PART A-4 METHODS in GEOMETRY to PROVE COMMON FACTOR in C^z & A^x

By Part A-2

We have verified that in Beal Conjecture C^z = A^x + B^y

C, A, & B are integers > 1 & 

C^z, A^x & B^y are COMPOSITE integers

..........

[1] By Rules in Geometry

If C^z & A^x are represented by Rectangle or Square Figures

with sides in integers value

then

Equal sides that have value > 1 projects the FACTORS

that are common in C^z & A^x &

prove COMMON FACTOR in C^z & A^x

which implies that

For Rectangles formed for C^z = 1*C^z & A^x = 1*A^x 

Equal sides represent integer 1 & sides that represent C^z & A^x

are UNEQUAL

Also sides that represent C^z & A^x represent 2 terms that are multiple of 2 or more FACTORS

which implies that

Rectangles formed for C^z = 1*C^z & A^x = 1*A^x

are meaningless to PROVE or DISPROVE COMMON FACTOR in C^z & A^x

.........

C^z/N = U  where N is integer & C^z is divisible by N gives

C^z = U*N & Rectangle or Square figures for C^z 

that have sides in integer value

............

For the value N =1 & N = C^z itself

C^z/N = U gives C^z = U*N = 1*C^z or C^z = U*N = C^z*1

which implies that

For the values N =1 & N = C^z  in C^z/N = U

Rectangles formed for C^z = U*N 

have sides that represent integer 1 or C^z itself &

that Rectangles are meaningless

to PROVE COMMON FACTOR in C^z & A^x

.........

For C^z/N = U where N is an integer > 1 & N < C^z

U is an integer > 1

which implies that

for proper values of N every FACTOR in C^z can be projected as U

such that C^z = U*N = F1*F2 where F1 & F2 are 2 factors in C^z

...........

Also by RULES in Mathematics

If a COMMON FACTOR is verified in 2 INTEGERS T1 & T2

by 2 Rectangle or Square Figures for that 2 integers or

any other methods

that VERIFIED COMMON FACTOR in T1 & T2

is VALID for ever & for all cases 

including the Rectangles formed for T1 = 1*T1 & T2 = 1*T2

...........

Ex: Case(1): Let T1 = 15 & T2 = 18

In the case of Rectangles formed for 15 = 3*5 & for 18 = 3*8

Side = 3 form as Equal sides for T1 = 15 & T2 = 18 & PROVES that

T1 = 15 & T2 = 18 have COMMON FACTOR 3

.........

Ex:Case(2)

In the case of  Rectangles formed for 15 = 1*15 & 18 = 1*18

Equal sides represent integer 1 &

there isn't  any PROOF or DISPROOF for 

COMMON FACTOR in T1 = 15 & T2 = 19

But COMMON FACTOR verified for T1 & T2 by Case(1) is VALID

for the Case(2) 

where Rectangles formed for T1 = 1*15 & T2 = 1*18

...............

Which implies that

For the values N > 1 & N < C^z

C^z/N = U gives C^z = U*N = F1*F2 where F1 & F2 are 2 Factors

in C^z &

Rectangles or Square Figures formed for

C^z = U*N = F1*F2 

are SUFFICIENT to verify COMMON FACTOR C^z & A^x

by COMPARING it with Rectangle or Square figures formed for A^x

& thus to PROVE that whether

C^z & A^x have COMMON FACTOR or NOT

..................

PART A - 3 SIMPLIFIED STEPS to PROVE BEAL CONJECTURE

[3-a] SIMPLIFIED STEP (1)

A PROOF for COMMON FACTOR in any of the 2 terms in

BEAL CONJECTURE C^z = A^x + B^y

is sufficient to PROVE BEAL CONJECTURE as follows:

...............

Let C^z & A^x have COMMON FACTOR = D

which implies that C^z = K2*D & A^x = K1*D where K1 & K2 are integers such that K2 > K1 (since C^z > A^z)

C^z = A^x + B^y gives

B^y = C^z - A^x = K2*D - K1*D = (K2 - K1)*D

(K2 - K1) is an integer which implies that C^z, A^x & B^y have COMMON FACTOR

...........

[3-a] SIMPLIFIED STEP (2)

C^z & C have the same PRIME FACTORS

Similarly A^x & A have the same PRIME FACTORS

Also B^y & B have the same PRIME FACTORS

which implies that 

a PROOF for COMMON FACTOR in C^z, A^x & B^y

automatically PROVES that C, A & B have COMMON PRIME FACTOR

that are in the COMMON FACTOR in C^z, A^x & B^y

...........

