From: Gareth McCaughan
Newsgroups: sci.math.research
Subject: Re: Square of a polynomial ?
Date: 05 Nov 1998 16:22:54 +0000
Bruno Langlois wrote:
> Let P a monic polynomial in Z[X].
> We suppose that :
> - the degree of P is even,
> - for all positive integer n, P(n) is a square in Z.
> Can we say that P is a square in Z[X] ?
Yes.
Use the binomial theorem to write root(p(X)) = q(X) + O(1/X),
so that p(X) = q(X)^2 + O(X^{n-1}), where deg p = 2n.
For X very large such that p(X) isn't equal to q(X)^2,
|p(X)-q(X)^2| >= q(X)^2 - [q(X)-1]^2 = 2q(X)-1.
This is impossible for large X, since we just showed the LHS
can't grow that fast.
Hence, for all large enough X, p(X) = q(X)^2. Hence the polynomial
p-q^2 has infinitely many zeros, and is therefore identically zero. []
--
Gareth McCaughan Dept. of Pure Mathematics & Mathematical Statistics,
gjm11@dpmms.cam.ac.uk Cambridge University, England.
==============================================================================
From: israel@math.ubc.ca (Robert Israel)
Newsgroups: sci.math.research
Subject: Re: Square of a polynomial ?
Date: 5 Nov 1998 17:13:05 GMT
In article <36419B77.1C1A@club-internet.fr>,
langlois bruno wrote:
>Let P a monic polynomial in Z[X].
>We suppose that :
>- the degree of P is even,
>- for all positive integer n, P(n) is a square in Z.
>Can we say that P is a square in Z[X] ?
Yes, it must be a square. If p(z) is monic with degree 2n,
consider the Laurent series of q(z) = z^n sqrt(p(z)/z^(2n))
(taking the principal branch of the square root)
on |z| > R = max{ |r|: p(r)=0 }.
If p(z) is not a square, q(z) is not a polynomial, and we have
q(z) = r(z) + a_k z^(-k) + O(|z|^(-k-1))
where r(z) is a polynomial with rational coefficients and a_k \ne 0.
If M is the least common multiple of the denominators of the coefficients
in r(z), M (q(z)- r(z)) can't be an integer when z is a sufficiently large
integer, so neither can sqrt(p(z)).
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
==============================================================================
From: Mark Sapir
Newsgroups: sci.math.research
Subject: Re: Square of a polynomial ?
Date: Thu, 05 Nov 1998 12:01:25 -0600
langlois bruno wrote:
>
> Let P a monic polynomial in Z[X].
> We suppose that :
> - the degree of P is even,
> - for all positive integer n, P(n) is a square in Z.
> Can we say that P is a square in Z[X] ?
It can be a nice high school olympiad problem. Here is a sketch of an
elementary solution: if f(x) has degree 2n, find a polynomial g(x) such
that g(x)^2 has the same first n+1 coefficients as f(x) (it is always
possible to do); g(x) will have ratioanal coefficients, which is not a
problem, just multiply the variable x by a proper number, and you get a
polynomial with integer coefficients. Then use the fact that the
distance between two integer squares a^2-b^2 cannot be much smaller than
2*min{a,b}.
--
Mark Sapir
Professor of Mathematics
Vanderbilt University
Nashville, TN 37240
msapir@math.vanderbilt.edu
(615)322-6657
==============================================================================
From: rusin@math.niu.edu (Dave Rusin)
Newsgroups: sci.math.research
Subject: Re: Square of a polynomial ?
Date: 5 Nov 1998 17:40:54 GMT
In article <36419B77.1C1A@club-internet.fr>,
langlois bruno wrote:
>Let P a monic polynomial in Z[X].
>We suppose that :
>- the degree of P is even,
>- for all positive integer n, P(n) is a square in Z.
>Can we say that P is a square in Z[X] ?
