God Equations Of Balance.
Include Murgu The Lost Fundamental as obtained from.

Murgu The Lost Fundamental OPERATOR come with Infinity Equations in One even wearing Science Issues in 2 Forms Absolute Truth Values.
    Murgu Infinity In One brought also Infinty to Infinty Equations with ScienceValue Absolute Truth as
Murgu Irrationals Integrators ,
Murgu Special Irrationals
Murgu Special Quadruplets
Murgu Quadruplets
and Murgu God Equations Of Balance.
Murgu God Equations Of Balance include Murgu The Lost Fundamental OPERATOR both forms and can be verry util in Polynomials and not Only.

To use Murgu The Lost Fundamental OPERATOR need any skill and here we will present a part of God Equations Of Balance.
𝔖(x + y)ⁿ = n!   
𝔖(x - y)ⁿ = n!    if (x - y) ≠ 0
𝔖(xⁿ) + 𝔖(yⁿ) =2 n!    but
𝔖(x + y)ⁿ = n!
𝔖(xⁿ) - 𝔖(yⁿ) =0    but
𝔖(x - y)ⁿ = n!
The Lost Fundamental Operator as Identitie need to be used with care, but for sure util in any polynomials, by using any proprieties, as.
𝔖(xⁿ) = 𝔖(yⁿ)
𝔖(xⁿ) = n 𝔖(y^(n-1))
𝔖(xⁿ⁺¹) = (n+1) 𝔖(yⁿ)
𝔖(xⁿ⁺¹) + 𝔖(yⁿ) = (n+2)n!
𝔖(xⁿ⁺¹) - 𝔖(yⁿ) = (n-1)n!
AND MORE :
Important
God Equations Of Balance are pillars for Fermat's Last Theorem Unique Solution.

God Equations Of Balance
are part of USA 2015 AEPDF and are extracted from - Murgu Infinity In One - than are,
Infinity To Infinity
The name was a Reply to Imagination Beauty
Euler God Equation
Used with care as tools bring miracle for Equations and polinomials.
A equation of powers variables can wear now also new forms.

Murgu Double False Redundancy Of Truth.

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Murgu Double False Redundancy Of Truth institute Logic Logical check , review and control. Murgu Double False Redundancy Of Truth can be a big impact not only for Mathematic and Logic but for all Science.
Translated in a simple mode mean Any Equations In Physics, gen (y=ax)n or more complex may have possible form (iy=iax). That mean Math Relativity. We can't, yet, make a case for, right now, but a remark for future.

We will make Analyze for starting with Fermat's Last Theorem , then Fermat's Equations.
If Fermat's Last Theorem Equations have a solution into Integers
Xⁿ + Yⁿ - Zⁿ = 0
(X,Y,Z) Integers Solution, Then Infinity (iX,iY,iZ)

Xⁿ + Yⁿ - Zⁿ = 0
0, important then W Integers because around of Zero play any Conservations Laws, then let us to remark ,
Xⁿ + Yⁿ - Zⁿ = W
can have multiple solutions into Integers (X,Y,Z,W)Integers. To return to Xⁿ + Yⁿ - Zⁿ = 0
1 Solution In Integers then Infinity iⁿ(Xⁿ + Yⁿ - Zⁿ) = 0
not simplify and then (iX,iY,iZ) Solutions.

It make an absolute rights to write
Murgu Double False Redundancy Of Truth Analytic Equations
Xⁿ + Yⁿ - Zⁿ = 0 mdfr1
(i(Xⁿ) + i(Yⁿ) - i(Zⁿ) = 0 ) mdfr2
(iX)ⁿ + (iY)ⁿ - (iZ)ⁿ = 0 mdfr3
but also
(1/i(Xⁿ) + 1/i(Yⁿ) - 1/i(Zⁿ) = 0 ) mdfr4
(X/i)ⁿ + (Y/i)ⁿ - (Z/i)ⁿ = 0 mdfr5

