[I] Essential COMMON UNIT needed to express 3 ENTITIES or Single Terms as E1 + E2 = E3
..... .... ....
[IA] By Rules in Mathematics
For E1 + E2 = E3
When E1 = 1 Kg & E2 = 1 Pound
Since Units are different for E1 & E2
E3 haven't a solution in single term
But Unit Pound can be converted to Unit Kg &
when Unit of E2 is converted to Kg [or Unit Kg to Unit Pound]
we have a solution for E3 in single Term
for E2 = 1 Pound = 0.454 Kg & E1 + E2 = E3 gives
1 kg + 0.454 Kg = 1.454 Kg
......... .....
But
For E1 = 1 Kg & E2 = 1 meter
E1 & E2 can't be converted to same Unit &
For E1 + E2 = E3
E3 haven't a solutions in single term
...... ....
For E1 = 3 & E2 = 5
E3 have a solution a solution in single term as E3 = 8
where integer 1 forms as Unit in 3 & 5 such that
3*1 + 5*1 = (3+5)*1 = 8*1 = 8
Which implies that
Integer 1 is available as Unit term with all integers
... .... ..
Which implies that
to ADD Two Entities E1 & E2 & to express its result as a Single Term E3 such that E1 + E2 = E3
there should be a COMMON Unit for E1, E2 & E3
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[IB] BASIC UNIT Term for INTEGERS &
SKELETAL Expression of INTEGERS
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In Number Series every new integer is generated by adding integer 1 to its nearest previous integer
which implies that
Basic Unit of every integer can be taken as integer 1 &
an integer T can be expressed by a Basic Skeletal Expression as
T = U + U + U + .... .... .... up to N number of Unit Terms where
Value of Unit Term U =1
Number of Unit Terms N = T itself &
T = U*N = 1*T
....... ....
Also
2 or more Basic Unit Term =1 can be added to form as Unit Term of Higher value where
T = N*U have
U = integer > 1
N = integer < T
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Ex: T = 6 = 1 + 1 +1 +1 +1 +1 where U =1 & N = 6
T = 6 = 2 + 2 +2 where U = 2 & N = 3
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[1C] EQUATION for SKELETAL EXPRESSION of an integer
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For T/N where N is an integer & T is divisible by N
T/N = U where U is an integer &
T/N = U implies T is divided to N number of UNIT terms of integer value = U &
T can be expressed by a SKELETAL EXPRESSION as
T = U + U + U + ... ..... ..... ..... up to N number of Unit Terms
Which implies that
T/N = U gives
Skeletal Expression of an integer for different values of Unit Term U
which implies that
There are 2 CASES for Skeletal Expressions of an integer based on value of Unit Term
Case (1): U = 1
For the value N = T itself
T/N = U = 1 where
T = U*N = 1*T
Case(2): U > 1
For the value N < T
T/N = U where U is an integer > 1
which implies that
for the Case (2} T is a COMPOSITE integer such that
T = U*N = F1*F2 where U =F1 & N = F2 are integers > 1 &
2 FACTORS in T
...... .....
Therefore
Prime integer 5 have only Case(1) Skeletal Expression
where T/N = U forms as 5/5 = 1 & gives Skeletal Expression as
5 = 1 + 1 + 1 + 1
where
Composite integer 6 have both Case(1) & Case(2) Skeletal Expressions
For T/N = U
6/6 = 1 forms as
6 = 1 + 1 + 1 + 1 +1
6/3 = 2 forms as
6 = 2 + 2 + 2
.. ..... ......
[1D] U - N graph for an integer based on Skeletal Expression & its
Rectangle or Square Figures formed to every integers for T = U*N
...... ...... .....
For T = U*N
By taking Term for U along X-Axis & Term for N along Y=Axis
every integer have a Rectangle or Square Figure that have sides in integer value
which implies that
There are Two cases for U-N Rectangle or Square Figures for integers
Case(1): Rectangles that have breadth = 1 & Length = T itself
formed for T = U*N where U = 1 & N = T itsel
Case(2): Rectangle or Square Figures
that have both sides as integers > 1 & < T
Formed for T = U*N = F1*F2 where F1 & F2 are Factors in T
........ ........ .....
which implies that
Prime Integers have only the Rectangle in the CASE(1) &
Composite integers can be represented by any of the Rectangle or Square figures in the CASE(1) or the CASE(2)
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[II] UNIT Term for Beal Conjecture C^z = A^x + B^y &
Skeletal Expression by keeping conditions of Beal Conjecture
..... .......
