From: Bill Dubuque
Newsgroups: sci.math
Subject: FLT for Polynomials
Date: 03 Feb 1998 10:26:58 0500
cc: mathhistorylist@enterprise.maa.org
Jim Propp wrote:
 Does anyone know how Tchebychev proved Fermat's Last Theorem for
 polynomials (equivalently, the nonexistence of a parametric solution)?

 I know of one proof of this result via Mason's abc theorem,
 but that is of course not how Tchebychev would have done it.

 I'm also interested in knowing about other proofs (especially simple ones).
Chebyshev's proof employs the theory of integration in finite terms,
in particular his result that if r,s,t are rational numbers then
x^r (a + b x^s)^t is integrable in elementary functions iff
t or (r+1)/s or their sum is an integer. This result is discussed
by Shanks (see my post below), who refers to Ritt for a proof of
Chebyshev's integrability criterion. I wouldn't be surprised if
there are simpler proofs known by now. Perhaps there is a
discussion in Manuel Bronstein's book, on which see his home page
http://www.inria.fr/safir/whoswho/Manuel.Bronstein/
Later Jim added:
 Henry Cohn has pointed out to me that the trick of factoring both sides
 of f^n + g^n = h^n over a suitable algebraic extension field (which, as
 you may recall, was the idea underlying Lam\'e's false proof), actually
 gives a valid oneparagraph proof of FLT for polynomials. Details are
 given in Michael Rosen's article "Remarks on the History of Fermat's Last
 Theorem 1844 to 1984", published in the book "Modular Forms and Fermat's
 Last Theorem" (see page 507).