Absurd or not , need to include mdfr1 as Integers possible solution and to remind
X² + Y² - Z² = 0
are Fermat's Last Theorem Except.
Then if mdfr1 have a solution , Then Infinity, and mdfr2 come with Infinity Irrational Solutions.
mdfr4 mean in the same , every n-times multyply with any Integers brought a new Solutions including
Xⁿ(Xⁿ + Yⁿ - Zⁿ) = 0 mdfr6
Yⁿ(Xⁿ + Yⁿ - Zⁿ) = 0 mdfr7
Zⁿ(Xⁿ + Yⁿ - Zⁿ) = 0 mdfr8

mdfr6, mdfr7 and mdfr8 maybe don't say much here , but are important for Fermat's Last Theorem Unique Solution. Part two of this Analytic is for Rationals and Integers as any times may can fall in and bring
Euler Murgu Equations 1 Equal 1
for i equal
1/(Zⁿ)
(X/Z)ⁿ + (Y/Z)ⁿ - 1 = 0 EME1E1
And we know EME1E1 CERTIFY Complet and accurate
Fermat's Last Theorem
Fermat's Last Theorem Unique Solution and Euler Murgu Equations 1 Equal 1 after it as imply a bit Double False Redundancy Of Truth.
As can see from both sides we started from absurd supositins and let we demonstrated FASLE, Then Double False , And Redundant folowin i those express a Truth - mean -

Fermat's Last Theorem Unique Solution
Fermat Murgu Impossible Equations Method.

If Apply to Fermat's Last Theorem Equations MTLF Direct
𝔖Xⁿ + 𝔖Yⁿ - 𝔖Zⁿ = n!
Function of which operator form apply we will get first or last component
{Xⁿ + Yⁿ - Zⁿ = 0}
which for 3 components(X,Y,Z) implied, is impossible, but weird to demonstrate.
For a Component, can have a Term 0, but for 3 not but a clear demonstration is hard.
𝔖Xⁿ can have a term 0 but for 3 is like
a + b - c =0
which into integer is clear possible. For our case a,b and c are Integers of form Tⁿ , that intuitive is IMPOSSIBLE.
REPEAT, is there an unclerity, then I Chose another Logical Way
a Logical Analytic for symmetric neighbors of Supposed by Absurd
{Xⁿ + Yⁿ - Zⁿ = 0}
Step 1:
{((X-1), (Y-1), (Z-1)); ((X-2), (Y-2), (Z-2)); ...} AND
{((X+1), (Y+1), (Z+1)); ((X+2), (Y+2), (Z+2)); ...} we will work only with
((X-1), (Y-1), (Z-1)); and ((X+1), (Y+1), (Z+1));
Step 2:
Apply adequate Operator (ascendend or descede)
𝔖(X-1)ⁿ + 𝔖(Y-1)ⁿ - 𝔖(Z-1) ⁿ = n!
𝔖(X+1)ⁿ + 𝔖(Y+1)ⁿ - 𝔖(Z+1) ⁿ = n!
Step 3:
for evryone , second component will be
n(Xⁿ + Yⁿ - Zⁿ) = 0
Step 4 :
Logical Analytic for
n(Xⁿ + Yⁿ - Zⁿ) = 0
passing of first impulse is
{(ⁿ√ _n)X, (ⁿ√ _n)Y, (ⁿ√ _n)Z}
are also SOLUTIONS . and are IRRATIONALS .
This is already first sign of Impossible
Step 5:
If now apply for next symmetric (n-1) Neighbors into The Left and into The Right Will get 2(n-1) new K_nI(Xⁿ + Yⁿ - Zⁿ) = 0
for I between 2 and n. We started from a supposed
(Xⁿ + Yⁿ - Zⁿ) = 0
NOW every conditional claim Irrationals Solutions . This Is IMPOSSIBLE when supposed
(Xⁿ + Yⁿ - Zⁿ) = 0
claim fix values into Integers for (X,Y,Z)
That Mean

n neighbors into the left and n into the right will have one term of Operator Zero.
For sure there can be more then one term 0. To chose one
((X-1), (Y-1), (Z-1)); or ((X+1), (Y+1), (Z+1));
((X-1)= U, (Y-1)=V, (Z-1)=W); For Sure
(Uⁿ + Vⁿ - Wⁿ) = B Then
((U+1)ⁿ + (V+1)ⁿ - (W+1)ⁿ) = 0
Make one of them Impossible for powers n>2 . For Power 3 and even 4 we can show: Here for 3
U³+3U²+3U+1 + V³+3V²+3V+1 - W³-3W²-3W-1 =0
B+3U²+3U+3V²+3V-3W²-3W+1 =0
and in the same time for
((X+1)=T, (Y+1)=L, (Z+1)=F)
B-3T²+3T-3L²+3L+3F²-3F-1 =0
IMPOSSIBLE
Don't forgot we can Apply Murgu The Lost Fundamental Operator to groups (U,V,W) and (T,L,F) making abstraction of B as know
𝔖Uⁿ + 𝔖Vⁿ - 𝔖Wⁿ = n! AND
𝔖Tⁿ + 𝔖Lⁿ - 𝔖Fⁿ = n!
Then C _n,I (Xⁿ + Yⁿ - Zⁿ) = 0
inside those operators mean POSSIBLE only IN IRRATIONAL.