[IIA] Minimum value of C & one among A & B
in Beal Conjecture C^z = A^x + B^y
..... .... ....
By Condition of Beal Conjecture
C, A, B, x, y & z are positive integers where x, y & z > 2
which implies that
A^x + B^y > 1
Therefore for A^x + B^y = C^z implies that
C is an integer > 1 & C^z is a COMPOSITE integer
.... ....
Also
C = integer > 1 & z = integer > 2 implies that
the least valid term for C^z = 2^3 = 8 &
A^x + B^y = C^z = integer 8 or more implies that
A & B can't be simultaneously = 1
therefore
Conditions of Beal Conjecture implies that
C = integer > 1 &
either A or B must be > 1
Let we take A > 1
..... .....
Thus we verified that
For Beal Conjecture C^z = A^x + B^y
C & A are integers > 1 & C^z & A^x are Composite Integers
.... .... ......
which is sufficient to PROVE that C^z, A^x & B^y have COMMON FACTOR &
PROOF for COMMON FACTOR in C^z, A^x & B^y
will provide an AUTOMATIC PROOF for Beal Conjecture
as explained in the following parts
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Also by conditions of Beal Conjecture
z = 2 + n where n is integer > 0
which implies that for Beal Conjecture z = integer 3 or more &
n = z-2
.... ..... ..... .......
[IIB] By keeping conditions of Beal Conjecture
SKELETAL EXPRESSION for C^z
...... ..... .....
By conditions of Beal Conjecture z = 2 + n where n is integer > 0 &
C^z = C^(2+n) = C^2*C^n = (C^2)*C^(z-2)
which implies that
For all values of C & z
C^2 forms as BASIC Unit Term for Beal Conjecture Term C^z &
SKELETAL EXPRESSION for Beal Conjecture term C^z forms a
C^z = C^2 + C^2 + C^2 + ..... ..... .... up to C^(z-2) number of Unit Terms where
U = C^2 = integer > 1 &
N = C^(z-2)
C^z = U*N where U = integer > 1 implies that
N = C^(z-2) = integer > C^z
Which further implies that
By keeping conditions of Beal Conjecture
C^z haven't a Skeletal Expression where U = 1 &
its Rectangle Figure formed C^z = U*N = 1*C^z
Also which implies that
in the case of U-N Rectangle or Square Figures
for Beal Conjecture Term C^z
only CASE(2) Rectangle or Square Figures are relevant
where
Rectangle or Square Figures for C^z have both sides > 1 & < C^z
for C^z = U*N = F1*F2
...... ..... ...
Also
By conditions of Beal Conjecture
for the Least Valid Term for C^z = 2^3 &
2^3 = 2^2 + 2^2 where
U = C^2 & N = 2 = C
Similarly for the next higher value of C = 3
least term for C^z = 3^3 &
3^3 = 3^2 + 3^2 + 3^2 where
U = 3^2 & N = 3 = C
Which implies that
By keeping conditions of Beal Conjecture
For C^z = U*N
U = 2^2 or more & N = 2 or more
which implies that
For C^z = U*N values of U & N are integers >1 & < C^z
.... ..... .....
[Similarly
as we have taken A^x as the verified term that have base number > 1
among A^x & B^y
By keeping conditions of Beal Conjecture
Also A^x haven't a Skeletal Expression for U = 1 & its Rectangle figure for A^x = 1*A^x]
....... ..... .......
[III] LEMMA-1
for BIFURCATION of an INTEGER to 2 INTEGERS
where T3 = T1 + T2 &
COMMON Unit Term U as multiple in T3, T1 & T2
where U is an integer
..... ..... ...... ...
