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c (a,b) c(b,a) = 1 = c(a,-a)

Source: 7th QEDMO problem 2 (14. - 17. 1. 2010) https://artofproblemsolving.com/community/c1512515_qedmo_200507

May 9, 2021
algebra

Problem Statement

Let c:Q−{0}→Q−{0}c: Q-\{0\} \to Q-\{0\} a function with the following properties (for all x,y,a,b∈Q−{0}x,y, a, b \in Q-\{0\} and x≠1x \ne 1): a) c(x,1−x)=1c (x, 1- x) = 1 b) c(ab,y)=c(a,y)c(b,y)c (ab,y) = c (a,y)c(b, y) c) c(y,ab)=c(y,a)c(y,b)c (y,ab) = c (y, a)c(y,b) Show that then c(a,b)c(b,a)=1=c(a,−a)c (a,b) c(b,a) = 1 = c(a,-a) also holds.
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