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INMO 2019 P6

Source:

January 20, 2019
functional equationalgebra

Problem Statement

Let ff be a function defined from ((x,y):x,y((x,y) : x,y real, xy≠0)xy\ne 0) to the set of all positive real numbers such that (i)f(xy,z)=f(x,z)⋅f(y,z) (i) f(xy,z)= f(x,z)\cdot f(y,z) for all x,y≠0x,y \ne 0 (ii)f(x,yz)=f(x,y)⋅f(x,z) (ii) f(x,yz)= f(x,y)\cdot f(x,z) for all x,y≠0x,y \ne 0 (iii)f(x,1−x)=1 (iii) f(x,1-x) = 1 for all x≠0,1x \ne 0,1 Prove that (a)f(x,x)=f(x,−x)=1 (a) f(x,x) = f(x,-x) = 1 for all x≠0x \ne 0 (b)f(x,y)⋅f(y,x)=1(b) f(x,y)\cdot f(y,x) = 1 for all x,y≠0x,y \ne 0
The condition (ii) was left out in the paper leading to an incomplete problem during contest.
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