But above relations mark clear
(Xⁿ + Yⁿ - Zⁿ) = 0 Impossible into Integers .
(n(Xⁿ) + n(Yⁿ) - n(Zⁿ)) = 0
Conditional coming for first neighbor express a TRUTH
(Xⁿ + Yⁿ - Zⁿ) = 0
have solutions only in Irrationals. And for every Z are Infinity. Fermat's Last Theorem Solved
Unique Solution Murgu FMIE
As Reveal The Domain of Solutions for and now offer to
Fermat's Last Theorem absolute right of
SECOND Fundamental Science.

Down will follow any logical which lead to same results.

Double False Redundancy Of Truth bring more light on.
If a solution into Integers then Infinity
kⁿ(Xⁿ + Yⁿ - Zⁿ) = 0
(X,Y,Z) Integers Solution, Then Infinity (kX,kY,kZ)

For k#(l)ⁿ k(Xⁿ + Yⁿ - Zⁿ) = 0
Solutions will be
Xⁿ√ _k , Yⁿ√ _k , Zⁿ√ _k
That mean , even need a heavy logic ,
Solution into Integers Come from Irrationals or Integers to Irrationals
Both alternative are Equivalents.
That mean IF
(Xⁿ + Yⁿ - Zⁿ) = 0
Solution in Integers Then Multiple Images in Irrationals
It make Euler Murgu Equations 1 Equal 1 an accurate and complete
Fermat's Last Theorem Certify

FMIE And Euler Murgu Equations 1 Equal 1 Make Fermat's Last Theorem
Second Science Fundamental
where for Physics Science Fundamental mean Science Fundament.
FMIE send Solutions In Irrationals (Infinity) and stand for Impossible Solution into Integers and Rationals.
DOWN is Demonstration

If luck back to mdfr6, mdfr7 and mdfr8, that mean also:
Iⁿ(Xⁿ + Yⁿ - Zⁿ) = 0

Murgu The Lost Fundamental Operator Solved
Fermat's Last Theorem
AND Make
Fermat's Last Theorem
SECOND Fundamental Science.

FMIE using Murgu The Lost Fundamental Operator Claim as
(Xⁿ + Yⁿ - Zⁿ) = 0
to have a solution into Integers ((X,Y,Z)-Integers) then also
n(Xⁿ + Yⁿ - Zⁿ) = 0
Xⁿ√ _k , Yⁿ√ _k , Zⁿ√ _k
This is also a refuse of (Xⁿ + Yⁿ - Zⁿ) = 0 into Integers, and if follow back Double False Redundancy Of Truth Infinity Refuses.
This is important even for Murgu The Lost Fundamental - into Integers for 3 components do not accept 0 terms.

(Xⁿ + Yⁿ - Zⁿ) = 0
Solution into Integer (see above), bring Infinity Absurd Solutions into Integers, as :
XⁿXⁿ+XⁿYⁿ - XⁿZⁿ
XⁿYⁿ+YⁿYⁿ - YⁿZⁿ
XⁿZⁿ+ZⁿYⁿ - ZⁿZⁿ
and all combinations possible from .
But also (see Double False Redundancy Of Truth),
(X/Z)ⁿ + (Y/Z)ⁿ - 1 = 0 EME1E1
COMPARED with

(X/Z)² + (Y/Z)² - 1 = 0 EME1E1 for P=2
which lead my in maybe 1993 , 94, 95 , or 96 to a geometric solution in which I make a connection to Geometric Media , from which will return down. Now to use it to Explain FMIE once more time.