By Geometry
T3, T1 & T2 can be expressed as areas of Rectangles for T3 = 1*T3
T2 = 1*T2 & T1 = 1*T1
therefore by geometry
T3 = T1 + T2 implies that
area for T3 can be bifurcated where ONE of the portion represents area for T1 & the remaining 2nd portion represents area for T2
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[IIIA] Basic Relation Equation between T1 & T3 in T3 = T1 + T2
.... .... ......
As a portion of area that represent T3
T1 = L/K*T3
where L & K are integers such that K > L ..... [since T3 > T1]
Also for the term L/K
Meaningless EQUAL FACTORS in Numerator L & Denominator K
can be avoided
which implies that for L/K
L & K are integers without COMMON PRIME FACTORS
..... ..... ......
Which further implies tha
even to EQUATE with INTEGER value of T1
For T1 = L/K*T3
T3 must be divisible by integer K such that
T1 = L/K*T3 = L*D where D is an integer & D = T3/K
..... ......
Also in this article
T3/K = D & T3/N = U are identical Equations where
N & K are just 2 INTEGERS such that T3 is divisible by N or K
Therefore
N & K can be substituted with each other
For convenience
Let we take T3/N = U for T3/K =D
so that Basic Relation Equation of T1 with T3 can be linked with already explained Skeletal Expression for T3
Therefore
T1 = L*(T3/N) = L*U where T3 is divisible by integer
...... ........ ......
Also
T3/N = U gives T3 = U*N and
T3 = U*N & T1 = L*U implies that
For 3 integers such that T3 = T1 + T2
T3 & T1 are Multiple of a Term of integer value = U= T3/N
...... ...... .....
[IIIB] BASIC Relation Equation between T2 & T3
where T3 = T1 + T2
..... ..... ...
T2 = T3 - T1 where T1 = L*(T3/N) = L*U
T2 = T3 - [L*(T3/N)] = T3*[1- (1/N)] = T3*[N/N - L/N]
=T3*[(N-L)/N]
N-L is an integer say M Then
T2 = M/N*T3 = M*(T3/N) = M*U
..... ....
[IIIC]Therefore
for 3 integers expressed as T3 = T1 + T2
Basic Relation Equations for T1 & T2 with T3 form as
T1 = L*U & T2 = M*U where U = T3/N & T3 = N*U
Which implies that
T3, T1 & T2 are Multiples of a COMMON Unit term U that have integer value
........ ....... ......
[IIIC] Based on SKELETAL Expression
Explanation for BASIC Relation Equations for T1 & T2 with T3
where T3 = T1 + T2
..... ....... .....
By Basic Relation Equations
T1 + T2 = L*(T3/N) + M*(T3)N) = (L+M)*(T3/N) = T3
which implies that L+M = N
Also
T3 = N*U = (L+M)*U = L*U + M*U = T1 + T2
where
T3 = N*U, T1 = L*U & T2 = M*U implies that for 3 integers expressed as T3 = T1 + T2
One of the Skeletal Expression for T3 = N*U is bifurcated such that
N = L+M where term for N is bifurcated to 2 integers L & M & Term for U is kept fixed
which implies that Skeletal Expressions for T1 & T2 are formed such that
T1 = L*U & T2 = M*U
... .... ....
which implies that
there are 2 cases for T3 = T1 + T2
CASE(1): Skeletal Expression of T3 is bifurcated where U > 1
which implies that
T3, T1 & T2 have COMMON FACTOR caused by the Common Unit term U such that
T3 = U*N = U*(L+M) = U*L + U*M = T1 + T2
CASE(2): Skeletal Expression of T3 is bifurcated where U =1
then Common Unit Term U = 1 don't cause COMMON FACTOR in T3, T1 & T2
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[IIID] Implication of Lemma-1 & COMMON FACTOR in Beal Conjecture Terms C^z, A^x & B^y
.... .... ......
For Beal Conjecture C^z = A^x + B^y
C^z, A^x & B^y are integers & Rules of Lemma-1 is applicable &
Basic Relation Equations for T1 = A^x & T2 = B^y with T3 = C^z
form as
A^x = L*(C^z/N) = L*U &
B^y = M*(C^z/N) = M*U
where C^z = N*U
Also
By conditions of Beal Conjecture C^z haven't a Skeletal Expression where U = 1
which implies that
Skeletal Expression of Beal Conjecture C^z have U = integer > 1 &
Bifurcation of the Skeletal Expression of C^z cause formation of Beal Conjecture C^z = A^x + B^y
where U > 1 cause COMMON FACTOR in C^z, A^x & B^y
for the expansion
C^z = U*N = F1*F2 = F1*(L+M) = F1*L + F1*M &
for proper values of F1, L & M
F1*L forms as A^x & F1*M forms as B^y
.... .... .....
