From: [Permission pending]
Newsgroups: sci.math
Subject: Is FLT easy in Z[x,y]?
Date: Fri, 9 DEC 94 01:47:44 0500
Hi. I am wondering if it is easy to show that there cannot be a polynomial
formula for exceptions to FLT, i.e.
is it easy to show that there are no nonconstant polynomials f(x,y) and g(x,y)
in Z[x,y] which can simultaneously have positive values for some x and y
and which satisfy an equation f(x,y)^n + g(x,y)^n = h(x,y)^n for some n greater
than 2?
For example for n = 2 such a formula is:
(x^2y^2)^2 + (2xy)^2 = (x^2+y^2)^2
How would you show this?
[sig deleted  djr]
==============================================================================
From: pan@cantor.math.psu.edu (Garth Payne)
Newsgroups: sci.math
Subject: Re: Is FLT easy in Z[x,y]?
Date: 10 Dec 1994 05:48:22 GMT
[Permission pending] writes:
>is it easy to show that there are no nonconstant polynomials f(x,y) and g(x,y)
>in Z[x,y] which can simultaneously have positive values for some x and y
>and which satisfy an equation f(x,y)^n + g(x,y)^n = h(x,y)^n for some n
>greater than 2?
This follows from the abc Theorem for polynomials.
%This is amstex.
Let $A$, $B$ , and $C$ be relatively prime polynomials in $x$ such that
$A+B+C=0$.
Let
$$d
=
\left
\matrix
A&A'\\
B&B'
\endmatrix
\right
=

\left
\matrix
A&A'\\
C&C'
\endmatrix
\right
=
\left
\matrix
B&B'\\
C&C'
\endmatrix
\right
\quad.
$$
Let $a=(A,A')$, $b=(B,B')$, and $c=(C,C')$.
Then $abcd$.
But $\deg(d)=\deg(AB)1$.
Hence,
$
\deg(a)+
\deg(b)+
\deg(c)
<
\deg(A)
+\deg(B)
$.
Similarly,
$
\deg(a)+
\deg(b)+
\deg(c)
<
\deg(A)
+\deg(C)
$ and
$
\deg(a)+
\deg(b)+
\deg(c)
<
\deg(B)
+\deg(C)
$.
Adding these three inequalities we obtain
$$
\deg(a)+
\deg(b)+
\deg(c)
<\frac23 % i.e., 2/3
(
\deg(A)
+\deg(B)
+\deg(C)
)
\quad .$$
Let
$A=\alpha^n(x)$,
$B=\beta^n(x)$,
and
$C=\gamma^n(x)$.
Then $\alpha^{n1}a$, etc.
Hence,
$(n1)(
\deg(\alpha)
+
\deg(\beta)
+
\deg(\gamma))
<
\frac23
n(
\deg(\alpha)
+
\deg(\beta)
+
\deg(\gamma)
)
$.
Hence,
$\frac{n1}n
<
\frac23
$,
hence,
$n<3$.
This proves Fermat's last theorem for polynomials.
\bye
Garth Payne
pan@math.psu.edu
==============================================================================
From: clong@cnj.digex.net (Chris Long)
Newsgroups: sci.math
Subject: Re: Is FLT easy in Z[x,y]?
Date: 11 Dec 1994 06:03:27 0500
[Permission pending] writes:
>is it easy to show that there are no nonconstant polynomials f(x,y) and
>g(x,y) in Z[x,y] which can simultaneously have positive values for some
>x and y and which satisfy an equation f(x,y)^n + g(x,y)^n = h(x,y)^n for
>some n greater than 2?
Shanks in his _Number Theory_ gives a very simple proof that for rational
k there do not exist rational functions f(t) and g(t) such that f(t)^k +
g(t)^k = 1 unless k = 1/q or 2/q for some nonzero integer q. It follows:
Consider y=(1x^k)^{1/k} and the integral
\int (1x^k)^{1/k} dx = \int y dx.
If x=f(t), and y=g(t), by the change of variable x=f(t) the integral
becomes
\int g(t)(df/dt) dt
and since this integrand is a rational function, the integral is
elementary. But by a theorem of Chebyshev if u, v, w are rational
numbers, then
\int x^u (A+Bx^v)^w dx
is integrable in terms of elementary functions if and only if (u+1)/v,
w, or (u+1)/v+w is an integer. It follows that 1/k or 2/k must be
an integer, say q. The result follows.

Chris Long, 265 Old York Rd., Bridgewater, NJ 088072618
Score: 0, Diff: 1, clong killed by a Harvard Math Team on 1
==============================================================================
From: clong@cnj.digex.net (Chris Long)
Newsgroups: sci.math
Subject: Re: Is FLT easy in Z[x,y]?
Date: 11 Dec 1994 06:10:15 0500
I forgot to mention that a proof of the theorem by Chebyshev mentioned can
be found in the sci.math ubiquitous Ritt _Integration in Finite Terms_.

