[1] PROOF for COMMON FACTOR in C^z & A^x of Beal Conjecture
C^z = A^x + B^y
which automatically PROVE COMMON FACTOR in C^z, A^x & B^y &
then PROVE Beal Conjecture as explained in this post
........ ......... ......
[1a]
By PART A [2] we have
For valid terms in Beal Conjecture C^z = A^x + B^y
C, A & B are integers > 1
therefore C^z, A^x & B^y are COMPOSITE integers
.... .....
which implies that
C^z can be expressed as multiple of 2 Factors in C^z such that C^z = F1*F2 where F1 * F2 are integers > 1 &
in Geometry
C^z = F1*F2 can be represented by Rectangle of Square Figures that have sides with integer value
...... .....
Also A^x & B^y can be represented similarly as Rectangles or Squares
......... ........ ......
[1b] VERIFICATION of COMMON FACTOR in C^Z & A^x by Geometry methods
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By Rules in Geometry,
If C^z & A^x are represented by Rectangle or Square Figures that have sides with integer value,
Then
Equal sides in that 2 Figures prove COMMON FACTOR in C^z & A^x
................
[1c] BIFURCATION of C^z to 2 INTEGERS T1 & T2
such that C^z = T1 + T2
......... ......
For N = an integer such that C^z is divisible by N
C^z/N = U where U is an integer
which implies that
C^z is divided to N number of equal Terms of integer value = U &
gives a Skeletal Expression for C^z as
C^z = U + U + U + .... ..... ..... up to N number of Equal Terms
where U forms as Unit Term &
C^z = U*N
where U = value of Unit Term & N = number of Unit Terms
which implies that
C^z/N gives C^z/N = U where C^z = U*N &
Then bifurcation of N to 2 integers such that N = L+M give
BIFURCATION of C^z to 2 integers
such that
C^z = U*N = U*(L+M) = U*L + U*M = T1 + T2
where U forms as Unit Term for C^z, T1 & T2
Which further implies that
For the value N =1
C^z/N = U = C^z itself &
There isn't BIFURCATION of C^z as C^z = T1 + T2 &
ONLY For the value N > 1
There is BIFURCATION of C^z as C^z = T1 + T2
...... ........
[1d] BIFURCATION of C^z to 2 integers such that C^z = A^x + B^y & Basic Relation Equations for A^x & B^y with C^z
Which give Rectangle or Square Figures for C^z, A^x & B^y
that have sides with integer value
....... .......
By Geometry
Beal Conjecture C^z = A^x + B^y implies that
C^z, A^x & B^y can be represented as Areas of Rectangle or Square Figures formed for C^z, A^x & B^y &
Area that represent C^z can be bifurcated such that
one portion represents the area of Rectangle for A^x &
the other portion represents the area of Rectangle for B^y
..... .......
As a portion of C^z or portion of area that represent C^z
A^x = L/N*C^z where L & N are integers such that L < N [since A^x < C^z]
Also
L & N are integers by avoiding meaningless EQUAL FACTORS in Numerator L & Denominator N of the term L/N
which implies that
L & N are integers without COMMON PRIME FACTORS &
even to equate with integer value of A^x, with its RHS part term L/N*C^z
C^z must be divisible by N
which implies that
A^x = L/K*C^z = L*(C^z/N) = L*U where U is an integer = C^z/N
.... ........
Also
B^y = C^z - A^x
= C^z - [L*(C^z/N)]
= C^z*[1- (L/N)] = C^z*[N/N - L/N] = [(N-L)/N]
N-L is an integer say M
Then
B^y = M*(C^z/N) = M*U &
BASIC RELATION EQUATIONS of A^x & B^y with C^z form as
A^x = L*U & B^y = M*U
where U is an integer = C^z/N & C^z = U*N
....... ........
Which implies that
For an integer N such that C^z is divisible by N & C^z/N = U
C^z = U*N & its related Basic Relation Equation A^x = L*U give Rectangle or Square Figures for C^z & A^x that have sides with value in integers &
that rectangle or square figures can be used for VERIFICATION of COMMON FACTOR in C^z & A^x by methods in geometry
..... ...... ......
Also
C^z/N = U implies that
For the value N = C^z value of U = 1 & C^z = U*N = 1*N = 1*C^z
Similarly
For the value of N = 1 value of U = C^z itself & C^z = U*N = C^z*1
which implies that
For the value of N = 1 & N = C^z itself
Rectangles formed for C^z = U*N are identical Rectangles that have breadth = 1 & Length = C^z itself
& meaningless to VERIFY COMMON FACTOR in C^z & A^x by methods in geography
since all Factors in C^z are available as multiple represented by One side
........