Murgu Identities Explain FMIE for Power 3.
FMIE via Murgu The Lost Fundamental Solved Fermat's Last Theorem
but for Science good will try to help FLT Proof also to EXTRACT all from NOW
SECOND Fundamental Science - Fermat's Last Theorem -

Via Fermat Murgu Quadruplets Table we get not only Zero not reach for Fermat Eq. ,
(X³ + Y³ - Z³) ≠ 0     in Integers an Rationals
but also can have ;
(X³ + Y³ - Z³) ≠3 or 4 or 5 or 13 ...
That mean FLT Proof took a demonstration way ,heavy ,and may be wrong as can demonstrate the path of elliptic.
(X/Z)² + (Y/Z)² - 1 = 0 EME1E1 AND
(X/Z)ⁿ + (Y/Z)ⁿ - 1 = 0 EME1E1
And work for Murgu Quadruplets lead My to Murgu Identities and Comparative Analytic for Power 3.
Murgu Identities we used for Power 3 and 2 and are:
(X+i+1)³ - 2((X+i)³ ) + (X+i-1)³ = 6(X+i)
For i=0 will Be (X+1)³ - 2((X)³ ) + (X-1)³ = 6(X)
Murgu Integers Cubic Dispersion Formula and
(X+1)² -2((X)²) + (X-1)² =2
Murgu Algebraic Power 2 Geometric Media
If for (X³ + Y³ - Z³) = 0
chose Y>X and Z>Y and Z>X wrote as
Y = X+a and Z=X +b a,b Integers b>a - can wrote
(X+a+1)³ - 2(Y³) + (X+a-1)³ = 6(X+a)
(X+b+1)³ - 2(Z³) + (X+b-1)³ = 6(X+b)
Then Fermat's Last Theorem for power 3 became
2(X³)+(X+a+1)³+(X+a-1)³-6(X+a)-(X+b+1)³ -(X+b-1)³+6(X+b)=0
Introducing 2(X³) from above Indentities
(X+1)³ +(X-1)³-6X+(X+a+1)³+(X+a-1)³-6(X+a)-(X+b+1)³ -(X+b-1)³+6(X+b)=0

This mean Any for Analytic. First in a childish analyze , but maybe TRUE -
6(X+b-a)=0 or a-b=X b-a=X
AND
(X+1)³ +(X-1)³+(X+a+1)³+(X+a-1)³-(X+b+1)³ -(X+b-1)³=0

That Impossible, like I said in 1994 or 95 or 96 .But that mean ALSO
(X+1)³ +(X-1)³+(Y+1)³+(Y-1)³-(Z+1)³ -(Z-1)³=0
or returning

(X+1)³ +(X-1)³+(Y+1)³+(Y-1)³-(Z+1)³ -(Z-1)³= 6(X+b-a) EQ.1.1.0

Instead of a CONSTANT as to assure a Geometric Media. For Powers 3 not longer a Geometric Media like for Power 2 may Be Possible.
(X+1)² -2((X)²) + (X-1)² =2
That mean Impossible

(X+1), (X-1) , and so for (Y+1), (Y-1) and (Z+1), (Z-1) are (X,Y,Z) neighbors we speak about. Now, If Apply Murgu The Lost Fundamental Operator to those neighbors for power 3, and extract every one, will get:
(X+1)³ = 3(X³) - 3(X-1)³ + (X-2)³ + 6
(X-1)³ = 3(X³) - 3(X+1)³ + (X+2)³ + 6
Making theirs summ
4((X+1)³ + (X-1)³) = 6(3(X³)) + (X-2)³ + (X+2)³ + 12
(Y+1)³ = 3(Y³) - 3(Y-1)³ + (Y-2)³ + 6
(Y-1)³ = 3(Y³) - 3(Y+1)³ + (Y+2)³ + 6
Making theirs summ
4((Y+1)³ + (Y-1)³) = 6(3(Y³)) + (Y-2)³ + (Y+2)³ + 12
(Z+1)³ = 3(Z³) - 3(Z-1)³ + (Z-2)³ + 6
(Z-1)³ = 3(Z³) - 3(Z+1)³ + (Z+2)³ + 6
Making theirs summ
4((Z+1)³ + (Z-1)³) = 6(3(Z³)) + (Z-2)³ + (Z+2)³ + 12
Then, Left Side Term of EQ.1.1.0 may be write as
4[((X+1)³ +(X-1)³+(Y+1)³+(Y-1)³-(Z+1)³ -(Z-1)³)] =
6(3(X³)) + (X-2)³ + (X+2)³ + 12 + 6(3(Y³)) + (Y-2)³ + (Y+2)³ + 12 - ( 6(3(Z³)) + (Z-2)³ + (Z+2)³ + 12)
into Integers and absurd or not X³ + Y³ - Z³ = 0