Yes. When I ran into this question last year I got this information from
Bruce Reznick:
>Michael Filaseta [...] contacted Schinzel, who'd know. The
>consensus is that the original result (f: Z -> Z^2 => f = g^2) is
>apparently well-known and has been discovered several times.
I don't have any references, however.
Reznick's solution was to compute Q = sqrt(P) = F + G formally where F is
polynomial and G(n)->0 as n->oo. By assumption the values of Q are
integers so the iterated differences are too; on the other hand high-enough
differences of F also vanish. Thus the differences of G are both
integral and small, hence zero, making G itself both polynomial and small,
hence zero (as n->oo, and hence identically). So Q is a polynomial.
Here's an alternative proof Neil Dummigan and I worked out: factor
P = P1^2*P2 with P2 square-free; then y^2=P2(x) is a non-singular curve
with infinitely many integral points, hence P2 is at most quadratic.
(Obviously P2 can't be linear with all P2(n) squares unless P2=constant.)
Completing the square makes this equivalent to an equation y^2=ax^2+b
(Pell's equation) and the premise becomes that it has O(N) solutions
with x Z, does it follow that
P = f o g for some polynomial g? I suppose something like Reznick's
solution will apply. Note that there is a distinction between integer
polynomials and polynomials all of whose values at integers are integers
(e.g. x(x+1)/2) so the question has to be phrased carefully.
Note that the proofs cannot be purely formal, e.g. P(X) = X^4+4 is not
a square in R[X] even though P(x) is a square in R for every x in
R, when R = Z/5Z or R is the real number field.
dave
==============================================================================
From: Drew Vandeth
Newsgroups: sci.math.research
Subject: Re: Square of a polynomial ?
Date: Fri, 06 Nov 1998 10:30:15 +1100
langlois bruno wrote:
> Let P a monic polynomial in Z[X].
> We suppose that :
> - the degree of P is even,
> - for all positive integer n, P(n) is a square in Z.
> Can we say that P is a square in Z[X] ?
There is a classical theorem which states that if a polynomial with
integral coefficients is an
$m$th power for every integral value of its argument, then it is the
$m$th power of a polynomial
with integral coefficients. This follows from Hilbert's irreducibility
theorem but can be
proven in an elementary way, see for example P\'olya-Szeg\"o, Problems
and theorems in analysis,
vol II, part VIII, Chapter 2, Springer-Verlag 1976.
--
Drew Vandeth
ceNTRe for Number Theory Research
Department of Mathematics
Macquarie University
Sydney, Australia
vandeth@mpce.mq.edu.au
==============================================================================
From: israel@math.ubc.ca (Robert Israel)
Newsgroups: sci.math.research
Subject: Re: Square of a polynomial ?
Date: 10 Nov 1998 22:23:18 GMT
In article <71tcso$od1$1@nnrp1.dejanews.com>, johnmitchell2100@my-dejanews.com writes:
|> In article <36419B77.1C1A@club-internet.fr>,
|> blang@club-internet.fr wrote:
|> > Let P a monic polynomial in Z[X].
|> > We suppose that :
|> > - the degree of P is even,
|> > - for all positive integer n, P(n) is a square in Z.
|> > Can we say that P is a square in Z[X] ?
|> Several readers have already supplied proofs that this is true. I just wanted
|> to mention that this question appeared on the "Aggregation" exam given to
|> students at the Ecole Normale de Jeune Filles (in Paris) around 1980. One of
|> the "jeunes filles" told an Argentinian mathematician I knew about the
|> problem, and he then asked me about it. I came up with a finite difference
|> solution that's essentially the same as Bruce Resnick's solution mentioned in
|> Dave Rusin's reply. Notice that with this solution (and perhaps with others),
|> you don't need the assumptions that P is monic and has even degree.
On the other hand, some of these solutions (including mine, which used
Laurent series) solve the harder question where "for all positive integer n"
is replaced by "for infinitely many positive integers n". For that one
you do need P to be monic and of even degree, otherwise counterexamples
are P(n) = n and P(n) = 2 n^2 + 1.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2