Ramanujan and the Quartic Equation 24+24+34+44+44 = 54
By Titus Piezas III
Keywords: diophantine equations, equal sums of like powers, fourth powers, algebraic identities, Ramanujan, Pythagorean triples, Eisenstein integers.
Contents:
I. Introduction
II. The 6-10-8 Connection
III. The Equation pk + qk + (p+q)k = rk + sk + (r+s)k (in Linear Forms)
IV. The Equation pk + qk + (p+q)k = rk + sk + (r+s)k (in Quadratic Forms)
V. Pythagorean Triples and a14 + a24 + a34 + a44 + a54 = a64
VI. Conclusion: Beyond Quartics
This paper is the sequel to “Ramanujan and the Cubic Equation 33 + 43 + 53 = 63” [1]. In that work, we answered (rather belatedly) Ramanujan’s question about finding quadratic form solutions to the diophantine equation a13 + a23 + a33 + a43 = 0, or a cubic quadruple, by giving a general algebraic identity. It is recommended that one reads that (See References) before this one to appreciate the context of our approach since now we are dealing with the quartic sextuple a14 + a24 + a34 + a44 + a54 = a64. In this paper, our objective is to generalize Ramanujan’s two quadratic form identities for these sextuples, among other things.
Ramanujan had a lot of beautiful equations and one of which, seemingly unrelated, in fact has a bearing to our present topic. This is his amazing 6-10-8 Identity which is given by,
64[(a+b+c)6 + (b+c+d)6 + (a-d)6 - (c+d+a)6 - (d+a+b)6 - (b-c)6][(a+b+c)10 + (b+c+d)10 + (a-d)10 - (c+d+a)10 - (d+a+b)10 - (b-c)10] = 45[(a+b+c)8 + (b+c+d)8 + (a-d)8 - (c+d+a)8 - (d+a+b)8 - (b-c)8]2
where ad = bc. We can express this more concisely by defining,
Fk = (a+b+c)k + (b+c+d)k + (a-d)k - (c+d+a)k - (d+a+b)k - (b-c)k
so 64F6F10 = 45F82.
This is a remarkable identity by most people’s standards. In the words of Berndt, primary editor of Ramanujan’s Notebooks, “it is one of the most fascinating identities we have ever seen”. One can easily appreciate its simplicity of form and the rather unexpected use of certain exponents. There doesn’t seem to be any identity quite like it prior to it.
There is now (well, from a few years back) something similar, namely Hirschhorn’s 3-7-5 Identity which was inspired by Ramanujan’s. Still letting ad = bc, define,
Hk = (a+b+c)k - (b+c+d)k - (a-d)k + (c+d+a)k - (d+a+b)k + (b-c)k
then 25H3H7 = 21H52.
However, we can still give others, one as linear forms and another as quadratic forms. Let c = a+b and define,
Pk = (ax+(b+c)y)k + (bx-(a+c)y)k + (cx-(a-b)y)k - (ax-(b+c)y)k - (bx+(a+c)y)k - (cx+(a-b)y)k
Qk = (ax2+2(b+c)xy+ay2)k + (bx2-2(a+c)xy+by2)k + (cx2-2(a-b)xy+cy2)k - (ax2-2(b+c)xy+ay2)k - (bx2+2(a+c)xy+by2)k - (cx2+2(a-b)xy+cy2)k
then,
64P6P10 = 45P82
64Q6Q10 = 45Q82
Similar forms will give analogous versions to Hirschhorn’s. So how did we find these new identities? Before answering that consider the diophantine equation,
a14 + a24 + a34 + a44 = a54
or a quartic quintuple. Euler (1707-1783) conjectured that this had a solution in integers and he was right, though it took about 150 years before the first one was found by R. Norrie in 1911,
304 + 1204 + 2724 + 3154 = 3534
and many more were found subsequently, though it is not known if there is a parametric solution. We now have examples of n positive nth powers equal to an nth power for n up to 8, skipping n = 6. Whether it will take another 150 years before we find n = 6 remains to be seen, and would be a disgrace considering the amount of computing power we now have available. For the diophantine equation,
a14 + a24 + a34 + a44 + a54 = a64
or a quartic sextuple, we do have parametric solutions, though only three are known so far, one discussed in Hardy and Wright’s book, “An Introduction To The Theory Of Numbers” (p.333),
(4a4-b4)4 + 2(4a3b)4 + 2(2ab3)4 = (4a4+b4)4
and two quadratic form identities by Ramanujan which we will give later. We will focus on the latter two and will later show how to find an indefinite number of such quadratic forms, one small example given by,
(12x2+120xy-36y2)4 + (24x2-96xy-72y2)4 + (36x2+24xy-108y2)4 + (16x2+48y2)4 + (63x2+189y2)4 = (65x2+195y2)4
and many more.
