Proof of Beal's conjecture

Proof of Beal’s conjecture
Solving of generalized Fermat equation xp
+ yq
= zw
N.Mantzakouras
e-mail: nikmatza@gmail.com
Athens - Greece
Abstract
The difference between of the Fermat’s generalized equation and Fermat’s regular equation is the
different exponents of the variables and the method of solution. As we will show, for the proof of
this equation to be complete, Fermat’s theorem will be must hold,as we know it has been proven.
There are only 10(+4) known solutions and all of them appear with exponent 2. This very fact
is proved here using a uniform and understandable method. Therefore, the solution is feasible
under the above conditions, as long as it we accepts that there is no solution if any other case,
occurs for which all exponents values are greater than 2. The truth of this premise is proved
in Theorem 6, based on the results of Theorem 4 and 5. The primary purpose for solving the
equation is to see what happens in solving of Diophantine equation a·x+b·y = c·z which refers to
Pythagorean triples of degree 1. This is the generator of the theorems and programs that follow.
I.1. Theorem 1 (Pythagorean triples 1st degree)
Let P1 be the set of Pythagorean triples and defined as P1 = {(x, y, z) | a, b, c, x, y, z ∈ Z − {0} and
a · x + b · y = c · z, {x, y, z} are pairwise relatively primes}. Let G1 be the set defined as: G1 = {(x =
k · (c · λ − b), y = k · (a − c), z = k · (a · λ − b)), (x = k · (b − c), y = k · (c · λ − a), z = k · (b · λ − a)), (x =
k · (c + b · λ), y = k · (c − a · λ), z = k · (α + b)) | k, λ ∈ Z+
}. We need to prove that the sets P1 = G1.
Proof.
Given a triad (a, b, c) such that abc ̸= 0 and are these positive integers, if we divide by y ̸= 0, we get
according to the set P1 then apply a ·(x/y) + b = c · (z/y) and we call X = x/y and Z = z/y. We declare
now the sets:
F1 = {(X, Z)} ∈ Q2
− {0} | a · X + b = Z · c, where a, b, c ∈ Z − {0}, and where X, Z ∈ Q − {0}

and
S1 =

(X, Z) ∈ Q2
− {0} | X = m − λ ∧ Z = m, where m, λ ∈ Q − {0}

The set F1 ∩ S1 has 3 points as a function of parameters m, λ and we have solutions for the corresponding
final equations,
F1 ∩ S1 =
a · (m − λ) + b = m · c ⇔ m = a·λ−b
a−c , a − c ̸= 0
m − λ = c·λ−b
a−c , a − c ̸= 0, y = k · (a − c), k ∈ Z+
x = c·λ−b
a−c · y ∧ z = c·λ−b
a−c · y, a − c ̸= 0
x = (c · λ − b) · k, y = k · (a − c), z = k · (a · λ − b), k ∈ Z+
, a − c ̸= 0
1

Therefore
F1 ∩ S1 = ⟨x = (c · λ − b) · k, y = k · (a − c), z = k · (a · λ − b), k ∈ Z+
, a − c ̸= 0⟩ (I)
Dividing respectively by x ̸= 0 we get the set and the relations we call Y = y/x and Z = z/x
F2 =

(Y, Z) ∈ Q2
− {0} | a + b · (y/x) = c · (z/x), where a, b, c ∈ Z − {0}, and where Y, Z ∈ Q − {0}

and
S2 =

(Y, Z) ∈ Q2
− {0} | Y = m − λ ∧ Z = m, where m, λ ∈ Q − {0}

Then as the type (I) we get the result
F2 ∩ S2 = ⟨x = (b − c) · k, y = k · (c · λ − a), z = k · (b · λ − a), k ∈ Z+
, b − c ̸= 0⟩ (II)
and finally dividing by z ̸= 0 similarly as before we call X = x/z and Y = y/z
F3 =

