Help generalizing Lebesgue Polynomial Identity
TPiezas
The Lebesgue Polynomial Identity is given by,
(a^2+b^2-c^2-d^2)^2 + (2ac+2bd)^2 +(2ad-2bc)^2 = (a^2+b^2+c^2+d^2)^2
This can be generalized to the form,
x1^2 + x2^2 +... + x_m^2 = (y1^2 + y2^2 +... + y_n^2)^2
It can be seen there are polynomial identities with integer
coefficients for all m = n.
Q1: But what if m < n?
I found that the one given by Lebesgue belongs to a family involving
normed algebras so that there are {m,n} for {3,4}, {5,8}, and {9,16}.
These can be extended to cover all n EXCEPT n = {3, 5, 8w+1}. The
first n is easy to dispose of, but the other two remain. To take the
second n:
Q2: Can we identically solve x1^2 + x2^2 + x3^2 + x4^2 =
(y1^2+y2^2+y3^2+y4^2+y5^2)^2 where the x_i are non-trivial
POLYNOMIALS in the y_i with integer coefficients? (Trivial would be
all but one x_i as equal to zero.)
P.S. We are not interested in numerical solns as, by the Lagrange Four-
Square Theorem, one can always express any positive integer as the sum
of four squares.
See http://sites.google.com/site/tpiezas/005b for more details.
- Titus
alainv...@gmail.com
Dear Friend,
A good and known way to generate equalities is considering partition
of integer polynomials.
Ex:two variables, 2^2 terms:
p(s,t)= a1s^2t^2+a2s^2t+a3st^2+a4st (1)
where a1s^2t^2 stands for all terms s^2n*t^2p , n,p positive integer.
Needed properties of (1) is that we can isolate every terms of it,
and that in usual operations (sum,product,integer power) some
terms remain distinct.
So calculating s^2t^2 terms of (p(s,t))^2 two ways, it yields
with p(s,t)=a,p(-s,t)=b,p(s,-t)=c,p(-s,-t)=d
4(a^2+b^2+c^2+d^2)
=(a+b+c+d)^2+(a-b-c+d)^2+(a-b+c-d)^2+(a+b-c-d)^2 (2)
Here, d=0 gives us (2a)^2+(2b)^2+(2c)^2
= (a+b+c)^2+(a-b-c)^2+(a-b+c)^2+(a+b-c)^2
Generally, 'truncature' works up to a given number of null terms:
with c=d=0 in (2) , we obtain double squares RHS.
This process works for n variables polynomials with 2^n terms
I did get 8 and 16 terms square equalities,
Best regards,
Alain