From: Bill Dubuque
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 (2x1)^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 oneline 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. (3n) deg(pqr) >= 3 so n < 3.
The proof of [1] is easy, requiring only a halfpage of highschool 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