(X, Y) ∈ Q2
− {0} | a · (x/z) + b · (y/z) = c, where a, b, c ∈ Z − {0}, and where X, Y ∈ Q − {0}

and
S3 =

(X, Y) ∈ Q2
− {0} | X = m − λ ∧ Y = m, where m, λ ∈ Q − {0}

.
F3 ∩ S3 = ⟨x = (c + b · λ) · k, y = k · (c − a · λ), z = k · (a + b), k ∈ Z+
, a + b ̸= 0} (III)
As a complement we can state that the parameter λ can be equal with λ = p/q, where p and q relatively
primes. Therefore P1 = G1 and the proof is complete.
General Solving xp
+ yq
= zw
1.2 Theorem 2 [4,5,6]
To be proven that the equation xq
+ yp
= zw
, {x, y, z, q, p, w} ∈ Z+
1can be equivalently transformed into 3
linear forms of equations according to the theorem of primary Pythagorean triads 1st
degree.
Proof
To we prove initially must accept the general solution of the equation ax + by = cz (1) and a, b, c ∈ Z,
which, as we know from Theorem 1 has solution
x = (c · λ − b) · k
y = (a − c) · k
z = (a · λ − b) · k
with x, y, z, k ∈ Z, λ ∈ Q
2

From the definition of the function we take
f(x, y, z) =

∃x, y, z ∈ C3
: A(x) = xq−1
, B(y) = yp−1
, C(z) = zw−1
∧ A(x)x + B(y)y = C(z)z

⊆ C3
with q, p, w ∈ Z+
∧ {q 1, p 1, w 1}.
Therefore we have the correspondence A(x) = xq−1
, B(y) = yp−1
, C(z) = zw−1
on the primary equa-
tion (1) and then we have the equivalence of the systems,
x = (C(z) · λ − B(y)) · k
y = (A(x) − C(z)) · k
z = (A(x) · λ − B(y)) · k
⇔
x = zw−1
· λ − yp−1

· k
y = xq−1
− zw−1

· k
z = xq−1
· λ − yp−1

· k
with x, y, z, k ∈ Z+
, λ ∈ Q+
(2)
Substituting x, y, z to solve the equation xq
+ yp
= zw
resulting
A(x)x+B(y)y = C(z)z ⇒ xq−1
· zw−1
· λ − yp−1

·k+yp−1
· xq−1
− zw−1

·k−zw−1
· xq−1
· λ − yp−1

·k = 0
Which seeing as applicable for each q, p, w ∈ Z+
1. Of course this extends to x, y, z ∈ R or C. The
general solution of
m · xq
+ n · yp
= d · zw
m, n, d ∈ Z ∧ {q 1, p 1, w 1}
of course if we apply the same method we developed, F is given by:
F(x, y, z) =





x = (d · C(z) · λ − n · B(y)) · k
y = (m · A(x) − d · C(z)) · k
z = (m · A(x) · λ − n · B(y)) · k





with x, y, z, m, n, d, k ∈ Z+
, λ ∈ Q+
(3)
It remains to see that finally produced the solutions of the equation with more great from unit powers
of unity with 3 equivalent forms
x = zw−1
· λ − yp−1

· k
y = xq−1
− zw−1

· k or
z = xq−1
· λ − yp−1

· k
x = yp−1
− zw−1

· k
y = zw−1
· λ − xq−1

· k or
z = yp−1
· λ − xq−1

· k
x = zw−1
+ λ · yp−1

· k
y = zw−1
− λ · xq−1

· k
z = xq−1
+ λ · yp−1

· k
(4)
In the analysis of this special case xq
+yp
= zw
we assume x, y, z, p, q, w ∈ Z=2, λ ∈ Q+
where λ = u/s, {u, s
co-primes) or λ ∈ Z+
.
It should be mentioned that for the solution of the simple equation, xq
+ yp
= zw
, which is reduced to
a pseudo-linear system as we have shown, we use the newer Gröbner basis method proposed in mathemat-
ica. In the next chapter we will discuss some of the elements of the method and how it is applied.
3

2.1 Method Gröbner [8]
For systems with equational constraints generating a zero-dimensional ideal, Mathematica uses a variant
of the CAD algorithm that finds projection polynomials using Gröbner basis methods. If the lexicographic
order Gröbner basis of contains linear polynomials with constant coefficients in every variable but the last
one (which is true ”generically”), then all coordinates of solutions are easily represented as polynomials in
the last coordinate. Otherwise the coordinates are given as Root objects representing algebraic numbers
defined by triangular systems of equations. Setting to True causes Mathematica to represent each coordi-
nate as a single numeric Root object defined by a minimal polynomial and a root number. Computing this
reduced representation often takes much longer than solving the system.
Correctness of the Algorithm
The correctness of the algorithm is based on the following ”Main Theorem of Gröbner Bases Theory:
F is a Gröbner basis ⇔ ∀
f1,f2∈F
RF [F, S− polynomial [f1, f2]] = 0
2.2 Examples
A Simple Set of Equations
We now show how Gröbner bases can be applied to solving systems of polynomial equations. Let us,
first, consider again the example:
f1 = xy − 2y
f2 = 2y2
− x2
F = {f1, f2}
The Gröbner basis G of F is
G :=