So what in the world is the connection between Ramanujan’s 6-10-8 Identity and his parametrizations of quartic sextuples? It turns out they both depend on the multiplicative properties of the simple algebraic form,
a2+ab+b2
a form intimately connected to Eisenstein integers, numbers which involve the complex cube roots of unity.
It’s almost a shame to give away the secret of the 6-10-8 Identity. To quote the poet Keats, it’s like unweaving a rainbow: things will be different once we know something too well. However, to quote David Hilbert, one of the foremost mathematicians of the 20th century, “Wir müssen wissen. Wir werden wissen.” (“We must know. We shall know.”) Very well, we’ll take the advice of the latter. To recall, we defined the expression Fk as,
Fk = (a+b+c)k + (b+c+d)k + (a-d)k - (c+d+a)k - (d+a+b)k - (b-c)k
Ramanujan knew that Fk = 0 for k = 2 and 4. (From this point on, k will be assumed to have those values unless indicated otherwise.) The author, in addition to quartic sextuples, was also looking at parametric solutions to,
a1k + a2k + a3k = b1k + b2k + b3k
and realized that Fk was just the special form,
a1k + a2k + (a1+a2)k = b1k + b2k + (b1+b2)k
The significance of this is that expressed in terms of the variables (p,q,r,s), the 6-10-8 Identity factors as,
64[p6+q6+(p+q)6-r6-s6-(r+s)6] [p10+q10+(p+q)10-r10-s10-(r+s)10]-45[p8+q8+(p+q)8-r8-s8-(r+s)8] = (p2+pq+q2-r2-rs-s2)P(z)
where P(z) is a homogeneous polynomial of degree 14. The problem is then reduced to finding expressions (p,q,r,s) such that,
p2+pq+q2-r2-rs-s2 = 0
or simply,
p2+pq+q2 = r2+rs+s2
the algebraic form mentioned earlier. One solution is given by Ramanujan’s version,
(b+c+d)2 + (b+c+d)(a-d) + (a-d)2 = (c+d+a)2 + (c+d+a)(b-c) + (b-c)2
for ad = bc, though this is just one out of many possible solutions with another two, the linear and quadratic forms given earlier, also satisfying it.
For Hirschhorn’s 3-7-5 Identity, it is,
25[p3+q3-(p+q)3-r3-s3+(r+s)3] [p7+q7-(p+q)7-r7-s7+(r+s)7]-21[p5+q5-(p+q)5-r5-s5+(r+s)5] = -525pqrs(p+q)(r+s)(p2+pq+q22-r2-rs-s2)2
hence is the same condition. In fact, the equation,
Rk = pk+qk+(p+q)k-rk-sk-(r+s)k = 0
factors as,
R2 = (p2+pq+q2-r2-rs-s2) = 0
R4 = (p2+pq+q2-r2-rs-s2) (p2+pq+q2+r2+rs+s2) = 0
therefore solutions to R2 and R4 automatically give solutions to the 6-10-8 and 3-7-5 identities. The connection to Ramanujan’s quartic sextuples which are explicitly,
(2x2+12xy-6y2)4 + (2x2-12xy-6y2)4 + (4x2-12y2)4 + (4x2+12y2)4 + (3x2 +9y2)4 = (5x2+15y2)4
(6x2-44xy-18y2)4 + (8x2+40xy-24y2)4 + (14x2-4xy-42y2)4 + (9x2+27y2)4 + (4x2 +12y2)4 = (15x2+45y2)4
can be seen considering that,
(2x2-12xy-6y2) + (2x2+12xy-6y2) = (4x2-12y2)
(6x2-44xy-18y2) + (8x2+40xy-24y2) = (14x2-4xy-42y2)
hence are sextuples of the form a4 + b4 + (a+b)4 + d4 + e4 = f4.
