2=0

[JackG]

Nice trick. Of course, anyone looking at the results for 1782^12 + 1841^12 and 1922^12 individually will think that it breaks Fermat's theorem, but entering (1782^12) + (1841^12) - (1922^12) shows that the two sides are not in fact equal. If you could just find values of a, b, and c so that their difference is greater than 12 decimal points away, however, I think that the calculator would fall for the trick because of rounding problems due to memory issues. At least it would work in Matlab, and I'm guessing the calculator stores data the same way.

So, anyone know some a, b, and c such that (c^n)/((a^n)+(b^n)-(c^n)) > 1*(10^13) ? Did I even do that right?