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

By: Don Blazys.

By: Don Blazys.

The Beal Conjecture can be stated as follows:

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

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

Proof:

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

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

Factoring (1) results in:

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

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

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

and it's newly discovered logarithmic consequences are represented, we now have both:

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

and:

[tex]\left(T\left(\frac{c}{T}\right)^{\left(\frac{\frac{\frac{z}{2}\ln(c)}{\ln(T)}-1}{\frac{\ln(c)}{\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.[/tex] ___________(5)

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

which

*clearly*means that [tex]T=c[/tex]

**must be allowable**.

We also note that the logarithms

*preventing*[tex]T=c[/tex] "

**cancel out**" and therefore

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

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

and:

[tex]\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.[/tex] _____________________(7)

[tex]T=c[/tex] is now

*clearly*allowable, and simplifying (6) and (7) shows that

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

[tex]c-b^y=a^x,[/tex] ___________________________________________________________(8)

and:

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

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

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