But a4 + b4 + (a+b)4 = 2(a2+ab+b2)2, a form that should be familiar by now. In general, ak + bk + (a+b)k = 2(a2+ab+b2)k/2 for k = 2 or 4 so it seems it might be worthwhile to investigate parametric solutions to the equation,
pk + qk + (p+q)k = rk + sk + (r+s)k
III. The Equation pk + qk + (p+q)k = rk + sk + (r+s)k (in Linear Forms)
Ramanujan gave other quartic identities, one the intriguing,
34 + (2x4-1)4 + (4x5+x)4 = (4x4+1)4 + (6x4-3)4 + (4x5-5x)4
and another as a family,
2(ab+ac+bc)2 = a4 + b4 + c4
2(ab+ac+bc)4 = a4(b-c)4 + b4(c-a)4 + c4(a-b)4
2(ab+ac+bc)6 = (a2b+b2c+c2a)4 + (ab2+bc2+ca2)4 + (3abc)4
where a+b+c = 0 and which he claimed would go on, though in his Notebooks he didn’t give the rest. In general, Ford’s theorem guarantees identities of the form,
2(a2+ab+b2)2m = P14 + P24 + P34
for some computable polynomials Pi. See Mathworld’s entry at http://mathworld.wolfram.com/FordsTheorem.html. Note that the first is equivalent to,
2(a2+ab+b2)2 = a4 + b4 + (a+b)4
and the second as,
2(a2+ab+b2)4 = a4(a+2b)4 + b4(-2a-b)4 + (a2-b2)4
While these are certainly interesting and worth mentioning, we are more concerned with the form pk + qk + (p+q)k = rk + sk + (r+s)k though it seems Ramanujan’s quartic family was headed virtually in the same direction. In fact, we will be meeting the expressions a+2b, 2a+b, and a-b again.
In Browning and Heath-Brown’s paper, Equal Sums of Three Powers, we find the identity,
(-23x+57y)4 + (40x+6y)4 + (17x+63y)4 = (-23x-57y)4 + (40x-6y)4 + (17x-63y)4
which Dickson attributes to V.G. Tariste (a typo “37” instead of “57” was corrected by this author, as well as minor sign changes). The identity, in fact, is valid for k = 2 or 4. It is not hard to see the pattern, nor the fact that it exploits opposite parity. However, we need not depend on parity, as can be shown by a variation found by the author,
(-23x-6y)k + (40x+63y)k + (17x+57y)k = (-23x-63y)k + (40x+57y)k + (17x-6y)k
There is really nothing special about these particular values as the general forms for these two are given by,
(ax+u1y)k + (bx+u2y)k + (cx+u3y)k = (ax-u1y)k + (bx-u2y)k + (cx-u3y)k
(ax-u2y)k + (bx+u3y)k + (cx+u1y)k = (ax-u3y)k + (bx+u1y)k + (cx-u2y)k
where u1 = a+2b, u2 = -(2a+b), u3 = -(a-b), and c = a+b. Note that the ui are the same expressions we encountered in Ramanujan’s quartic family! We have spared the reader the algebraic details but these were derived using the same method in the paper [1] where we simply expand the equation for k = 2 and 4, collect powers of (x,y), and solve the resulting system of equations equated to zero for the unknown ui.
As sums of one side of the equation, they are,
(ax+u1y)k + (bx+u2y)k + (cx+u3y)k = (ak + bk + (a+b)k)(x2+3y2)k/2
(ax-u2y)k + (bx+u3y)k + (cx+u1y)k = (ak + bk + (a+b)k)(x2+3xy+3y2)k/2
However, we can still generalize our system and we can have our first general identity,
Identity 1 (Id.1)
(ax+v1y)k + (bx+v3y)k + ((a+b)x+(v1+v3)y)k = (ax+w1y)k + (bx+w3y)k + ((a+b)x+(w1+w3)y)k
where v1, v3 are free variables with the two unknowns, w1, w3, given by,
w1 (a2+ab+b2) = (a2-b2)v1+(a2+2ab)v3
w3 (a2+ab+b2) = (2ab+b2)v1-(a2-b2)v3
with the form (a2+ab+b2) appearing again. Id.1 can then give a 6-parameter (a,b,x,y,v1,v3) solution to the 6-10-8 and 3-7-5 identities. Judicious choice of v1, v3 will give the two particular versions given earlier.
