Problem

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Tags: functional equation, algebra



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