Chris Long, 265 Old York Rd., Bridgewater, NJ 088072618
Score: 0, Diff: 1, clong killed by a Harvard Math Team on 1
==============================================================================
From: [Permission pending]
Newsgroups: sci.math
Subject: Re: Is FLT easy in Z[x,y]?
Date: Thu, 15 DEC 94 16:11:52 0500
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].
[sig deleted  djr]
==============================================================================
From: rusin@washington.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: Is FLT easy in Z[x,y]?
Date: 16 Dec 1994 17:46:44 GMT
In article <[identifier deleted]>
[Permission pending] wrote:
>>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 would really be interested in an algebraicgeometry interpretation of this.
I'm never sure about the accuracy of statements in algebraic geometry since
even the basic definitions seem to shift from generation to generation, but
I think the following is accurate:
A variety is _rational_ if it is birationally equivalent to affine space. An
example is the circle x^2+y^2=1. If you set t=y/(x+1) (a rational function
of x and y) then almost every point on the circle is of the form
x=(t^21)/(t^2+1), y=(2t)/(t^2+1) (which are rational functions of t).
A variety is _unirational_ if there is a rational mapping from affine space
onto [a dense subset of] the variety; we don't assume a rational inverse
this time. There are unirational varieties which are not rational. There
are also nonempty varieties which are not rational, of course (for
example, the nonsingular cubics).
Thus the statement that the curve u^n+v^n=1 has no solutions in
rational functions u and v (equivalent to your statement above) is
the statement that this is not a unirational curve.
This is a situation that arises fairly frequently: a solution to a
problem is shown to correspond to a point on a variety; a family of
solutions corresponds to a rationally paramaterized curve on that
variety; one likes to find the latter but often it does not exist
and so one finds the former, which then appears as an adhoc
construction. This for example describes the situation in the
inverse problem of Galois theory (given G is there a K with G=Gal(K/Q)?)
Incidentally, I believe the usual adjectival form of "algebraic geometry"
is "algebrogeometric".
dave
==============================================================================
Date: Sat, 17 Dec 1994 15:38:47 GMT
From: [Permission pending]
To: rusin@washington.math.niu.edu
Newsgroups: sci.math
Subject: Re: Is FLT easy in Z[x,y]?
rusin@washington.math.niu.edu (Dave Rusin) writes:
>In article <[identifier deleted]>
>[Permission pending] wrote:
>>>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 would really be interested in an algebraicgeometry interpretation of this.
>Thus the statement that the curve u^n+v^n=1 has no solutions in
>rational functions u and v (equivalent to your statement above) is
>the statement that this is not a unirational curve.
But, I was asking the question about polynomials in C[x], *not* Z[x].
[sig deleted  djr]
==============================================================================
Date: Sun, 18 Dec 94 23:48:17 CST
From: rusin (Dave Rusin)
To: [Permission pending]
Subject: Re: Is FLT easy in Z[x,y]?
Right, the statement that u^n+v^n=1 has no solution in nonconstant
(I forgot to say that before) complex rational functions
is the statement that this (complex) variety is not unirational. This
has nothing to do with Z[x], except of course that an integral solution
would also be a complex one. Naturally the nature of this variety is
of some interest but its rationality has nothing to do with solutions
of the Fermat equation in rational integers.
Am I misunderstanding the question?
dave
==============================================================================
Date: Mon, 19 Dec 1994 22:40:46 0500 (EST)
From: [Permission pending]
Subject: Re: Is FLT easy in Z[x,y]?
To: rusin@math.niu.edu
>Right, the statement that u^n+v^n=1 has no solution in nonconstant
>(I forgot to say that before) complex rational functions
>is the statement that this (complex) variety is not unirational.
OH. I had a misconception about what a "rational" mapping is.
Apparently you mean a ratio of two polynomials. I thought
it was a function that mapped rationals to rationals.
by the way, if you happened to look at that proof of a^n + b^n = c^n, n>2
being impossible in C[x], I wonder if it looked right? I assume the Net
would have pounced on it like 1000 hungry vultures if it were not, but
this close to Christmas I don't know.
[sig deleted  djr]
==============================================================================
From: wgd@zurich.ai.mit.edu (Bill Dubuque)
Newsgroups: sci.math.research,sci.math
Subject: poly FLT, abc theorem, Wronskian formalism [was: Entire solutions of f^2+g^2=1]
Date: 17 Jul 96 04:13:12
"Harold P. Boas" wrote to sci.math.research on 7/3/96:
:
:Robert Israel wrote:
:>
:> Alan Horwitz writes:
:> > 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.
:>
:> I've also seen this before, in fact 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
==============================================================================
Proof of FLT for polynomial rings (sketch by Rusin). I think the ABC=>FLT
implication is the sci.math FAQ.
==============================================================================
Wronskian proof: Assume f^n+g^n=h^n, gcd(f,g)=1. Apply d/dx:
[ f g ] [f^(n1)] [ h ]
[ ] [ ] = h^(n1) [ ]
[ f' g'] [g^(n1)] [ h']
Multiply by adjunct of left matrix. Then use the resulting conclusion
h^(n1)  gcd( det*f^(n1), det*g^(n1) )
to deduce h^(n1)  det=f g'  g f'.
Now, we may assume d := deg f = deg h >= deg g; then we have
(n1) d <= deg(f g'  g f') <= d + deg(g) 1 < 2d1
so (n3)d <= 1; esp, n < 3.
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