IV. The Equation pk + qk + (p+q)k = rk + sk + (r+s)k (in Quadratic Forms)
It is in the parametric solutions to the above equation that we can find the generalization to Ramanujan’s sextuple identities. For particular examples, we have,
(-23x2+57xy-23y2)k + (40x2+6xy+40y2)k + (17x2+63xy+17y2)k = (-23x2-57xy-23y2)k + (40x2-6xy+40y2)k + (17x2-63xy+17y2)k
(-23x2-6xy-23y2)k + (40x2+63xy+40y2)k + (17x2+57xy+17y2)k = (-23x2-63xy-23y2)k + (40x2+57xy+40y2)k + (17x2-6xy+17y2)k
Compare to the linear ones given previously. The similarity can be better appreciated considering the general forms are,
(ax2+2u1xy+amy2)k + (bx2+2u2xy+bmy2)k + (cx2+2u3xy+cmy2)k = (ax2-2u1xy+amy2)k + (bx2-2u2xy+bmy2)k + (cx2-2u3xy+cmy2)k
(ax2-2u2xy+amy2)k + (bx2+2u3xy+bmy2)k + (cx2+2u1xy+cmy2)k = (ax2-2u3xy+amy2)k + (bx2+2u1xy+bmy2)k + (cx2-2u2xy+cmy2)k
where the ui are the same, namely u1 = a+2b, u2 = -(2a+b), u3 = -(a-b), and c = a+b, with the variable m an additional free parameter (where m = 4, then the variable y scaled down in the example). If we take the sum of one side of the equation of the 1st version and let m = -3 we have,
Identity 2 (Id. 2)
(ax2+2u1xy-3ay2)k + (bx2+2u2xy-3by2)k + (cx2+2u3xy-3cy2)k = (ak + bk + (a+b)k)(x2+3y2)k
This turns out to be the “template” identity that is the basis for Ramanujan’s quartic sextuples! If ak + bk + (a+b)k + dk + ek = fk, then obviously ak + bk + (a+b)k = fk -dk - ek, so the tempplate is equivalent to,
(ax2+2u1xy-3ay2)k + (bx2+2u2xy-3by2)k + (cx2+2u3xy-3cy2)k + dk(x2+3y2)k + ek(x2+3y2)k = fk(x2+3y2)k
for the appropriate k. For example, given the equation 24 + 24 + 44 + 34 + 44 = 54, then, u1 = 6, u2 = -6, u3 = 0. Using these values, we have,
(2x2+12xy-6y2)4 + (2x2-12xy-6y2)4 + (4x2-12y2)4 + (3x2 +9y2)4 + (4x2+12y2)4 = (5x2+15y2)4
which is just Ramanujan’s identity! The problem of finding quadratic form solutions to quartic sextuples, at least for certain forms, is then reduced to simply finding “seeder” equations a4 + b4 + (a+b)4 + d4 + e4 = f4, though we can obviously generalize this for suitable quartic n-tuples, like the septuple,
24 + 44 + 64 + 64 + 64 + 74 = 94
And just like the linear forms, we can find an even more general identity. This involves solving the equation,
(ax2+v1xy+v2y2)k + (bx2+v3xy+v4y2)k + ((a+b)x2+(v1+v3)xy+(v2+v4)y2)k = (ax2+w1xy+w2y2)k + (bx2+w3xy+w4y2)k + ((a+b)x2+(w1+w3)xy+(w2+w4)y2)k
for k = 2 and 4. Since this is quite hard to solve, an assumption was made that quadratic forms q1 and q2 have the same discriminant, and q3 and q4 also as another pair. This gives linear relations that can reduce the complexity of our system of equations. With v1 and v3 as free variables, it turns out there are three linear solutions, though since a pair involves a mere transposition of variables, there are really just two:
Identity 3 (Id.3) First solution:
v12-4av2 = f1, v32-4bv4 = f1, w12-4aw2 = f1, w32-4bw4 = f1,
w1 (a2+ab+b2) = (a2-b2)v1 + (a2+2ab)v3
w3 (a2+ab+b2) = (2ab+b2)v1 - (a2-b2)v3
f1 = (bv1-av3)2/(a2+ab+b2)
Not surprisingly, w1 and w3 of Id. 3 (quadratic forms) are the same for Id. 1 (linear forms).