−2x2
+ x3
, −2y + xy, −x2
+ 2y2

By the fact that F and G generate the same ideal, F and G have the same solutions. The elimination prop-
erty of Gröbner bases guarantees that, in case G has only finitely many solutions, G contains a univariate
polynomial in x. (Note that, here, we use the lexicographic order that ranks y higher than x. If we used
the lexicographic order that ranks x higher than y then, correspondingly, the Gröbner basis would contain
a univariate polynomial in y.) In fact, the above Gröbner basis is ”reduced”, i.e. all polynomials in the
basis are reduced modulo the other polynomial in the basis. It can be shown that reduced Gröbner bases
(with finitely many solutions) contain exactly one univariate polynomial in the lowest indeterminate.
If we look at this example using the mathematica, and the method Groebner Basis summarized in the
following program and for two variables {x, y}
4

polys = {x∗
y − 2y, 2y∧
2 − x∧
2} ;
gb=GroebnerBasis[polys, {x, y, z}, Method=”Buchberger” ]
soly1=NSolve[gb[[1]]=0,{y}]
soly2=NSolve[gb[[2]]=0,{x}]
Reduce[polys=0,{y,x,z},Backsubstitution=True]//N
NSolve[polys = 0, {y, x}]
Results.

−2y + y3
, −2y + xy, x2
− 2y2

{{y = −1.41421}, {y = 0.}, {y = 1.41421}}
{{x = 2.}}
(y = 0.x = 0.)||(y = −1.41421x = 2.)||(y = 1.41421x = 2.)
{{y = −1.41421, x = 2.}, {y = 1.41421, x = 2.}, {y = 0., x = 0.}, {y = 0., x = 0.}}
Then we have to solve the generalized system generated from the equation xq
+ yp
= zw
and which written
before as
x = zw−1
· λ − yp−1

· k
y = xq−1
− zw−1

· k
Z = xq−1
· λ − yp−1

· k
with k ∈ Z+, λ ∈ Q+
In language mathematica can be written as a partial but total solutions for triplets {x, y, z} as one of
the sets of solutions x3
+ y3
= z3
and for the values k = r = 1, written
q:=3;p:=3;w:=3;λ:=1;k=1;
eqs1={x-k* zw−1
λ + k∗
yp−1
, y − k∗
xq−1
+ k∗
zw−1
, z − k∗
xq−1
∗ λ + k∗
yp−1
};
gb=GroebnerBasis[eqs1,{x,y,z},Method-”Buchberger”]
soly1=NSolve [gb[[1]] == 0, {z}, Integers]
soly2=NSolve [gb[[2]] == 0, {x}, Integers]//N
soly3=NSolve [gb[[3]] == 0, {y}, Integers]//N
NSolve[eqs1 == 0, {x, y, z}]
5

Results.

−z + z3
, z + 2yz − z2
, −y + y2
+ z − z2
, x + y − z

{{z → −1.}, {z → 0}, {z → 1.}}
{{y → 1. − 1.z} , {y → z}}
{{x → 1., y → 0., z → 1.}, {x → −1, y → 1, z → 0.},
{x → 0., y → −1, z → −1}, {x → 0., y → 0., z → 0.}}
This analysis does not reflect all solutions but an elementary part of them with r = k = 1. For the case
g = p = w = 3 we obtain complex or real roots with at least one root equal to zero. Of course we can
take many different powers for {q, p, w} with q = p = w 3 and we will see that the program follows the
same logic as for the values q = p = w = 3. This is the computational procedure of the proof of Fermat’s
theorem. Of course we can take different values for {q, p, w} and we will notice that only in some cases
where one of them is at least equal to 2, it has integer solutions.
3. Theorem 3 [2,3,6]
In the general solution of xq
+ yp
= zw
, where x, y, z ∈ Z=2 and p, q, w ∈ Z=2 but and x, y, z co-
primes numbers after final simplification by the common factor of GCD[x, y, z] will eventually obtain 4
variables {x′
, y′
, z′
} which will be the primes among then, and variables related with one of the 4 general
linear relations, x′
+ y′
· λ = z′
, or λ · x′
+ y′
= z′
or λ · z′
+ y′
= x′
or λ · z′
+ x′
= y′
independent of the
exhibitors and which are equivalent forms of powers equivalent to the general form.
Proof
Initially each Diophantine equation of the form xq
+ yp
= zw
is equivalent to the form x · xq−1
+ y · yp−1
=
z · zk−1
⇔ a · x + b · y = c · z, a ≡ xq−1
∧ b ≡ yp−q
∧ c ≡ zw−1
.
To prove this we must first accept the general solution according to theorem (1) of the equation a·x+b·y = c·z
(1) with a, b, c ∈ Z+
.
As we know from Theorem 1 If you divide the variables {y, z} with x, {x, z} with y and {x, y} with z
and we get:
x = (c · λ − b) · k y = (c · λ − a) · k x = (c + b · λ) · k
y = (a − c) · k or x = (b − c) · k or z = (a + b) · k with k ∈ Z, λ ∈ Q
z = (a · λ − b) · k z = (b · λ − a) · k y = (c − a · λ) · k
In conjunction with Theorem 2 we will apply the same method and construct a relevant function that will
link all these variables.
So we have 4 cases: i) For the first case in general form for x ̸= 0 the function will has relationship
6

