From: Osher Doctorow [EMAIL PROTECTED], Wed. Nov. 28, 2000 3:10PM First-order LBP analysis of Fermat's Last Theorem (FLT for short in what follows - see my contribution from the last few days) asserts that first-order LBP maximum entropy equations/functions are the only admissible or acceptable kinds, which means that functions must be linear (in either independent or dependent variables), quadratic, cross-product (like kxy), simple exponential (like a exp(bx) with b real or complex), or have vanishing or "neglect-able" 3rd and higher partial derivatives (the latter mostly applies to mathematical physics as in general relativity). It turns out that only x + y = z and x^^2 y^^2 = z^^2 (where ^^ denotes exponentiation) are admissible under 1st-order LBP and not the higher degree equations, which "proves" FLT under lst-order conditions. Notice that in general elements x, z, y can be arbitrary objects (tensors, vector, pseudovectors, bivectors, etc.), but in FLT they're usually taken as (positive) integers, and in LBP they are taken as continuous (e.g., all nonnegative reals) on the nonnegative real line unless some other connected domain is specified. Second-order LBP analysis of FLT is instructive for its applicability to other Mersenne/prime equations. Here we use second-order LBP maximum entropy, which permits multiplication, addition, or subtraction of elements of 1st-order LBP, but not division, not inversion except subtraction, not limits unless they enter into the previous types, and not composition of functions. It turns out that x^^n + y^^n = z^^n needs to have two of the quantities x, y, or z "independent", and without loss of generality I will choose x and y to be "independent" and z = f(x, y). However, if z is a function of x and y, then z must be a two-variable polynomial (with terms like kxy, kx^^3y^^5, etc.) since multiplication of x and y times themselves only yield these (they do not yield fractional or rational powers of x and/or y without inverse and/or limiting operations). Therefore, z^^n must be a two-variable polynomial unless it is a constant. However, there is no known equation in mathematics for which x^^n + y^^n = k^^n (constant) for n > 2 integer (for n = 2 it is a circle, for n = 1 it is a line). Therefore, we have an identity x^^n + y^^n = x^^n + y^^n by setting equal powers of x and y equal in pairs. An identity (in this sense) is a trivial equation and is excluded from LBP since trivial equations make no statements about functions and hence none about their LBP entropy. This gives a second-order LBP proof of FLT. Alternatively, notice that if we took partial derivatives with respect to x, symbol Dx, of both sides of x^^n + y^^n = z^^n, we would obtain n! = Dx...x z^^n where Dx...x is the nth partial, and since an nth partial is also a function of x and y, we would either end up with n! = n! (trivial) or n! = two-variable polynomial in x and y, which by equating terms with equal powers on the left and right sides yields n! = n!. This second method does not involve the fact that there is no known function of form x^^n + y^^n = k^^n for k constant and x, y independent and n > 2 integer. We can also exclude n! = x^^m + y^^m and similar types from it by going back to Leibniz' rule and using induction on z = f(x,y). Osher Doctorow _________________________________________________________________________ Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers
