |
1 |
-1 |
3 |
|
-499 |
5 |
1 |
25 |
5 |
0 |
1 |
1 |
|
-547 |
11 |
5 |
13 |
1 |
3 |
2 |
4 |
|
-643 |
7 |
1 |
23 |
3 |
2 |
1 |
7 |
|
-883 |
13 |
1 |
17 |
-4 |
1 |
-3 |
1 |
|
-907 |
13 |
9 |
19 |
3 |
2 |
3 |
5 |
Notice that the ri are conveniently small. The same phenomenon occurs for negative d with h(d) = 6 or 9 (and perhaps beyond) with most of the ri only a single digit. The situation is quite different for positive d. The (a,b,c) were found using Keith Matthews’ website “Some BCMath/PHP number theory programs” given in the references.
B. Class number 6 (51)
While for d with h(d) = 3 it was easy enough to test them since there were only three forms, for h(d) > 3 one necessarily has to test more. For example, d = -87 with h(d) = 6 has (1,1,22), (2,1,11), (2,-1,11), (3,3,8), (4,3,6), (4,-3,6). It turns out that all d with h(d) = 6 or 9 have at least three forms, (1,p,q), (a,b,c), and (a,-b,c), that are perfect of the type specified by Theorem 1 or 2 and it is tempting to speculate if this is in fact true for all h(d) = 3m. For brevity, only the first ten d for h(d) = 6 will be given.
d |
a |
b |
c |
r1 |
r2 |
r3 |
r4 |
|
-87 |
4 |
3 |
6 |
0 |
2 |
1 |
3 |
|
-104 |
3 |
2 |
9 |
3 |
0 |
1 |
2 |
|
-116 |
5 |
2 |
6 |
2 |
1 |
2 |
2 |
|
-152 |
6 |
4 |
7 |
1 |
2 |
2 |
3 |
|
-212 |
6 |
2 |
9 |
3 |
0 |
2 |
1 |
|
-244 |
5 |
4 |
13 |
1 |
2 |
1 |
6 |
|
-247 |
4 |
3 |
16 |
4 |
0 |
1 |
3 |
|
-339 |
7 |
5 |
13 |
3 |
1 |
2 |
4 |
|
-411 |
7 |
3 |
15 |
-3 |
2 |
-1 |
3 |
|
-424 |
10 |
4 |
11 |
-3 |
2 |
-2 |
1 |
The situation with positive d is slightly more complicated. The class number for negative d can be specified by the number of its reduced binary quadratic forms but for positive d the reduced form first is defined as,
Definition 1 (Gauss). “A quadratic form ax2+bxy+cy2 with positive discriminant d = b2-4ac is reduced if Öd > 0 and b >|Öd-2|a||.” (McMath)
Based on this definition alone, since (a,b,c), and (-a,b,-c) have the same d, then the two can be seen as equivalent reduced forms. Using this, then all 42 (a,b,c) for positive d with h(d) = 3 have, again, at least three forms, (1,p,q), (a1,b1,c1), and (a2,b2,c2), that are perfect of the type specified by Theorem 1 or 2. For example, d = 733 has (1,27,-1), (3,23,-17), (3,25,-9). If the last is given as (-3,25,9), then all three are perfect. For brevity, only one (a,b,c) and only the first fifteen d will be given. (The list of positive d with h(d) = 3 can be found in the Online Encyclopedia of Integer Sequences though unfortunately it does not have one for h(d)=6.)
A. Class number 3 (42)
|
d |
a |
b |
c |
r1 |
r2 |
r3 |
r4 |
|
229 |
3 |
11 |
-9 |
-6 |
-9 |
13 |
5 |
|
257 |
2 |
13 |
-11 |
1 |
-1 |
1 |
12 |
|
316 |
-3 |
14 |
10 |
2 |
-1 |
-2 |
-6 |
|
321 |
-2 |
15 |
12 |
6 |
-8 |
-11 |
3 |
|
469 |
-3 |
17 |
15 |
3 |
-3 |
-4 |
-2 |
|
473 |
-2 |
19 |
14 |
2 |
-2 |
-3 |
-5 |
|
568 |
3 |
20 |
-14 |
2 |
-1 |
1 |
18 |
|
733 |
3 |
23 |
-17 |
2 |
-1 |
1 |
21 |
|
761 |
2 |
25 |
-17 |
-7 |
-11 |
17 |
6 |
|
892 |
-3 |
26 |
18 |
6 |
-9 |
-14 |
2 |
|
993 |
-2 |
29 |
19 |
17 |
-27 |
-43 |
10 |
|
1016 |
-5 |
24 |
22 |
6 |
-7 |
-9 |
2 |
|
1101 |
-5 |
29 |
13 |
2 |
-1 |
-3 |
-9 |
|
1229 |
-5 |
27 |
25 |
5 |
0 |
-1 |
-27 |
|
1257 |
-2 |
33 |
21 |
291 |
17 |
-1 |
-4980 |
While most of the ri for negative d with h(d) = 3,6,9 were only single digits, for positive ones these can get quite large, with the largest ri for d = 2857, (2,51,-32) with r1 = 3,326,866 and r2 = -127,404.
