FORTH GO

Fermat’s Last Theorem may have already been proved, but the proof is so dense amateurs like myself still foolishly search for a simple proof. My latest foray was centered around a trigonometric mappings.

That is, start with aⁿ + bⁿ = cⁿ, divide through by cⁿ and then map the resulting equation to the trig identity sin² x + cos² x = 1. So (a/c)ⁿ maps to sin² x and (b/c)ⁿ maps to cos² x, which means we need to find an angle x that sin^2/n x and cos^2/n x are both rational. If so, then we can raise them both to the nth power to get the trig identity and multiply through by the Least Common Multiple of the demoninators to get a solution to Fermat’s equation.

So we actually need to prove that sin^2/n x and cos^2/n x cannot be both rational for a given x, except for trivial cases like x = 0. That’s about as far as I got, but the presentation has a nice feature in that it suggests that n = 1 and n = 2 are special since those are the only values when 2/n is an integer. Likewise, they are the only values of n for which Fermat’s equation has non-trivial solutions. Also the trivial solutions to Fermat’s equation, a = 0 and b = 0, correspond nicely to the trivial solutions in the trig representation where x is 0 or Ï€/2.

I tried a Taylor expansion of sin^2/n x and cos^2/n x, but I couldn’t expand them at 0 since the exponents would result divisions in by 0. I expanded around Ï€/4, which has the nice property of sin Ï€/4 = cos Ï€/4 = sqrt(1/2). Still nothing obvious panned out. Simplifying infinite series is not a specialty of mine.

I was hoping to reduce each series to the form U ± V, where at least one of U or V can be proved irrational. In that case, I could prove that U + V and U - V couldn’t both be rational. If both U + V and U - V were rational, then their sum (2U) and their difference (2V) would be rational.

Some of our azaleas bloom in both Spring and Fall if the Summer is not too dry.