0018: Article 8 (Sagan's Identity)





Sagan’s Identity and Ramanujan’s 6-10-8 Identity


By Tito Piezas III




Prologue:  Back in 2008, I dedicated an algebraic identity I found to the astronomer and science writer Carl Sagan (1934-1996).  I read his "Broca's Brain" and "The Dragons of Eden" in my late teens, and read and eventually saw "Contact".  Since I already dedicated one article I wrote on Degen's Eight-Square Identity to the novelist Katherine Neville, author of the amazing book "The Eight", I thought it was only fitting I name one of the identities I found after Sagan.  After all, it has billions and billions of solutions.  (You need to be a Sagan fan to understand the previous remark.) 


This article was originally posted in one of my Geocities pages.  Since Yahoo unfortunately closed that domain (last October 2009), I'm transfering it, with some edits, to my Google Sites pages since today, November 7, 2009, is the first Carl Sagan Day.



I. Introduction

II. Ramanujan’s 6-10-8 Identity

III. Hirshhorn’s 3-7-5 Identity

IV. Sagan’s Identity

V. Conclusion: Going Beyond



I. Introduction


The remarkable Ramanujan 6-10-8 Identity 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.  Its form and use of high exponents is certainly unusual.  This can be more concisely expressed as, define,


Fk: = (a+b+c)k + (b+c+d)k + (a-d)k - (c+d+a)k - (d+a+b)k - (b-c)k


If ad = bc, then 64F6F10 = 45F82.  Note also that the terms can be related,


(b+c+d) + (a-d) = (a+b+c),

(c+d+a) + (b-c) = (d+a+b),


as well as F2 = F4 = 0. These details will be important later.  On the other hand, Sagan’s Identity, named by this author after the astronomer and science writer Carl Sagan (1934-1996), is given by the multi-grade equation,


1 + 5k + (3+2y)k + (3-2y)k + (-3+3y)k + (-3-3y)k = (-2+x)k + (-2-x)k + (5-y)k + (5+y)k


for k = 1,3,5,7 where x2-10y2 = 9.


Such equations appear in the context of the Prouhet-Tarry-Escott Problem.  While Ramanujan’s 6-10-8 Identity involves even exponents and Sagan’s has odd ones, and seem to be of very different forms, it can be shown that both fundamentally have their basis in the same simple system of equations, call this S4, as ak+bk+ck = dk+ek+fk,  for k = 2,4.



II. Ramanujan’s 6-10-8 Identity


Though employing high exponents, it has its roots on the much simpler S4.  First, some preliminaries.


Theorem 1:  If ak+bk+ck = dk+ek+fk for k = 2,4.  Define {x,y} = {a2+b2+c2,  a4+b4+c4}. Then,


3(a8+b8+c8-d8-e8-f8) = 4(a6+b6+c6-d6-e6-f6)(x)                          (eq.1)


6(a10+b10+c10-d10-e10-f10) = 5(a6+b6+c6-d6-e6-f6)(x2+y)            (eq.2)


One can combine eq.1 and eq.2.


Theorem 2:  If ak+bk+ck = dk+ek+fk for k = 2,4.  Then,


32(a6+b6+c6-d6-e6-f6)(a10+b10+c10-d10-e10-f10) = 15(n+1)(a8+b8+c8-d8-e8-f8)2


where n is defined by a4+b4+c4 = n(a2+b2+c2)2.


This is the more general case of Ramanujan’s 6-10-8 Identity. 


Proof:  Define Ek := ak+bk+ck – (dk+ek+fk). So eq.1 and eq.2 become,


3E8 = 4E6(x)                 (eq.1)

6E10 = 5E6(x2+y)          (eq.2)


Square eq.1, multiply eq.2 by E6, and then eliminate E62 between eq.1 and eq.2 to get:


32E6E10 = 15E82 + 15E82y/x2




32E6E10 = 15E82 (1+y/x2)

32E6E10 = 15E82 (1+n)


(End of proof)


Corollary:  If ak+bk+ck = dk+ek+fk for k = 2,4 where a+b = c. Then


64(a6+b6+c6-d6-e6-f6)(a10+b10+c10-d10-e10-f10) = 45(a8+b8+c8-d8-e8-f8)2


This is Ramanujan’s 6-10-8 Identity phrased differently. 


Proof:  It is easily shown that,


a4+b4+c4 = n(a2+b2+c2)2


when a+b = c, then n = 1/2. Substituting n = 1/2 into Theorem 2, one gets this corollary.  (End of proof)


Thus, as claimed, this remarkable identity has its roots on the properties of S4 with the variables subject to certain constraints.



