Sagan’s Identity and Ramanujan’s 6108 Identity
By Tito Piezas III
Prologue: Back in 2008, I dedicated an algebraic identity I found to the astronomer and science writer Carl Sagan (19341996). 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 EightSquare 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 6108 Identity III. Hirshhorn’s 375 Identity IV. Sagan’s Identity V. Conclusion: Going Beyond
I. Introduction
The remarkable Ramanujan 6108 Identity is given by:
64[(a+b+c)^{6}+(b+c+d)^{6}+(ad)^{6}(c+d+a)^{6}(d+a+b)^{6}(bc)^{6}][(a+b+c)^{10}+(b+c+d)^{10}+(ad)^{10}(c+d+a)^{10}(d+a+b)^{10}(bc)^{10}] = 45[(a+b+c)^{8}+(b+c+d)^{8}+(ad)^{8}(c+d+a)^{8}(d+a+b)^{8}(bc)^{8}]^{2}
where ad = bc. Its form and use of high exponents is certainly unusual. This can be more concisely expressed as, define,
F_{k}: = (a+b+c)^{k} + (b+c+d)^{k} + (ad)^{k}  (c+d+a)^{k}  (d+a+b)^{k}  (bc)^{k}
If ad = bc, then 64F_{6}F_{10} = 45F_{8}^{2}. Note also that the terms can be related,
(b+c+d) + (ad) = (a+b+c), (c+d+a) + (bc) = (d+a+b),
as well as F_{2} = F_{4} = 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 (19341996), is given by the multigrade equation,
1 + 5^{k} + (3+2y)^{k} + (32y)^{k} + (3+3y)^{k} + (33y)^{k} = (2+x)^{k} + (2x)^{k} + (5y)^{k} + (5+y)^{k}
for k = 1,3,5,7 where x^{2}10y^{2} = 9.
Such equations appear in the context of the ProuhetTarryEscott Problem. While Ramanujan’s 6108 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 S_{4}, as a^{k}+b^{k}+c^{k} = d^{k}+e^{k}+f^{k}, for k = 2,4.
II. Ramanujan’s 6108 Identity
Though employing high exponents, it has its roots on the much simpler S_{4}. First, some preliminaries.
Theorem 1: If a^{k}+b^{k}+c^{k} = d^{k}+e^{k}+f^{k} for k = 2,4. Define {x,y} = {a^{2}+b^{2}+c^{2}, a^{4}+b^{4}+c^{4}}. Then,
3(a^{8}+b^{8}+c^{8}d^{8}e^{8}f^{8}) = 4(a^{6}+b^{6}+c^{6}d^{6}e^{6}f^{6})(x) (eq.1)
6(a^{10}+b^{10}+c^{10}d^{10}e^{10}f^{10}) = 5(a^{6}+b^{6}+c^{6}d^{6}e^{6}f^{6})(x^{2}+y) (eq.2)
One can combine eq.1 and eq.2.
Theorem 2: If a^{k}+b^{k}+c^{k} = d^{k}+e^{k}+f^{k} for k = 2,4. Then,
32(a^{6}+b^{6}+c^{6}d^{6}e^{6}f^{6})(a^{10}+b^{10}+c^{10}d^{10}e^{10}f^{10}) = 15(n+1)(a^{8}+b^{8}+c^{8}d^{8}e^{8}f^{8})^{2}
where n is defined by a^{4}+b^{4}+c^{4} = n(a^{2}+b^{2}+c^{2})^{2}.
This is the more general case of Ramanujan’s 6108 Identity.
Proof: Define E_{k} := a^{k}+b^{k}+c^{k} – (d^{k}+e^{k}+f^{k}). So eq.1 and eq.2 become,
3E_{8} = 4E_{6}(x) (eq.1) 6E_{10} = 5E_{6}(x^{2}+y) (eq.2)
Square eq.1, multiply eq.2 by E_{6}, and then eliminate E_{6}^{2} between eq.1 and eq.2 to get:
32E_{6}E_{10} = 15E_{8}^{2} + 15E_{8}^{2}y/x^{2}
Or,
32E_{6}E_{10} = 15E_{8}^{2} (1+y/x^{2}) 32E_{6}E_{10} = 15E_{8}^{2} (1+n)
(End of proof)
Corollary: If a^{k}+b^{k}+c^{k} = d^{k}+e^{k}+f^{k} for k = 2,4 where a+b = c. Then
64(a^{6}+b^{6}+c^{6}d^{6}e^{6}f^{6})(a^{10}+b^{10}+c^{10}d^{10}e^{10}f^{10}) = 45(a^{8}+b^{8}+c^{8}d^{8}e^{8}f^{8})^{2}
This is Ramanujan’s 6108 Identity phrased differently.
Proof: It is easily shown that,
a^{4}+b^{4}+c^{4} = n(a^{2}+b^{2}+c^{2})^{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 S_{4} with the variables subject to certain constraints.