Identity 4 (Id.4) Second solution:
v12-4av2 = g1, v32-4bv4 = g1, w12-4aw2 = g2, w32-4bw4 = g2,
w1 (3(a+b)2) = (3a2+4ab-b2)v1 + 2(a2+2ab)v3
w3 (3(a+b)2) = 2(2ab+b2)v1 + (-a2+4ab+3b2)v3
g1 = (3a+b)(a+3b)(-bv1+av3)2/(3(a+b)4)
g2 = (a2-14ab+b2)(-bv1+av3)2/(9(a+b)4)
Both are 6-parameter (a,b,x,y,v1,v3) solutions as quadratic forms to the 6-10-8 and 3-7-5 identities. Note that for the first solution, both pairs of qi share the same discriminant and that the square-free part, namely (a2+ab+b2), factors over Ö(-3) while for the second solution, one pair has this as (a2-14ab+b2) which factors over Ö3.
As a sum of one side of the equation, Id. 3 gives us,
Identity 5 (Id.5)
(ax2+v1xy+v2y2)k + (bx2+v3xy+v4y2)k + ((a+b)x2+(v1+v3)xy+(v2+v4)y2)k = (ak+bk+(a+b)k)(ax2+v5xy+v6y2)k
v1, v3 (free variables)
v5 = ((2a+b)v1+(a+2b)v3)/(2(a2+ab+b2))
v2 = ((a+b)v12+2bv1v3-av32)/(4(a2+ab+b2))
v4 = (-bv12+2av1v3+(a+b)v32)/(4(a2+ab+b2))
v6 = (v12+v1v3+v32)/(4(a2+ab+b2))
which is the broader version of the template (Id. 2) given earlier so we have given all vi explicitly. By setting v5 = 0 which will define our v1 and v3, after simplication Id.5 reduces to Id.2.
Since it was mentioned earlier that we can use this approach to find parametrizations for quartic n-tuples, one may wonder if it applies to n< 6. Unfortunately, the equation a4 + b4 + (a+b)4 + d4 = e4 is equivalent to,
2(a2+ab+b2)2 + d4 = e4
and the diophantine equation 2r2 + s4 = t4 has been proven by T. Nagell to have no solution in the integers so there is no such quintuple. If a parametrization exists, then it would be for other forms.
While we now know how to generate an indefinite number of quadratic form solutions to quartic sextuples, the problem remains that one has to find appropriate equations first to serve as a “seed”. To facilitate finding these, we can take advantage of certain observations. The smallest sextuple is 24 + 24 + 44 + 34 + 44 = 54 and it’s hard to miss the Pythagorean triple 32 + 42 = 52 tucked away at the side. (This is not to say that appropriate equations we can use for our “template” must always involve such a triple since equations like 64 + 84 + 144 + 44 + 94 = 154 used by Ramanujan in his second identity will also do. The important condition is c = a+b.)
If we do limit ourselves to equations which involve these triples and since they have a complete parametrization known since classical times, namely,
(x2-y2)2 + (2xy)2 = (x2+y2)2
what would happen if we insert it into sextuples? It turns out that the equation,
a4 + b4 + (a+b)4 + (mn)4 + ((m2-n2)/2)4 = ((m2+n2)/2)4
has the factor,
mn(m2-n2) = 2(a2+ab+b2)
so given (m,n) of a Pythagorean triple if we can find rational (a,b), then we also find a sextuple. The factor is equivalent to the forms,
2mn(m2-n2) = (a+2b)2 + 3a2
2mn(m2-n2) = (2a+b)2 + 3b2
2mn(m2-n2) = (a-b)2 + 3(a+b)2
or simply finding the representation of 2mn(m2-n2) as p2+3q2. Note that these are the expressions a+2b, 2a+b, and a-b again. Finally, this is also equivalent to expressing it as a sum of three squares,
mn(m2-n2) = a2 + b2 + (a+b)2
The first six primitive sextuples of such form are,
24 + 24 + 44 + 34 + 44 = 54
44 + 224 + 264 + 214 + 284 = 354
24 + 444 + 464 + 394 + 524 = 654
124 + 244 + 364 + 164 + 634 = 654
24 + 324 + 344 + 134 + 844 = 854
164 + 224 + 384 + 134 + 844 = 854
each of which can give a quadratic form identity. While the first three involve the Pythagorean triple (3r)2 + (4r)2 = (5r)2 for r = 1, 7, 13, we are not necessarily limited to this as proven by the rest which involves 162 + 632 = 652 and 132 + 842 = 852.