III. Hirshhorn’s 3-7-5 Identity


Inspired by the 6-10-8, there's also Hirshhorn's 3-7-5 Identity:


25[(a+b+c)3-(b+c+d)3-(a-d)3+(c+d+a)3-(d+a+b)3+(b-c)3][(a+b+c)7-(b+c+d)7-(a-d)7+ (c+d+a)7-(d+a+b)7+(b-c)7] = 21[(a+b+c)5-(b+c+d)5-(a-d)5+(c+d+a)5-(d+a+b)5+(b-c)5]2


again with ad = bc.  In fact, the two can be combined:


Theorem:  If ak+bk+ck = dk+ek+fk, k = 2,4 where a+b+c = d+e+f = 0, then,


64(a6+b6+c6-d6-e6-f6)(a10+b10+c10-d10-e10-f10) = 45(a8+b8+c8-d8-e8-f8)2  


25(a3+b3+c3-d3-e3-f3)(a7+b7+c7-d7-e7-f7) = 21(a5+b5+c5-d5-e5-f5)2         


Proof:  By doing the substitution {a,b,c,d,e,f} = {p, q, -p-q, r, s, -r-s}, all three equations have the common factor p2+pq+q2-(r2+rs+s2) = 0, so the problem is reduced to finding expressions {p,q,r,s} that satisfy this.



IV. Sagan’s Identity


Compared to the ones above, this has the very different form:


1 + 5k + (3+2y)k + (3-2y)k + (-3+3y)k + (-3-3y)k = (-2+x)k + (-2-x)k + (5-y)k + (5+y)k,   for k = 1,3,5,7


where x2-10y2 = 9.


This has billions and billions (Sagan's favorite catchphrase) of solutions.  In fact, there is an infinite number of them, whether rational, or integral if the condition is to be solved as a Pell-like equation.  Again, this identity can be shown to depend on the properties of S4, but with its variables subject to different constraints.  As before, some preliminaries,


Sinha’s Theorem:  If,


(a+3c)k + (b+3c)k + (-a-b+2c)k = (c+d)k + (c+e)k + (2c-d-e)k,  for k = 2,4, then,


ak + (a+2c)k + bk + (b+2c)k + (-c+d+e)k = (a+b-c)k + (a+b+c)k + dk + ek + (3c)k,  for k = 1,3,5,7


excluding the trivial case c = 0.  (Sinha stated his theorem differently, but in this article it was re-phrased it to make it simpler.)  Note that the variables of S4, if expressed as {xi, yi}, have the constraint x1+x2+x3 = 2(y1+y2+y3) = 8c, excluding c = 0, hence all solutions applicable to the 6-10-8 and 3-7-5 Identities must be avoided. 


Sinha gave only one solution in terms of a binary quadratic form (with discriminant D = 70), but this author found others with a different square-free discriminant (D = 10), one of which is,


(y+6)k + (3y+4)k + (4y-2)k = (x+y-1)k + (-x+y-1)k + (2y+6)k,   for k = 2,4


where x2-10y2 = 9.


This obeys the constraints of Sinha's theorem.  By solving for {a,b,c,d} and constructing the terms of the k = 1,3,5,7 system, one can recover Sagan's Identity after some minor algebraic tweaking.  (This author knows of only four solutions as binary quadratic forms that satisfy Sinha’s theorem, all with discriminant D that, square-free and unsigned, is either D = {10, 70}.  Are there others with different D?) 



V. Conclusion: Going Beyond


The next step after S4 would be S5 and S6, or ak+bk+ck+dk = ek+fk+gk+hk,  for k = 1,3,5, or 2,4,6.  There are solutions for these, one of which for the latter is the beautifully simple result given by Chernick and Escott:


If, 1k + 2k + 3k = uk + vk + (u+v)k  for k = 2,4, then, 


(-u+7)k + (u-2v+1)k + (-3u-1)k + (3u+2v+1)k = (u+7)k + (-u+2v+1)k + (3u-1)k + (-3u-2v+1)k,   for k = 1,2,4,6.


This system depends on solving u2+uv+v2 = 7 in the rationals, which is easily done.  Analogous general identities similar to the 6-10-8 can be found using even higher powers, but none are known that have the simple form found by Ramanujan.


If we define S7 and S8 as ak+bk+ck+dk+ek = fk+gk+hk+ik+jk, for k = 1,3,5,7, or 2,4,6,8, there are several parametric solutions known for the former, but only one for the latter, the beautifully simple,


(a+c)k + (a-c)k + (3b+d)k + (3b-d)k + (4a)k = (3a+c)k + (3a-c)k + (b+d)k + (b-d)k + (4b)k


for k = 1,2,4,6,8, where,


a2+12b2 = c2 

12a2+b2 = d2


This was found by Letac and Sinha and can be reduced to an elliptic curve, hence has an infinite number of solutions.  Is there a solution to S8 in terms of binary quadratic forms?  No one knows. Then there’s S9, S10, etc….



P.S Beautiful music from Sting, “Shape of My Heart”:





– End –



© Dec 2010,  Tito Piezas III

Written:  Sept. 7, 2008

Revised:  Nov. 7, 2009 (Carl Sagan Day)

You can email author at tpiezas@gmail.com



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