 I'd still like to know the history of this result though. (I think I've
 seen it attributed to both Tchebychev and Liousville, though more often
 to the former.)
Ribenboim's book "13 Lectures on Fermat's Last Theorem" has its
entire last chapter devoted to generalizations of FLT over
various algebraic structures (rings of polys, matrices, entire
functions; ordinals, theta functions, nonassociative aspects, etc).
For the polynomial case of FLT, i.e. over K[t], K a field, Ribenboim
refers to an analytic proof by Liouville in 1879 for K = C, and an
algebraic proof by Korkine in 1880 for K any field of characteristic 0,
both in C.R. Acad. Sci. Paris. He includes a proof by Greenleaf [0]
which works if char(K) doesn't divide n; the proof proceeds by descent
and employs unique factorization in the style of Lame. Probably this is
the same proof you refer to in Rosen's article. Ribenboim also mentions
a slick proof communicated by Pierre Samuel: If f^n+g^n = h^n, where
f,g,h in K[t] are coprime nonconstant poly's, then (f/h,g/h) is a
generic point of Fermat's curve X^n+Y^n = 1. Hence K(f/h,g/h) < K(t).
By Luroth's theorem, Fermat's curve is rational, that is of genus 0,
and this implies n = 1 or 2.
Perhaps someone with access to all the algebraic proofs (Korkine,
Greenleaf and Rosen) can tell us how they compare.
I've appended an old post of mine which provides further discussion.
Bill Dubuque
[0] Greenleaf, Newcomb. On Fermat's equation in C(t).
Amer. Math. Monthly 76 1969 808809.
MR 40 #4204 (Reviewer: Fred Gross) 10.13
Subject: poly FLT, abc theorem, Wronskian formalism [was: Entire solutions of f^2+g^2=1]
From: wgd@zurich.ai.mit.edu (Bill Dubuque)
Date: 1996/07/17
Newsgroups: sci.math.research,sci.math
"Harold P. Boas" wrote to sci.math.research on 96/7/3:
:
: Robert Israel wrote:
: >
: > Alan Horwitz wrote to sci.math.research on 96/7/1:
: >  I am interested in all entire solutions f and g to f^2 + g^2 = 1.
: >  I remember seeing this somewhere, but I cannot recall where.
: >  In particular, I want to know if the entire function f + g
: >  can have finitely many zeroes.
: >
: > I've also seen this before; I recall assigning it as homework to one
: > of my classes, but I don't recall the source. The solutions are ...
:
: Robert B. Burckel gives some history about this problem in his
: comprehensive book An Introduction to Classical Complex Analysis,
: volume 1 (Academic Press, 1979). In Theorem 12.20, pages 433435,
: he shows that the equation f^n+g^n = 1 has no nonconstant entire
: solutions when the integer n exceeds 2; when n=2, the solution
: is as given by R. Israel in his post. ... (papers of Fred Gross)
Note that the rational function case of FLT follows trivially from
Mason's abc theorem, e.g. see Lang's Algebra, 3rd Ed. p. 195 for a
short elementary (highschool level) proof of both. Chebyshev also
gave a proof of FLT for poly's via the theory of integration in
finite terms, e.g. see p. 145 of Shanks' "Solved and Unsolved Problems
in Number Theory", or Ritt's "Integration in Finite Terms", p. 37.
The Chebyshev result is actually employed as a subroutine of Macsyma's
integration algorithm (implemented decades ago by Joel Moses). Via abc
a related result of Dwork is also easily proved: if A,B,C are fixed
poly's then all coprime poly solutions of A X^a + B Y^b + C Z^c = 0
have bounded degrees provided 1/a + 1/b + 1/c < 1. Other applications
in both number and function fields may be found in Lang's survey [3].
Mason's abc theorem may be viewed as a very special instance of a
Wronskian estimate: in Lang's proof the corresponding Wronskian
identity is c^3 W(a,b,c) = W(W(a,c),W(b,c)), thus if a,b,c are
linearly dependent then so are W(a,c),W(b,c); the sought bounds
follow upon multiplying the latter dependence relation through by
N0 = r(a) r(b) r(c), where r(x) = x/gcd(x,x').
More powerful Wronskian estimates with applications toward
diophantine approximation of solutions of LDEs may be found in
the work of the Chudnovsky's [1] and C. Osgood [2]. References
to recent work may be found (as usual) by following MR citations
to these papers in the MathSci database.
I have not seen mention of this Wronskian view of Mason's abc theorem.
Although elementary, it deserves attention since it connects the abc
theorem with the general unified viewpoint of the Wronskian formalism
as proposed by the Chudnovsky's and others.
[1] Chudnovsky, D. V.; Chudnovsky, G. V.
The Wronskian formalism for linear differential equations
and Pade approximations. Adv. in Math. 53 (1984), no. 1, 2854.
86i:11038 11J91 11J99 34A30 41A21
[2] Osgood, Charles F.
Sometimes effective ThueSiegelRothSchmidtNevanlinna bounds, or better.
J. Number Theory 21 (1985), no. 3, 347389. 87f:11046 11J61 12H05
[3] Lang, Serge
Old and new conjectured Diophantine inequalities. Bull. Amer. Math. Soc.
(N.S.) 23 (1990), no. 1, 3775. 90k:11032 11D75 1102 11D72 11J25
Bill Dubuque
Note: for the entire thread follow this link:
http://search.dejanews.com/dnquery.xp?QRY=%7Eg%28sci.math.research%29entire%20solutions&svcclass=dnold
==============================================================================
From: Nick Halloway
Newsgroups: sci.math
Subject: Re: FLT for Polynomials
Date: Wed, 4 Feb 1998 09:23:40 0800
On 3 Feb 1998, Bill Dubuque wrote:
> Jim Propp wrote:
>  Does anyone know how Tchebychev proved Fermat's Last Theorem for
>  polynomials (equivalently, the nonexistence of a parametric solution)?
> 
>  I know of one proof of this result via Mason's abc theorem,
>  but that is of course not how Tchebychev would have done it.
> 
>  I'm also interested in knowing about other proofs (especially simple ones).
>  Henry Cohn has pointed out to me that the trick of factoring both sides
>  of f^n + g^n = h^n over a suitable algebraic extension field (which, as
>  you may recall, was the idea underlying Lam\'e's false proof), actually
>  gives a valid oneparagraph proof of FLT for polynomials.
From my web page:
Someone mentioned to me that an analog of FLT is true in C[x] as well,
i.e. there are no relatively prime polynomials a(x), b(x) and c(x) in C[x]
with a^n + b^n = c^n for n>2.
I think this is true. You can show it by adapting the false proof of
FLT that is in the FAQ.
It's only necessary to show it for n an odd prime and for n = 4.
The proof is similar for n = 4 to n and odd prime, so I will just
show it for odd primes.
proof by induction on the sum of the degrees of a(x) and b(x).
Suppose the sum of the degrees is 2. Then a^p + b^p factors as
(a+b)(a+rb)...(a+r^(p1)b), where r is a primitive p'th root of unity.
These factors are all relatively prime polynomials of degree 1.
But c^p has repeated factors. So it won't work for this case.
Now assume the sum of the degrees is m. Again factor a^p + b^p as
(a+b)(a+rb)...(a+r^(p1)b). These factors are relatively prime.
But c^p has factors with multiplicity divisible by p. So each factor a+(r^i)b
must be a p'th power. So we have
a+b = d^p,
a+rb= e^p,
a+(r^2)b = f^p
So b = (e^p  d^p)/ (r1), and
a = (e^p  rd^p) / (1r)
So f^p = e^p (r+1)  rd^p
Since neither r+1 nor r are 0, and d(x) and e(x) are relatively prime,
this violates the induction hypothesis.
This also shows it for polynomials in Z[x], since polynomials that are
relatively prime in Z[x] are also relatively prime in C[x].
==============================================================================
From: Allen Adler
Newsgroups: sci.math
Subject: Re: FLT for Polynomials
Date: 04 Feb 1998 15:09:58 0600
A polynomial solution of a^n+b^n=c^n can be viewed as a rational mapping
from the projective line to the Fermat curve of degree n. There
are no nonconstant rational mappings from a curve to a curve of
higher genus. The genus of the Fermat curve is (n1)(n2)/2,
which is > 0 for n>2. QED.
Allan