Proofs of Beal's Conjecture, Fermat's Last Theorem and Pythagora's Theorem

James Constant

math@coolissues.com

        Assume

        z^m=x^a+y^b to be proved                                                                                 (1)

where z,x,y,m,a,b are specified numbers.

        What constitutes proof of (1)? Unlike the Pythagorean theorem, (1) has no known geometric proof and it has no algebraic proof because it is a single equation with six unknowns. There are sets of z,x,y and sets of m,a,b from which we are to find some sets z,x,y,m,a,b as "solutions", if any. In (1) all we have is a presumed single equality and some stated conditions on sets of z,x,y and m,a,b. Therefore, proof here are answers to the questions, first, does equality (1) necessarily exist for some numbers and, second, if it necessarily exists for some numbers does it sufficiently exist for specified set(s) of z,x,y,m,a,b. For example, we know from geometry that (1) exists for Pythagora's Theorem (PT) and inquire whether it exists for Fermat's Last Theorem (FLT) and for Beal's Conjecture (BC). BC states if (1) exists with positive integers z,x,y,m,a,b and m,a,b>2 then z,x,y have a common factor f>1. Thus, Beal's equality (1) exists only if (x^a+y^b)^1/m is a positive integer z. FLT is the negation of (1) when z,x,y,m are integers and m=a=b>2. PT is (1) with z,x,y algebraic rational or irrational numbers and m=a=b=2.

        We can proceed in two ways, first, try to prove or disprove that the equality in (1) necessarily exists and, second, assume the equality in (1) necessarily exists and try to prove or disprove the sufficiency of specified set(s) of z,x,y,m,a,b. I will try both ways.

Proof of FLT

        I first prove that the equality in (1) does not exist for FLT.

        Write FLT and PT as

        (x/z)^m + (y/z)^m = 1 = (x/z)² + (y/z)²         x/z<1 y/z<1                                      (2)

in which the right hand equality necessarily exists for PT and the left hand equality is to be proven for FLT. Since m>2, (x/z)^m<(x/z)², (y/z)^m<(y/z)², and (x/z')^m + (y/z')^m<1 and FLT is proven because the left hand equality in (2) cannot exist in the region z'z.

        More generally, in all regions, assume the following equality exists

                                                z',z,x,y integers                                    (3)

which can be rewritten as

                                                 z',z,x,y integers                                    (4)

from which observe that both sides of (4) are integers when

                                                       j = integer                                          (5)

which can be rearranged as

        x²(x^m-2-j) + y²(y^m-2-j) = 0                                                                              (6)

which is satisfied when j=x^m-2=y^m-2 and x=y. Equation (4) can now be restated as

                                                 z',z,x,y integers                                     (7)

from which it follows that z' cannot be an integer because z,x,y are coprime primitive Pythagorean triplets and, therefore (z/x)^2/m is an irrational number. The equality in equation (3), therefore, cannot exist. Accordingly, FLT is proven because the left hand equality in equation (2) cannot exist in the region z'z and because z' is an irrational number in all regions.

Disproof of BC

        Next write BC and PT as

        x^a/z^m + y^b/z^m = 1 = (x/z)² + (y/z)²                                                                  (8)

in which the right hand equality necessarily exists for PT and the left hand equality is to be proven for BC. Since x^a/z^m<1 when am and y^b/z^m<1 when bm, x^a/z^m + y^b/z^m <1 and BC is disproven because the left hand equality in (8) cannot exist in the region z'z, am and bm.

        More generally, in all regions, by replacing x^m by x^a and y^m by y^b in equations (2) and (3) and following the procedure of equations (3)-(7), I conclude that x^a-2=y^b-2 and

                x^a-2=y^b-2             z',z,x,y integers                                     (9)

from which it follows that z' cannot be an integer since z,x,y are coprime primitive Pythagorean triplets and, therefore, (z/x)^2/m is an irrational number, and x^a/m is a rational or irrational number. The equality in replaced equation (3), therefore, cannot exist. Accordingly, BC is disproven because the left hand equality in replaced (2) cannot exist in the region z'z,am and bm. and because z' is an irrational number in all regions.¹ Since Beal's equality does not exist with positive integers z,x,y,m,a,b and m,a,b>2 there is no need to answer the question whether z,x,y have a common factor f>1.

Alternative Disproof of BC and Proof of BC

        Next, I assume that the equality in (1) necessarily exists and disprove the sufficiency of specified set(s) of z,x,y,m,a,b.

        Write BC as

        u = x^a/z^m + y^b/z^m                     u<1 or u>1                                                  (10)

which also is FLT when a=b=m. Now take logarithms

        lnu=i=ln[x^a/z^m + y^b/z^m]          0<u<1 or 0<u<2                                           (11)

in which the inequalities are regions of convergence of i when the logarithm is expanded as an infinite series. The first inequality in (11) is obtained for all values of 0<u<1 by using the comparison test with exponential series e^k with k1. The second inequality is obtained for finite values of 0<u<2 by using the ratio test. If u<1, since u=eⁱ, i<0 and u are determinate irrational for all values of 0<u<1. If u>1, i>0 and u are indeterminate for finite values of 0<u<2, and I and u are indeterminable for all finite values of u>2. Thus, BC is disproved and FLT is proved for all specified values of z,x,y,m,a,b.

Proof of PT

        The situation is somewhat different for PT. Here, we need not try to prove or assume existence of the equality and prove the sufficiency of numbers z,x,y.

        Write PT as

        1 = (x/z)² + (y/z)²                 x/z<1 y/z<1                                                     (12)

and take logarithms

        ln1=i=ln[(x/z)² + (y/z)²]          i=0                                                             (13)

        Since u=eⁱ and i=0 there is no logarithmic expansion restraining values of u and, therefore, PT is proved determinate rational or irrational for all values of z,x,y.

Copyright © 2008 James Constant

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