On the other hand
For the value of N > 1 & N < C^z
C^z/N = U
gives value of U = integer > 1 & < C^z which implies that U is a Factor in C^z
.... ........
which further implies that
For proper values of N every Factor in C^z can be projected as U where C^z = U*N = F1*F2 such that F1 F2 are 2 Factors in C^z
Therefore
Rectangle or Square Figures for C^z = U*N = F1*F2
are SUFFICIENT to VERIFY & CONFIRM that whether C^z & A^x have COMMON FACTOR or NOT by methods in Geometry as given below
....... ....
C^z = U*N & A^x = L*U where U is an integer = C^z/N
implies that
for valid terms in Beal Conjecture C^z = A^x + B^y
for different values of N
C^z/N = U & C^z = U*N give all Rectangle or Square Figures for C^z &
For the Rectangle or Square Figures formed for C^x = U*N
A^x forms as an INNER AREA Rectangle or Square Figure in the Area of Rectangle or Square Figure for C^z
by sharing one of the side as COMMON SIDE such that C^z = U*N &
A^x = L*U
...... ...... .....
Which implies that
There are 2 CASES for the Rectangle or Square Figures formed for
C^z = U*N & A^x = L*U
For the CASE(1): U = 1 &
For the CASE(2): U is an integer > 1
For the CASE(2): where U = integer > 1
A^x = L*U & B^y = M*U where C^z = N*U implies that
C^z, A^x & B^y have COMMON FACTOR that is VERIFIED by the EQUAL SIDE formed for the term U
For the CASE(1) where U = 1
Rectangles formed for A^x, B^y & C^z have Breadth = 1 & Lengths
A^x, B^y & C^z where that Rectangles DOESN'T PROVE or DISPROVE that C^z, A^x have COMMON FACTOR by methods in Geometry
where EQUAL SIDES of 2 Rectangles for C^z & A^x verify COMMON FACTOR
.... ........
[1d] By comparing the Rectangle or Square Figures for C^z & A^x for the CASE(1) where U = 1 & for the CASE(2) where U > 1
we can PROVE that C^z & A^x have COMMON FACTOR as follows
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For Verification of COMMON FACTOR in C^z & A^x by methods in geometry by comparing Length of sides of Rectangles for C^z & A^x
the Rectangles for C^z & A^x must have sides in integer values
which implies that
Rectangles formed for the CASE(1) where breadth represent the term U =1 can be taken as ORIGINAL &
Rectangles formed for the CASE(2) where side that represent the term U = integer > 1
can be taken as Modified Rectangle of the Rectangle in the CASE(1)
....... ....
Let Rectangle PQRS represents C^z = U*N for the CASE(1) where U =1 which implies that N = C^z itself such that
Area of the Rectangle = 1* C^z = C^z
For the Rectangle PQRS
Let side PR = U =1 & side PQ = N = C^z
Therefore PR*PQ = C^z
.... ........ ......
Let Rectangle or Square P1Q1R1S1 represents C^z = U*N for the CASE(2) where U = integer > 1
which implies that U & N are 2 Factors in C^z such that
Area of the Rectangle or Square = U*N = F1*F2 = C^z
Let side P1R1 = U = F1 &
the side P1Q1 = N = F2
which implies that
Since both Rectangles have the same area = C^z &
sides have values in integer
side PR = U = 1 is increased as side P1R1 = U = integer > 1 &
side PQ =C^z is reduced to P1Q1 = F2
so that PR*PQ = P1R1*P1Q1 = C^z
which implies that
P1R1 > PR where PR = 1 & P1R1 = F1 . .......... .. [eq1] &
P1R1*P1Q1 = C^z ............ .. [eq2]
Also
PQ*PR = C^z = P1Q1*P1R1 where PR = 1 implies that
PQ = P1Q1*P1R1 ............. [eq3] &
PQ > P1Q1 where PQ = C^z & P1Q1 = F2 ........... [eq4]
[Also
when Rectangle PQRS that have side PR = U = 1 is Modified to Rectangle P1Q1R1S1
side PR = 1 is increased to side P1R1 = integer > 1 &
side PQ = C^z is decreased to PIQ1
to form Rectangle P1Q1R1S1
which implies that
PIR1 > PR &
PQ < P1Q1]
...........