III. Hirshhorn’s 375 Identity
Inspired by the 6108, there's also Hirshhorn's 375 Identity:
25[(a+b+c)^{3}(b+c+d)^{3}(ad)^{3}+(c+d+a)^{3}(d+a+b)^{3}+(bc)^{3}][(a+b+c)^{7}(b+c+d)^{7}(ad)^{7}+ (c+d+a)^{7}(d+a+b)^{7}+(bc)^{7}] = 21[(a+b+c)^{5}(b+c+d)^{5}(ad)^{5}+(c+d+a)^{5}(d+a+b)^{5}+(bc)^{5}]^{2}
again with ad = bc. In fact, the two can be combined:
Theorem: If a^{k}+b^{k}+c^{k} = d^{k}+e^{k}+f^{k}, k = 2,4 where a+b+c = d+e+f = 0, then,
64(a^{6}+b^{6}+c^{6}d^{6}e^{6}f^{6})(a^{10}+b^{10}+c^{10}d^{10}e^{10}f^{10}) = 45(a^{8}+b^{8}+c^{8}d^{8}e^{8}f^{8})^{2}
25(a^{3}+b^{3}+c^{3}d^{3}e^{3}f^{3})(a^{7}+b^{7}+c^{7}d^{7}e^{7}f^{7}) = 21(a^{5}+b^{5}+c^{5}d^{5}e^{5}f^{5})^{2}
Proof: By doing the substitution {a,b,c,d,e,f} = {p, q, pq, r, s, rs}, all three equations have the common factor p^{2}+pq+q^{2}(r^{2}+rs+s^{2}) = 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 + 5^{k} + (3+2y)^{k} + (32y)^{k} + (3+3y)^{k} + (33y)^{k} = (2+x)^{k} + (2x)^{k} + (5y)^{k} + (5+y)^{k}, for k = 1,3,5,7
where x^{2}10y^{2} = 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 Pelllike equation. Again, this identity can be shown to depend on the properties of S_{4}, but with its variables subject to different constraints. As before, some preliminaries,
Sinha’s Theorem: If,
(a+3c)^{k} + (b+3c)^{k} + (ab+2c)^{k} = (c+d)^{k} + (c+e)^{k} + (2cde)^{k}, for k = 2,4, then,
a^{k} + (a+2c)^{k} + b^{k} + (b+2c)^{k} + (c+d+e)^{k} = (a+bc)^{k} + (a+b+c)^{k} + d^{k} + e^{k} + (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 rephrased it to make it simpler.) Note that the variables of S_{4}, if expressed as {x_{i}, y_{i}}, have the constraint x_{1}+x_{2}+x_{3} = 2(y_{1}+y_{2}+y_{3}) = 8c, excluding c = 0, hence all solutions applicable to the 6108 and 375 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 squarefree discriminant (D = 10), one of which is,
(y+6)^{k} + (3y+4)^{k} + (4y2)^{k} = (x+y1)^{k} + (x+y1)^{k} + (2y+6)^{k}, for k = 2,4
where x^{2}10y^{2} = 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, squarefree and unsigned, is either D = {10, 70}. Are there others with different D?)
V. Conclusion: Going Beyond
The next step after S_{4} would be S_{5} and S_{6}, or a^{k}+b^{k}+c^{k}+d^{k} = e^{k}+f^{k}+g^{k}+h^{k}, 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, 1^{k} + 2^{k} + 3^{k} = u^{k} + v^{k} + (u+v)^{k} for k = 2,4, then,
(u+7)^{k} + (u2v+1)^{k} + (3u1)^{k} + (3u+2v+1)^{k} = (u+7)^{k} + (u+2v+1)^{k} + (3u1)^{k} + (3u2v+1)^{k}, for k = 1,2,4,6.
This system depends on solving u^{2}+uv+v^{2} = 7 in the rationals, which is easily done. Analogous general identities similar to the 6108 can be found using even higher powers, but none are known that have the simple form found by Ramanujan.
If we define S_{7} and S_{8} as a^{k}+b^{k}+c^{k}+d^{k}+e^{k} = f^{k}+g^{k}+h^{k}+i^{k}+j^{k}, 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} + (ac)^{k} + (3b+d)^{k} + (3bd)^{k} + (4a)^{k} = (3a+c)^{k} + (3ac)^{k} + (b+d)^{k} + (bd)^{k} + (4b)^{k}
for k = 1,2,4,6,8, where,
a^{2}+12b^{2} = c^{2} 12a^{2}+b^{2} = d^{2}
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 S_{8} in terms of binary quadratic forms? No one knows. Then there’s S_{9}, S_{10}, etc….
P.S. Beautiful music from Sting, “Shape of My Heart”:
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© Dec 2010, Tito Piezas III Written: Sept. 7, 2008 Revised: Nov. 7, 2009 (Carl Sagan Day) You can email author at tpiezas@gmail.com