[3-b] SIMPLIFIED STEP to PROVE BEAL CONJECTURE

 Based on simplified STEPS [3-a] (1) & (2)

.............

in this Article 

A proof for COMMON FACTOR in Beal Conjecture Terms C^z & A^x

is given to PROVE BEAL CONJECTURE as follows,

...................

(1) A PROOF for COMMON FACTOR in C^z & A^x automatically

PROVES that C^z, A^x & B^y have COMMON FACTOR.

..........

(2) C & C^z have SAME PRIME FACTORS,

     A & A^x have SAME PRIME FACTORS &

     B & B^y have SAME PRIME FACTORS

Therefore

A proof for COMMON FACTOR in Beal Conjecture Terms C^z & A^x

proves that C^z, A^x & B^y have COMMON FACTOR

which implies that 

in BEAL CONJECTURE C^z = A^x + B^y

C, A & B have COMMON PRIME FACTORS that are in

the COMMON FACTOR in C^z, A^x & B^y

which is the PROOF FOR BEAL CONJECTURE

     

[Part [ A-2 ] LEAST VALUES of IMPORTANT TERMS in BEAL CONJECTURE

[2A] LEAST value of C in C^z

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By conditions of Beal Conjecture A, B & C are
positive integers > 0. Which implies
A^x + B^y > 1^z 
Therefore for valid terms in Beal Conjecture C^z = A^x + B^y,
C must be integer > 1

i,e. C = integer 2 or more
Therefore C^z is a composite integer

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[2B] LEASET value of  VALID Term of  C^z, A & B 

in BEAL CONJECTURE C^z =  A^x + B^y

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We verified above in [2A]

for BEAL CONJECTURE C in C^z must be integer 2 or more,

Also

for Beal Conjecture  z must be +ve integer > 2 
Therefore for Beal conjecture value of C^z = 2^3 or more 
Which implies for Beal Conjecture
1^x + 1^y > C^z 

since 1^x + 1^y = 2 & C^z = 2^3 or more
 A & B can't be simultaneously = 1

which implies A & B are integers > 1 = 2 or more

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Also it is already proved that for,

Q^n = P^m + 1 where P,Q, m, & n are positive integers,

the only solution is 3^2 = 2^3 + 1 [Proved by Preda Mihailescu for Catalan Conjecture]

But equation  3^2 = 2^3 +1 is invalid equation for Beal Conjecture as x, y & z must be integers > 2

[2C ] Therefore for valid terms in Beal Conjecture C^z = A^x + B^y,

C, A & B are integers > 1 = 2 or more.

Which implies in BEAL Conjecture C^z = A^x + B^y

C^z, A^x & B^y are COMPOSITE INTEGERS

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Least Value for C^z = 2^3 = 8

Similarly,

For Skeletal Expression of C^z = C^2 + C^2 + C^2 + ...... ...... ......... up to C^(z-2) Number of Unit Terms

C^z = N*U = [C^2]*[C^(z-2)]

where value of Unit Term  U = C^2 & Number of Unit terms N = C^(z-2)

Least Value for C^2 = 4 &

Least Value for C^(z-2) = 2^(3-2) = 2

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PART 1 of ROOF for BEAL CONJECTURE

[I] BEAL CONJECTURE states that

if C^z = A^x + B^y

where A, B, C, x, y & z are positive integers &

x, y & z are integers > 2

then A, B & C have COMMON PRIME FACTOR

...... ........ ......

[IB] EXAMPLES for BEAL CONJECTURE C^z = A^x + B^y

...... ....... ........

(1) 2^13 = 4^6 + 16^3 .... [Common prime factor 2 ]

(2) 9^4 = 3^6 + 18^3 .... .. [Common prime factor 3

(3) (2*49^2)^3 = 49^6 + 7^13 .... [Common prime factor 7]

........ ......... ...

[IC] By the following method or similar methods

several VALID terms for BEAL CONJECTURE can be generated.

... .... ......

By FLT 

For x*y = n where x, y & n are integers > 2

A^n + B^n = K where K can't be expressed as K = C^n

Then

K^n*(A^n + B^n) = K*K^n &

[(K*A)^y]^x + [(K*B)^x]^y = K^[(x*y)+1]

gives Terms for A^x + B^y = C^z

where K cause Common Prime Factor in A, B & C

Ex: n = x*y = 3*4 = 12  where A = 2 & B = 3

Then

(2^3)^4 + (3^4)^3 = K & multiplying by K^3*4 = K^12

[(K*2)^3]^4 + [(K*3)^4]^3 = K^12+1

forms as terms for Beal Conjecture C^z = A^x + B^y

where K cause Common prime Factor in A, B & C

... .... .....