A^x = L*U forms as INNER AREA Rectangle or Square Figures in Area of Rectangle or Square figures for C^z = U*N
which implies that
For the CASE(1)
A^x = L*U forms as INNER AREA Rectangle PDRE to Rectangle PQRS
by sharing the side PR = U = 1 as COMMON SIDE
such that D is a point on the side PQ & E is a point on the side RS
For Rectangle PDRE
Its area = PD*PR = A^x where PR = 1
which implies that PD = A^x such that ......... [eq5] &
Also
For the Rectangle or square P1Q1R1S1 that represent C^z = U*N where U = integer >1 & U & N are 2 Factors in C^z such that C^z = F1*F2
A^x = L*U where U is an integer > 1 forms
as an INNER AREA Rectangle or Square Figure P1D1R1E1 to
Rectangle or Square P1Q1R1S1
by sharing side P1R1 = U as COMMON SIDE for both figures for C^z & A^x where P1R1 = integer > 1
.... ........
Therefore
By Rectangle PDRE & Rectangle P1D1R1E1
A^x = PR*PD = P1R1*P1D1 ........... [eq6]
PR = 1
which implies that
PD = P1D1*P1R1 where P1R1 is an integer > 1 which implies that
PD > P1D1 .......... [eq7]
.... ......
Therefore
by the Rectangle or square Figures in the CASE(2) where U > 1
C^z = PIR1*P1Q1 & A^x = P1R1*P1D1 .......... [eq8]
.... ......
Also Area Rectangle PDRE = Area Rectangle or Square P1D1R1F1
which implies that
PR*PD = P1R1*P1D1 where PR =1 which implies that
PD = P1R1*P1D1 ............... [eq9]
...........
[When Rectangle PQRS of area C^z is modified to Rectangle P1Q1R1S1
side PR = U = 1 is increased to side P1R1 = U = integer > 1 &
side PQ = N = C^z is decreased to side P1Q1 = F2
So that the Area = C^z is kept the same for both Rectangles
Similarly
When Rectangle PDRE of area A^x is modified to Rectangle P1D1R1E1
side PR = U =1 is increased to side PIR1 = U = integer >1 &
side PD = A^x is decreased to side PID1
so that the Area = A^x is kept the same for both Rectangles
which implies that
P1Q1 < PQ &
P1D1 < PD ]
... .... ....
Therefore by the rectangle or Square Figures for C^z & A^x
for the CASE(1) & CASE(2)
we have the following equations & values
By Rectangle PQRS & its modified Rectangle P1Q1R1S1
C^z = PQ = P1R1*P1Q1 &
PQ > P1Q1
By the Rectangle PDRE & its modified Rectangle P1D1R1F1
A^x = PD = P1R1*P1D1 &
PD > P1D1
Therefore
C^z/A^x = PQ/PD = P1R1*P1Q1/P1R1*P1D1 = P1Q1/P1D1
where P1Q1 < PQ
P1D1 < PD
Also
PQ, P1Q1, PD &P1D1 represent integer values
...... .......
which implies that
By dividing C^z/A^x
Numerator C^z = PQ is converted to a lesser valued integer PIQ1 = F2 &
Denominator A^x = PD is converted to a lesser valued integer PID1
By Rules in Mathematics
If an integer T1 is divided by another integer T2 &
T1 & T2 are converted to lesser valued integer by that process of Division
then T1 & T2 have COMMON FACTOR in Numerator Term = T1 & Denominator Term = T2
.... .....
Therefore
C^z/A^x = PR*PQ/PR*PD = PQ/PD = P1Q1/P1D1
where PQ & P1Q1 are integers such that P1Q1 < PQ &
PD & P1D1 are integers such that P1D1 < PD
implies & PROVE that
for Beal Conjecture C^z = A^x + B^Y
C^z & A^x have COMMON FACTOR
..........
Example:
For Beal Conjecture 9^4 = 18^3 + 3^6
9^4 = U*N = 9^3*9 & A^x = 18^3 = U*L = 8*9^3 gives
C^z/A^x = 9^4/18^3 = 9/8
which implies U = 9^3 is common factor in C^z = 9^4 & A^x = 18^3
.... ................
For Beal Conjecture C^z = A^x + B^y
Proof for COMMON FACTOR in C^z, A^x & B^y based on the PROOF for COMMON FACTOR in C^z & A^x &
PROOF for BEAL CONJECTURE based on the PROOF for COMMON FACTOR in C^z, A^x & B^y
are given in PART A- 6