4[((X+1)³ +(X-1)³+(Y+1)³+(Y-1)³-(Z+1)³ -(Z-1)³)] =
6( (X-2)³ + (X+2)³ + (Y-2)³ + (Y+2)³ - (Z-2)³ - (Z+2)³ + 12)
(I hope not a callculus error , if is is small) We Get 3 Conditionals as
X³ + Y³ - Z³ = 0 into Integers for power 3
(X+1)³ +(X-1)³+(Y+1)³+(Y-1)³-(Z+1)³ -(Z-1)³= 6(X+b-a)   EQ.1.1.0

2[((X+1)³ +(X-1)³+(Y+1)³+(Y-1)³-(Z+1)³ -(Z-1)³)] =
3( (X-2)³ + (X+2)³ + (Y-2)³ + (Y+2)³ - (Z-2)³ - (Z+2)³ + 12)   EQ.1.1.1

( (X-2)³ + (X+2)³ + (Y-2)³ + (Y+2)³ - (Z-2)³ - (Z+2)³ + 12) = 4(X+b-a)   Eq.1.1.2

THEN:
(X+1)³ +(X-1)³+(Y+1)³+(Y-1)³-(Z+1)³ -(Z-1)³= 6(X+Z-Y)   EQ.1.1.0

( (X-2)³ + (X+2)³ + (Y-2)³ + (Y+2)³ - (Z-2)³ - (Z+2)³ + 12) = 4(X+Z-Y)   Eq.1.1.2

2[(X+1)³ +(X-1)³+(Y+1)³+(Y-1)³-(Z+1)³ -(Z-1)³] = 3[( (X-2)³ + (X+2)³ + (Y-2)³ + (Y+2)³ - (Z-2)³ - (Z+2)³ + 12)]   EQ.1.1.3


IMPOSSIBLE
AND WE CAN VERIFFY.
TAKE THE CALCUL FOR THE BEGINING OF THIS div , MAYBE I MAKE ANY ... SMALL ANYWAY.
Forgot
X³ + Y³ - Z³ = 0
Now we need to get it by there.
The Start 2[6X+2(X)³ +6Y+2(Y)³ - ...] ...
will lead us to
X³ + Y³ - Z³ = F_(X,Y,Z) complicate dependence of (X)², (Y)², (Z)² and (X,Y,Z)
Normal and Instead of F_(X,Y,Z) = 0
We used a Time F_(X,Y,Z) = 0 , then been waiting it as answer.
The Answer is IMPOSSIBLE.
Anyway, even not correct by now, but only for curiosity you can analyze it.
Maybe will meet Complex Solution
I Love Complex But Here is not the case.
X³ + Y³ - Z³ = 0 is, over all about, if Sum of 2 Numbers with any property can be equal with another one which wear same porop.

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But now can bring any importance for Analytic in Math
F_(X,Y,Z) = I    with I Integer
Fermat's Last Theorem Proof even atack this can't say any.
Fermat Murgu Quadruplets treat
F_(X,Y,Z) = 1   
and brought which with your accept, I will Say
Murgu Quadruplets as New Math Beauty

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Also can use there The Fact , Power 2 Linear Forms writin (Y=X+A) and (Z=X+B)

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as Plaing Math , as for Our Problem The Answer is definitely IMPOSSIBLE.
This incursion was to extract any new from Fermat's Last Theorem
We SOLVED It for General case via FMIE.

This incursion was to extract any new from Fermat's Last Theorem
We SOLVED It for General case via FMIE.

Maybe I make any small errors but not of essence , and I don't re ceck as My intention was only provocation for young Scientist to Luck back to Proof good part, and reopen.
You can see a function used in Proof and this time not including any Parameters.
We solved Fermat's Last Theorem via FMIE and also Euler Murgu Equations 1 Equal 1.

FMIE Sent Fermat's Last Theorem Solutions in Irrational
and Certify
Not Solutions in Integers and Rational.
FMIE and Euler Murgu Equations 1 Equal 1 but also
Murgu Fermat's Last Theorem GEOMETRIC SOLUTION
PLACE Fermat's Last Theorem Solutions into concentric Circles with Radius
Zⁿ
excluding as all 3 (X,Y,Z) to be Integers or Rationals for any Value Z Integer or Rational.
Fermat's Last Theorem
get all rights for
Second Fundamental In Science .

Fermat's Last Theorem Geometric Solution

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Repeat FMIE and Euler Murgu Equations 1 Equal 1 solved Fermat's Last Theorem and
SENT IT AS Second Fundamental Science
but our interst is to extract as more can for.

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Than here we are luking for any new and will see any small errors you can repair.
We are luking for Conditionals Equations , which may be utils for future.

