MathDB

4

Part of 2003 Tuymaada Olympiad

Problems(2)

functional equation with continous function

Source: tuymaada 2003

3/11/2006
Find all continuous functions f(x)f(x) defined for all x>0x>0 such that for every xx, y>0y > 0 f(x+1x)+f(y+1y)=f(x+1y)+f(y+1x). f\left(x+{1\over x}\right)+f\left(y+{1\over y}\right)= f\left(x+{1\over y}\right)+f\left(y+{1\over x}\right) .
Proposed by F. Petrov
functionlimitreal analysisalgebrafunctional equation
Set of prime divisors of a_{n+1}=f(a_n), f polynomial.

Source: Tuymaada 2003, day 2, problem 4.

5/5/2007
Given are polynomial f(x)f(x) with non-negative integral coefficients and positive integer a.a. The sequence {an}\{a_{n}\} is defined by a1=a,a_{1}=a, an+1=f(an).a_{n+1}=f(a_{n}). It is known that the set of primes dividing at least one of the terms of this sequence is finite. Prove that f(x)=cxkf(x)=cx^{k} for some non-negative integral cc and k.k.
Proposed by F. Petrov
[hide="For those of you who liked this problem."] Check [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=62259]this thread out.
algebrapolynomialcalculusintegrationnumber theory proposednumber theory