Science Forums 2

[Don Blazys]

To: Qfwfq.
I apologize for my continuous failure to post my equations using LaTex. My "computer skills" are rudimentary at best, so I beg you to remain both patient and curious and to bear with me as I am doing my best to improve those skills under somewhat difficult circumstances and conditions. There is nothing that I want more than to have other mathematicians understand what I am presenting, and I really, really appreciate it when good folks such as yourself and CraigD take the time to contribute to this discussion.

If you can find the time to visit my website (donblazys.com), then please do so. (You can also find it by doing a "Google search" on "cohesive terms".) It was put together by the computer teachers at the school where I work and contains a lot more information on the equation (identity) that we are discussing here.

Now, here's why this really is a "cool new discovery", why it does indeed have "profound implications" for mathematics, and why I named this thread "A Mathematical Emergency".

You see, at this very moment, students throughout the world are being taught that in order to render the terms in the equation:

Ta^x+Tb^y=Tc^z

relatively prime or "co-prime", we must first divide by T, then "cross out" the T's. However, doing so results in:

(T/T)a^x+(T/T)b^y=(T/T)c^z=a^x+b^y=c^z,

which is wrong and inconsistent with both the "Beal Conjecture", and all observed results, because it falsely implies that x, y, and z can all be greater than 2 after the largest possible common factor T has been cancelled and the terms are co-prime.
Now, the proper, correct and effective way to render the terms "co-prime" is as follows:
Instead of "crossing out" the T's, we actually use them to derive at least one "cohesive term" so that it can be firmly established that the cancelled common factor T>1. This gives us three possibilities, one of which is:

(T/T)a^x+(T/T)b^y=(T/T)c^z=

T(c/T)^((zln(c)/(ln(T))-1)/(ln(c)/(ln(T))-1)).

Factoring does not involve the cancelled variable T, and results in:

((T/T)a^(x/2))^2+((T/T)b^(y/2))^2=((T/T)c^(z/2))^2=

(T(c/T)^(((z/2)ln(c)/(ln(T))-1)/(ln(c)/(ln(T))-1)))^2,

where it is now clear that before we do anything else, we must first let z=1 in the unfactored case, and z=2 in the factored case, so that we can "cross out" and eliminate the logarithms that are preventing us from letting T=c, which common sense tells us must be allowable. This gives us both:

(T/T)a^x+(T/T)b^y=(T/T)c=T(c/T),

and

((T/T)a^(x/2))^2+((T/T)b^(y/2))^2=((T/T)c)^2=(T(c/T))^2,

where T=c is now clearly allowable. Simplifying, we note that our co-prime equations:

a^x+b^y=c,

and

(a^(x/2))^2+(b^(y/2))^2=c^2

are now perfectly consistent with both the "Beal Conjecture", and all observed results, because the implication that x, y and z can all be greater than 2 after the largest possible common factor T has been cancelled no longer exists.
Thus, by using a "cohesive term" to guarantee that the common factor T>1, we prevent the T's from being "crossed out" prematurely, and automatically prove both the "Beal Conjecture", and "Fermat's Last Theorem" (which involves only the "special case" where x=y=z).
Had mankind learned how to properly represent and eliminate common factors to begin with, (using at least one "cohesive term" to ensure that any common factor is "non-trivial") problems such as the BC and FLT would never have surfaced!
That, to me, is profound.

Don.