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Help generalizing Liouville's Polynomial Identity

• To: mathgroup at smc.vnet.net
• Subject: [mg103596] Help generalizing Liouville's Polynomial Identity
• From: TPiezas <tpiezas at gmail.com>
• Date: Tue, 29 Sep 2009 07:40:23 -0400 (EDT)

Hello all,

"Liouville's polynomial identity" is given by
http://mathworld.wolfram.com/LiouvillePolynomialIdentity.html and can
be concisely encoded as,

6(x1^2 + x2^2 + x3^2 + x4^2)^2 = Sum(x_i +/- x_j)^4

To determine the number of terms of the summation, since we are to
choose 2 objects from 4, then this is Binomial[4,2] = 6. But as there
are 2 sign changes, then total is 2 x 6 = 12 terms, given explicitly
in the link above.  Going higher, and choosing 3 objects out of n
variables, I found that,

60(x1^2 + x2^2 + ... + x7^2)^2 = Sum(x_i +/- x_j +/- x_k)^4

is true. The RHS contains 4 x 35 = 140 terms. (If the x_i are non-
zero, one cannot express the square of the sum of less than 7 squares
in a similar manner.)

Question: If we take the next step,

a(x1^2 + x2^2 + ... + x_n^2)^2 = Sum(x_i +/- x_j +/- x_k +/- x_m)^4

for some positive integer "a", then what is the least n, if we are to
choose 4 objects at a time out of the x_n?

I believe the RHS is a simple matter for Mathematica to calculate, and
one can incrementally test n = 5,6,7,...etc until a neat identity is
found.

- Titus

• Follow-Ups:
  • Re: Help generalizing Liouville's Polynomial Identity
    • From: Daniel Lichtblau <danl@wolfram.com>