[seqfan] Sums of squares of polynomials
franktaw at netscape.net
franktaw at netscape.net
Thu Dec 4 20:49:12 CET 2008
There's a problem that's been bugging me, off and on, for quite
a while. I can't seem to find any real handle on it.
Not every polynomial with integer coefficients can be represented
as the sum of squares. Any such polynomial must obviously have
even degree; almost as obvious, the coefficients of odd powers
of x must be even. An additional constraint is that the polynomial
must be non-negative on the entire real number line (which
actually implies that the degree is even). I don't know if these
conditions are sufficient, but this is not the main focus of my
question.
The question is, given such a polynomial, how many squares are
required to represent it as the sum of squares? At first it seems
obvious that there is no upper bound -- the polynomial
1 + x^2 + x^6 + x^14 + ... + x^(2^n-2)
seems to require n squares. But this is not so clear; if we start
with
(x - x^(2^(n-1)-1)^2
we can then have two squares of polynomials with degree
2^(n-2), and from there it gets complicated.
In spite of these difficulties, I strongly suspect that a sum of
monomials with sufficiently fast growing exponents will require
an unbounded number of squares.
So the next question is, if we limit our attention to polynomials
of degree 2n, what is the maximum number of squares required?
a(0) = 4, from the sum of 4 squares theorem. I don't know
how to prove that a(n) is finite for n > 0, but again I strongly
suspect that it is.
It is easy to see that a(1) >= 5: just look at
x^2 + 7
There can be only one square that is not a constant, and that
must be x. It then requires 4 squares to get the 7.
But that's the sum of what I know: a(0) = 4, a(1) >= 5, I
can't prove a(n) is finite for n > 1, and I can't prove that the
sequence is unbounded. Any light anybody can shed on the
problem would be greatly appreciated.
Franklin T. Adams-Watters
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