An Elementary Proof Of Both The Beal Conjecture And Fermat's Last Theorem.

An Elementary Proof Of Both The Beal Conjecture And Fermat's Last Theorem.

By: Don Blazys

 

The Beal Conjecture can be stated as follows:

 

For positive integers: [math]a, b, c, x, y, [/math] and [math]z[/math], if [math]a^x+b^y=c^z[/math],

and [math]a, b[/math] and [math]c[/math] are co-prime, then[math]x, y[/math] and [math]z[/math] are not all greater than [math]2[/math].

 

Proof:

 

Letting all variables herein represent positive integers, we form the equation:

 

[math]c^z-b^y=a^x[/math].__________________________________________________________(1)

 

Factoring (1) results in:

 

[math]\left(c^\frac{z}{2}+b^\frac{y}{2}\right)\left(c^\frac{z}{2}-b^\frac{y}{2}\right)=a^x[/math].______________________________________________(2)

 

Here, it will be assumed that the terms in (1) and (2) are co-prime,

and that the only "common factor" they contain is the "trivial" unity,

which can not be defined in terms of itself, and must therefore be defined as:

 

[math]1=\left(\frac{T}{T}\right)[/math], where [math]T>1[/math]._________________________________________________(3)

 

Re-stating (1) and (2) so that the "trivial common factor" [math]1=\left(\frac{T}{T}\right)[/math]

and its newly discovered logarithmic consequences are represented,

we now have both:

 

[math]T\left(\frac{c}{T}\right)^{\left(\frac{\frac{z\ln©}{\ln(T)}-1}{\frac{\ln©}{\ln(T)}-1}\right)}-\left(\frac{T}{T}\right)b^y=\left(\frac{T}{T}\right)a^x[/math] __________________________________(4)

 

and:

 

[math]\left(T\left(\frac{c}{T}\right)^{\left(\frac{\frac{\frac{z}{2}\ln©}{\ln(T)}-1}{\frac{\ln©}{\ln(T)}-1}\right)}+\left(\frac{T}{T}\right)b^\frac{y}{2}\right)\left(\left(\frac{T}{T}\right)c^\frac{z}{2}-\left(\frac{T}{T}\right)b^\frac{y}{2}\right)=\left(\frac{T}{T}\right)a^x[/math].__________(5)

 

At this point, we note that the definition of unity in (3) implies: [math]1=\left(\frac{T}{T}\right)=\left(\frac{c}{c}\right)[/math],

which clearly means that [math]T=c[/math] must be allowable.

We also note that the logarithms preventing [math]T=c[/math] "cancel out" and therefore

cease to exist if and only if [math]z=1[/math] in (4), and [math]z=2[/math] in (5), which gives us both:

 

[math]T\left(\frac{c}{T}\right)-\left(\frac{T}{T}\right)b^y=\left(\frac{T}{T}\right)a^x[/math] ___________________________________________(6)

 

and:

 

[math]\left(T\left(\frac{c}{T}\right)+\left(\frac{T}{T}\right)b^\frac{y}{2}\right)\left(T\left(\frac{c}{T}\right)-\left(\frac{T}{T}\right)b^\frac{y}{2}\right)=\left(\frac{T}{T}\right)a^x[/math].____________________(7)

 

[math]T=c[/math] is now clearly allowable, and simplifying (6) and (7) shows that

the original equations, as stated in (1) and (2), are now:

 

[math]c-b^y=a^x[/math],__________________________________________________________(8)

 

and:

 

[math]\left(c+b^\frac{y}{2}\right)\left(c-b^\frac{y}{2}\right)=c^2-b^y=a^x[/math],_______________________________________(9)

 

which proves not only the Beal Conjecture, but Fermat's Last Theorem

(which is only the special case where [math]x=y=z[/math]) as well.

Edited March 19, 200916 yr by Don Blazys