Example. Let m = 19, n = 7, F(m,n) = 2mn(m2-n2) = 2(19)(7)(192-72). But F(m,n) is also equal to,
F(m,n) = (4+2*142)2 + 3(4)2
F(m,n) = (2*4+142)2 + 3(142)2
F(m,n) = (4-142)2 + 3(4+142)2
so we find (a,b) as (4, 142), thus 44 + 1424 + 1464 + (19*7)4 + 1564 = 2054. Of course, (19*7)2 + 1562 = 2052.
It should be pointed out that the representation of 2mn(m2-n2) as p2+3q2 may be in more than three ways, and hence give us more than one (a,b), as can be seen with the last two of the six examples above, with (a,b) as (2,32) and (16,22), and may partly explain some of the “clustering of sums” discussed in a separate article (and which also includes a list of the first 41 sextuples).
In this paper, we discussed certain quartic identities which we used as generalizing principles to better understand some of Ramanujan’s equations, like his 6-10-8 Identity and formulas for equal sums of like powers. In the process, we attained the objective of generalizing these so we can now, among other things, generate an indefinite number of formulas for the quartic sextuple,
a14 + a24 + a34 + a44 + a54 = a64
using Identity 2 (or Id. 5), up from only three known previously. And regarding his 6-10-8 Identity, while we can now find new versions of it, the question remains how he found it in the first place. It was obviously not by factoring the 16th degree polynomial like what we did. Nor could it have been how Hirschhorn found the 3-7-5 Identity, otherwise he would have listed down both in his notebooks. As usual, we have the benefit of Ramanujan making his way though the mathematical jungle, and we are just walking on the cleared footpath.
There are other questions we can ask. First, is there a quadratic form identity for quartic quintuples a14 + a24 + a34 + a44 = a54? Using the general method of expanding such forms and collecting powers of (x,y), we can actually have a system of eight equations in eight unknowns (since two of the vi, v1 and v3, will be assumed to be free variables) and hence resolvable to a single resolvent. What is smallest degree of its factors? The high degrees involved simply make it beyond a brute force approach.
Second, are there other kinds of quadratic forms for quartic sextuples? In this paper, we depended on the condition c = a+b, but is it possible there might be other conditions we can use or even a general identity that uses no conditions at all, like the one for cubics?
Third, why do Eisenstein integers figure so prominently? To recall, the form a2+ab+b2 was ubiquitous. But this factors over the Eisenstein integers as (a-bw)(a-bw2), where w is a complex cube root of unity. And a simple linear substitution of a = p+q, b = -p+q makes it p2+3q2. The multiplicative properties of this form was used by Euler to find the complete rational parametrization of a3+b3+c3+d3 = 0. So what makes it so special? It is because Eisenstein integers have a geometric interpretation as a triangular lattice? Note that they are close cousins to the Gaussian integers (this in turn is a square lattice) which are connected to the form p2+q2, involved in the complete parametrization of Pythagorean triples a2+b2 = c2. And finally, Gaussian and Eisenstein integers are the most symmetrical lattices in 2 dimensions, having 4-fold and 6-fold rotational symmetry, respectively. It would be nice if all of these were somehow connected.
Exponents, however, do not stop with the fourth. Ramanujan did not have any diophantine formulas for equal sums of like powers for fifth powers and above. Fortunately, it seems the torch has been passed and others have taken up the task of finding quadratic form identities (if we limit ourselves to these) for higher powers, with a quintic octuple found by G. Xeroudakes and A. Moessner in 1958 and independently (as a general case) by L.J. Lander in 1968 for k = 1, 3, 5,
(x2-4xy-9y2)k + (3x2+16xy+17y2)k + (3x2+8xy+y2)k + (x2+12xy+23y2)k + (-x2-8xy-3y2)k + (-x2+13y2)k + 2(-3x2-12xy-21y2)k = 0
It turns out this has a connection to the quartic identities discussed here. But again, that is another story.
--End--
© 2005
Titus Piezas III
Sept. 20, 2005
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