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 (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