Ex: C^z = 9^4 = N*U = (9)*(9^3) &
N = 9 = L+M = 1+8 gives
9^4 = N*U = (9)*(9^3) = (1+8)*9^3 = 1*9^3 + 8*9^3 = 3^6 + 18^3 & Beal Conjecture 9^4 = 3^6 + 18^3 is formed
where U = 9^3 cause COMMON FACTOR in C^z, A^x & B^y
..... .... ...
Similarly Also
For Beal Conjecture 2^13 = 4^6 + 16^3
2^13 = 2^12*(2) = 2^12*(1+1) = 1*2^12 + 1^2^12 = 4^6 = 16^3
U = 2^12 causes Common Factor in C^z, A^x & B^y
.... .... ....
For Beal Conjecture [2*(49)^2]^3] = 49^6 + 7^13
bifurcation of N = 2^3 = (1+7) gives
A^x = 1*49^6 & B^y = 7*(49)^6 =7*13 where
U = 49^6 = 7^12 causes Common Factor in A^x, B^y & C^z
.... .....
[IIIE] Common Factor Verification for C^z, A^x & B^y
for the cases C^z = U*N, A^x = U*L & B^y = U*M
where U > 1 & U = 1
.... ...... ....
By RULES in mathematics/Geometry
COMMON FACTOR is verified in 2 integers T1 & T2 by 2 Rectangle or Square Figures for T1 & T2 that have sides in integer values
where EQUAL SIDES prove Common Factor in T1 & T2
then that PROVED Common Factor is VALID for all Figures for T1 & T2
which implies that
Beal Conjecture Term C^z is a Composite integer that
essentially have Rectangle or Square Figures for C^z = U*N = F1*F2 & U = F1 is integer > 1
where Basic Relation Equations A^x = L*U = L*F1 &
B^y = M*U = M*F1 verify Common Factor in C^z, A^x & B^y
Also
Without considering conditions of Beal Conjecture
C^z, A^x & B^y can be represented by Rectangles for C^z = 1*C^z, A^x = 1*A^x & B^y = 1*B^y
where equal sides represents integer 1 &
1 is NOT considered for Common Factor
but by rules in Geometry as said above
Common Factor VERIFIED by Basic Relation Equations for
C^z = U*N = F1*F2, A^x = F1*L & B^y = M*F1 are VALID for
the Rectangles formed for C^z = 1*C^z, A^x = 1*A^x & B^y = 1*B^y
..... .......... ...
Therefore
By conditions of Beal Conjecture,
particularity of value of U for C^z &
Lemma-1 we have PROVED that
for Beal Conjecture C^z = A^x + B^y
C^z, A^x & B^y have COMMON FACTOR
........ ..........
[IIIF] COMMON FACTOR in C^z, A^x & B^y is the CAUSE for COMMON PRIME FACTOR in A, B & C
as explained in the PROOF for BEAL Conjecture
given below
...... ..... ....
C^z & C have same COMMON PRIME FACTORS,
A^x & A have same COMMON PRIME FACTORS &
B^y & B have same COMMON PRIME FACTORS
which implies that
PRIME FACTORS that are in COMMON FACTOR in C^z, A^x & B^y form as COMMON PRIME FACTORS in C, A & B
Ex: For 9^4 = 3^6 + 18^3
9^3 is Common Factor in C^z = 9^4, A^x = 3^6 & B^y = 18^3 &
3 forms as Common Prime Factor in C = 9 A = 3 & B = 18
.... ..... .....
Therefore in this article
we have PROVED that
For BEAL CONJECTURE C^z = A^x + B^y
C^z, A^x & B^y have COMMON FACTOR &
C, A & B have COMMON PRIME FACTOR
therefore
we have PROVED BEAL CONJECTURE
.... ...... .... .....
