From: Bill Dubuque <wgd@berne.ai.mit.edu>
Newsgroups: sci.math,rec.puzzles
Subject: Re: Russian 38 puzzle
Date: 25 Sep 1998 21:28:32 -0400

John Scholes wrote: (I've generalized the expt from 10 to n > 1)
|
| Find all real p,q,a,b so  (2x-1)^2n - (ax+b)^2n = (x^2+px+q)^n  for all x.

By Mason's abc theorem [1], if the polys share no root, their degrees are
smaller than the total count of all their distinct roots, so  2n < 1+1+2.
But n >= 2 so the polys must all share the root x=1/2. The rest is easy.

Mason's abc theorem is very powerful, e.g. see below the one-line proof
of the polynomial version of Fermat's Last Theorem (FLT).

-Bill Dubuque

[1] Mason's abc Theorem (1984).  Let a(x), b(x), c(x) in C[x]
be coprime polynomials with  a + b = c.  Then

  max deg{a,b,c} <= N(abc)-1,  N(p) = number of distinct roots of p in C.

[2] Corollary (summing the above over a,b,c)

     deg(abc) <= 3 (N(abc)-1)

This elementary result yields much power:  witness this trivial proof
of FLT for polynomials:  if  p^n + q^n = r^n  with p,q,r coprime, by [2]

  n deg(pqr) = deg(abc) <= 3 N(abc) - 3 <= 3 deg(pqr) - 3

i.e.  (3-n) deg(pqr) >= 3  so  n < 3.

The proof of [1] is easy, requiring only a half-page of high-school algebra,
see, for example:  Lang, Serge.  Algebra. 3rd Edition. 1993. S.IV.7 p.194.

For more on Mason's abc Theorem, e.g. a wronskian viewpoint, and
further info on FLT for polynomials, search the DejaNews usenet
newsgroup archive of sci.math for "mason abc theorem": follow URL

http://www.dejanews.com/dnquery.xp?QRY=mason%20abc%20theorem&groups=sci.math&ST=PS