Correct solutions from members of the Bradley family came from Lena Folwaczny, Tim Roberts. Correct solutions from outside the Bradley University community were received from Jens Voβ, Lou Cairoli, Dan Dima, Jérôme Lefebvre, Paul Lee.
We are given that
(1) x y = z, (2) y z = x, (3) z x = y.
It is not a good idea to use the logarithm function since we don't know whether the numbers in these equations are positive. So let's raise the first equation to the xz power, the second to the xy power, and the third to the yz power. This gives
(1') x xyz = z xz, (2') y xyz = x xy, (3') z xyz = y yz;
but z xz = (z x) z = y z = x, and similarly x xy = y, y yz = z, so
(1'') x xyz = x, (2'') y xyz = y, (3'') z xyz = z.
Put n = xyz, then x n = x, y n = y, z n = z from which we easily conclude that x = ±1, y = ±1, and z = ±1. Considering the possible combinations yields the only solutions to be
x = y = z = 1 and x = y = z = -1.
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