Russian 38 puzzle
John Scholes
(x^2+px+q)^10 for all x.
NOTES
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--
John Scholes
Werner Aumayr
-(a/2+b)^20=(..)^10
both sides to be 0
hence -2b=a and q=-p/2-1/4
With this the equation forms to
(1-b^20)(2x-1)^20 = (1/2^20)(2x-1)^10(x+p+1/2)^10
valid for all x
hence p=-1
and b^20 = 1-1/2^20
answer:
p = -1, q = 1/4, b^20 = 1 - 1/2^20, a=-2b
John Scholes <jsch...@kalva.demon.co.uk> wrote in article
<Kr8y6CA2...@kalva.demon.co.uk>...
David Moews
John Scholes <jsch...@kalva.demon.co.uk> wrote:
|Find all real p, q, a, b such that we have (2x-1)^20 - (ax+b)^20 =
|(x^2+px+q)^10 for all x.
The left-hand side factors as the product of (2x-1)-w(ax+b), where w
ranges over the 20th roots of 1. But since the right-hand side has only
two distinct roots---x0 and x1, say---there must be unequal w and w'
with (2 x0 - 1) - w (a x0 + b) = (2 x0 - 1) - w' (a x0 + b) = 0.
Then 2 x0 - 1 = a x0 + b = 0, so x0 = 1/2 and a/2 + b = 0, giving
ax+b=k(2x-1) for some k. The leading coefficient gives the equation
2^20(1-k^20)=1, so (real) k is plus or minus (1-2^(-20))^(1/20), and then,
since the left-hand side will equal (x-1/2)^20, the equation will be
satisfied if x^2+px+q=(x-1/2)^2=x^2-x+1/4. The two solutions are then
(p,q,a,b)=(-1,1/4,(2^20-1)^(1/20),-(1-2^(-20))^(1/20)) and
(p,q,a,b)=(-1,1/4,-(2^20-1)^(1/20),(1-2^(-20))^(1/20)).
--
David Moews dmo...@xraysgi.ims.uconn.edu
Bill Dubuque
|
| 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