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Geometric Fermat's Last Theorem Solution started in 1994 or 95 from a simple observation and I can't say was complete then as not longer is on Internet but for sure I hit the base as I remind were using Geometric Media.
The observation was simple and had a correct base. As or if note:
Xⁿ = A² , Yⁿ = B² , Zⁿ = C²
Can Wrote A² + B² - C² = 0 ECF.2
As Logical corespondence into Pythagora Theorem , but I wasn't able then to demonstrate Pythagora Theorem is Fermat's Last Theorem Except .
We already spoke a bit about it above, but to repeat : Pythagorean Triples Pay Respect for Geometric Media as only and every Pythagorean Triple Group may need to confirm formula:
(A+1)² + (A-1)² + (B+1)² + (B-1)² - (C+1)² - (C-1)² = 2
Can Name It Pythagorean Triples Rule and come from Geometric Media identities:
(A+1)² + (A-1)² = 2(A)² +2
(B+1)² + (B-1)² = 2(B)² +2
(C+1)² + (C-1)² = 2(C)² +2

Also need to know and accept logic , now real form for Fermat's Last Theorem Equations is
{Xⁿ + Yⁿ - Z = 0}
and for Power 2 A² + B² - C² = 0 where
C² any Time can be an Simple Integer W .
(Those can wear a second case but don't change solution.)
Returning to ECF.2 but in old form as don't know we need at the last one to be Irrational , now
A² + B² - C² = 0 ECF.2
became or said
X^n/2 ,Y^n/2 , Z^n/2 = 0 Solutions
For n=2k this send us back to intial problem .

Infinity Repetition of coming back to base? , this is a face of
Double False Redundancy Of Truth
and offer absolute logical power to ALL This Material, even for our First Impossible
Murgu The Lost Fundamental Operator Applied to Fermat's Last Theorem REFUSE into Integers and Rationals a Zero Term.
FMIE is Key Method for Fermat's Last Theorem - as send direct to Solutions Domain.

A logical verification via Geometric Media for :
(X^n/2+1)² + (X^n/2-1)²+(Y^n/2+1)² + (Y^n/2-1)²- (Z^n/2+1)² - (Z^n/2-1)² =2

For n Prime Numbers >2 this is Impossible.
For Integers Divisible with 2
THE ANSWER
Subtle and hard to catch , is The Same as bring us back to Start.
Logic and Infinity Repetition For
For n Prime Numbers >2 this is Impossible.
May Be enough to say
Fermat's Last Theorem have also a Geometric Solution
But for n=3 , we demonstrated in FMIE Method using Murgu Integers Cubic Dispersion Formula Identities and Pythagorean Triples Geometric Media.

For Divisible with 2 , to exclude all doubth , can make analyse for n=4 .
Murgu Integers Quadratic Dispersion Formula is :
(X⁴+i + 1) + (X⁴+i - 1) - 2((X+i)⁴) =(12(((X+i)²))+2)

Dispersion Formula from Liniar Dependence for n=3 , A Constant for n=2 Pythagorean Triples Evolved to binomial Dependence .
Analytic of Murgu Integers Quadratic Dispersion Formula make IMPOSSIBLE into Integers and Rationals
(Xⁿ + Yⁿ - Zⁿ) = 0
for n divisible with 2

And Make Geometric Solution accurate and correct
Fermat's Last Theorem solution
Beggin with 2015

Now became a need any remarks
Power 2 also a Dispersion formula
(X²+1)² + (X²-1)² -2(X²) =2
(X+i+1)² + (X+i-1)² - 2((X+i)²) =2
and can be compared now with Murgu Integers Cubic Dispersion Formula
(X+i+1)³ -2((X+i))³ + (X+i-1)³ =6(X+i)
AND Murgu Integers Quadratic Dispersion Formula
(X⁴+i + 1) + (X⁴ +i - 1) - 2((X+i)⁴) =(12(((X+i)²))+2)

Comparing right terms 2, 6(X+i) , (12((X²))+2)
may lead to future study as also valid arguments for Future Science Logic. Then

(X⁴ + 1) + (X⁴ - 1) - 2(X⁴) =(12((X²))+2)
Murgu Integers Quadratic Dispersion Formula
Became an Important Math Tool
But also Murgu Pythagorean Triples Conditional.
For Power 4 already exist a solution which confirm, but , Murgu Integers Quadratic Dispersion Formula can bring second one.