Before going to quaternary forms, several points can be discussed. First, some of our observations can be summed up as the following conjecture:
“Let d, positive or negative, be a fundamental discriminant with class number h(d) = 3m. Then among the reduced binary quadratic forms of the number field Q(Öd), three will be perfect of the type specified by Theorem 1 or 2.”
For negative d these will be of form (1,p,q), (a,b,c), and (a,-b,c) while for positive d will be (1,p,q), (a1,b1,c1), and (a2,b2,c2), taking note that for the latter, (a,b,c) and (-a,b,-c) are considered equivalent. This author does not know if the precise wording should be “at least” or “exactly” three. For d = -87 which has h(d) = 6 (and a few others tested) it was the latter. It should be good to know if indeed the conjecture is true.
Second, given a perfect (a,b,c) so that F(x1,x2)*F(y1,y2) = F(z1,z2), then do the zi always have a formula in terms of the xi and yi? For example, it seems one of the non-monic (a,b,c), for d = -47 with h(d) = 5, is perfect though neither of type theorem 1 or 2. If indeed it is perfect and has a parametrization, then it must be using assumptions distinct from the ones we used to derive theorem 1 and 2.
Finally, recall that perfect forms are a generalization of a property possessed by the Bramagupta-Fibonacci two-square identity. But this is just the first in a series of identities, with the next given by the Euler four-square. If we go to quaternary quadratic forms and limit ourselves to the diagonal case (one without cross terms),
F(x1,x2,x3,x4) := ax12+bx22+cx32+dx42
which we can abbreviate (a,b,c,d), then for what (a,b,c,d) is there such that,
(ax12+bx22+cx32+dx42)(ay12+by22+cy32+dy42) = az12+bz22+cz32+dz42 (eq.3)
or the product of two diagonal quaternary quadratic forms is of like form? For brevity, perhaps these can be called perfect diagonal forms. It turns out that using a result established by Ramanujan4 in 1916, we can find there are at least 54. Note that (1,1,1,1) by Lagrange’s Four-Square Theorem can suffice to represent all positive integers. But this is not the only possible form, with a few others in Ramanujan’s list of 54 cases being (1,1,1,n) for n = 2 to 7. Thus, given the left hand side of eq.3, one is assured of finding zi in the integers when (a,b,c,d) is in the list as the product will always be a positive integer. For (1,1,1,1) this is just Euler’s four-square identity and the zi have a known bilinear expression in terms of the xi and yi. It might be interesting to know if the other 53 (a,b,c,d) have zi that are similar bilinear expressions in terms of the xi and yi.
And while this list is complete as a list of diagonal forms representing all positive integers, eq.3 is a distinct problem, as it seeks only to find (a,b,c,d) that are perfect, and not that it is able to represent all positive integers. Are there then other (a,b,c,d) that can satisfy eq.3 not in the list?
--End--
Update: May 24, 2006,
Given the diagonal quadratic form:
F(v1,v2,…vn) := a1v12+a2v22+…+anvn2
we can define the perfect diagonal form as a F(v1,v2,…vn) with all variables integral such that,
F(x1,x2,…xn)F(y1,y2,…yn) = F(z1,z2,…zn)
for some constants ai but for all xi and yi one can always find zi. The case n=2 has been studied by Arnold, Aicardi, and Timorin. For n=4, we said the ai could be any of the 54 quaternaries of Ramanujan. It turns out there is an infinite more as the form (1,1,m,m) for any integer m will do. This can be shown by the identity,
(a2+b2+mc2+md2)(e2+f2+mg2+mh2) = (z12+ z22+mz32+mz42) (eq.4)
where,
z1 = (ae-bf-mcg-mdh), z2 = (af+be+mch-mdg), z3 = (ag-bh+ce+df), z4 = (ah+bg-cf+de),
with additional solutions given by exchanging the variables a,b and other pairs. This is just a quaternary version of Brahmagupta’s lemma and can be extended to the octonary case n=8, with the ai as (1,1,1,1,m, m, m, m). I did work on sums of squares identities a few years ago and saw a connection. For unit ai such identities are only for n = 1,2,4,8, but if allow general ai, then for what other n is there? And for n = 4,8 what other ai are there?
Footnotes:
© Titus Piezas III
May 19, 2006
titus_piezasIII@yahoo.com (Pls. remove “III”)
www.oocities.org/titus_piezas/ramanujan.html
References: