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Postby bill » Wed Feb 24, 2016 1:59 pm

The identity (T^2-S^2)^2+(2st)^2=(T^2+S^2)^2 is true for any and all T and S as can easily be proven. Let A=(T^2-S^2) and B=(2ST) and C=(T^2+S^2). Now we have (A+B)^2=A^2+2AB+B^2=D^2 and (A-B)^2=A^2-2AB+B^2=E^2. Adding both we obtain A^2+B^2=(D^2+E^2)/2=C^2 From this we see two necessary and sufficient conditions as follows:
1. All terms except the A^n and B^n must cancel.
2. There must be both an A^n and B^n

For all odd n, by the binomial theorem, the B^n term cancels out.
For all even n greater than 2 there exist terms other then A^n and B^n.

Therefore there can be no A^n+B^n=C^n with n greater than 2. QED
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