Re: polynomials whose values are all perfect squares
- From: quasi <quasi@xxxxxxxx>
- Date: Sun, 23 Sep 2007 15:23:33 -0400
On Sun, 23 Sep 2007 13:21:22 -0500, Robert Israel
<israel@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:
quasi <quasi@xxxxxxxx> writes:
On Fri, 21 Sep 2007 20:09:33 -0500, Robert Israel
<israel@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:
quasi <quasi@xxxxxxxx> writes:
Let f be an n-variate polynomial with integer coefficients, regarded
as a function from Z^n to Z. Suppose every element of f(Z^n) is the
square of an integer. Must f = g^2 for some n-variate polynomial g
with integer coefficients?
See the sci.math.research thread "Square of a polynomial?" from
November 1998,
<http://groups.google.ca/group/sci.math.research/browse_frm/thread/1a25a2c11
<f419ecb>
Thanks.
Some very interesting responses in that thread.
As far as I can see, it gives a "yes" answer to my question only for
the case of _univariate_ integer polynomials.
However, for my question, the polynomial f is allowed to be
multivariate.
Since the multivariate case can be specialized (in infinitely many
ways) to the univariate case by selecting values for all but one
variable, it's almost certain that the answer to the multivariate
question is also "yes".
But to prove it will require some further work.
I'll think about it, but if anyone else sees how to finish it, please
do.
It seems to me an induction on n should work.
Write your polynomial as f(x,y) where x is the first variable
and y the other n-1. The degree of f in x must be even (since f
takes no negative values on Z^n), say 2m. Write
f(x,y) = sum_{j=0}^{2m} a_j(y) x^j, where a_j(y) are polynomials
in y with integer coefficients. We must have a_{2m}(y) >= 0
for all y in Z^(n-1).
Given the univariate case, you can say that for every
y in Z^(n-1), f(x,y) has a square root g_y(x) that is a
polynomial in x with integer coefficients, and degree half
that of f(.,y): say
g_y(x) = sum_{j=0}^m b_j(y) x^j, where b_j(y) is an integer
for each y in Z^(n-1). Now b_m(y)^2 = a_{2m}(y), so
a_{2m} is a polynomial in n-1 variables with integer coefficients
whose values on Z^(n-1) are all squares.
By the induction hypothesis, b_m(y) (if we choose the signs right)
is equal to a polynomial with integer coefficients. Now b_{m-1}(y)
satisfies a_{2m-1}(y) = 2 b_{m-1}(y) b_m(y), which makes b_{m-1}(y)
equal to a rational function. Proceeding in this way for each b_j
from m-1 down to 0, we find that there is a rational function (say
A(x,y)/B(x,y) in lowest terms) of x and y such that
g_y(x) = A(x,y)/B(x,y) for y in Z^(n-1). But if B(x,y)
was non-constant, f(x,y) = A^2/B^2 wouldn't be a polynomial.
Thanks.
I'll need some time to look this over, however I'm low on time for the
next few days. The overall plan of attack seems right, though.
Presumably the argument should work in the same way for perfect k'th
powers, where k is an arbitrary fixed integer greater than 1, no?
quasi
.
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- From: Robert Israel
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