f(x, y, z) =





x = (C(z) · λ − B(y)) · k
y = (A(x) − C(z)) · k
z = (A(x) · λ − B(y)) · k





with k ∈ Z+
, λ ∈ Q+
(2)
therefore dividing by k resulting





x/k = (C(z)λ − B(y))
y/k = (A(x) − C(z))
z/k = (A(x) · λ − B(y))





⇔





x/k = (C(z)λ − B(y))
y/k = (A(x) − C(z))
x/k − z/k = −(A(x) · λ − λ · C(z))





⇔





x/k = (C(z)λ − B(y))
y/k = (A(x) − C(z))
x/k − z/k = −λ · y/k





⇔ x + λ · y = z (3)
if we divide the triad {x, y, z} with the greatest common divisor ε = GCD[x, y, z], then it is clearly arises
the final triad {x′
= x/ε, y′
= y/ε, z′
= z/ε} and the relationship. x′
+ λ · y′
= z′
.
ii) If we divide the variables {x, z} with y, y ̸= 0 and therefore will occur in accordance with the basic
relationship and certainly using of Theorem 1 that the relationship meets these data are:
y = (c · λ − a) · k
x = (b − c) · k with x ∈ Z, λ ∈ Q
z = (b · λ − a) · k
Of course we will apply the same method and the function f will be given by the relationship in general
form:
f(x, y, z) =





y = (C(z) · λ − A(y)) · k
x = (B(x) − C(z)) · k
z = (B(x) · λ − A(y)) · k





with more generally k ∈ Z+, λ ∈ Q+
by the sequence





y/k = (C(z) · λ − A(y))
x/k = (B(x) − C(z))
z/k = (B(x) − λ − A(y))





⇔





y/k = (C({z) · λ − A(y))
x/k = (B(x) − C(z))
y/k − z/k = −(B(x) · λ − λ · C(y))





⇔





x/k = (C(z) · λ − B(y))
y/k = (A(x) − C(z))
y/k − z/k = −λ · x/k





⇔ y + λ · x = z (4)
7

Similarly If we divide the triad {x, y, z} by the greatest common divisor ε = GCD[x, y, z], then it is clearly
that the final triad {x′
= x/ε, y′
= y/ε, z′
= z/ε} and relationship is obtained λ · x′
+ y′
= z′
.
iii) Continuing, in the same way and dividing by z variables {x, y} with z which z ̸= 0 will occur in
accordance with the basic relationship and certainly using Theorem 1 that the relationship meets these
data will be:
x = (c + b · λ) · k
z = (a + b) · k κ, λ ∈ Q
y = (c − a · λ) · k
By the same method and to the display function f which is given by the general form:
f(x, y, z) =





x = (C(z) + λ · B(y)) · k
z = (A(x) + B(y)) · k
y = (C(z) − λ · A(x)) · k





more generally with k ∈ Z+
, λ ∈ Q+
,
where λ = u/s, (u, s co-primes) or λ ∈ Z+
by the sequence





x/k = (C(z) + λ · B(y))
z/k = (A(x) + B(y))
y/k = (C(z) − λ · A(y))





⇔





x/k = (C(z) · λ − B(y))
z/k = (A(x) + B(y))
x/k − y/k = (B(y) · λ + λ · A(x))





⇔
⇔





x/k = (C(z) · λ − B(y))
z/k = (A(x) + B(y))
x/k − y/k = λ · z/k





⇔ y + λ · z = x (5)
These resulted from the initial a · x + b · y = z assuming that x
z = m and y
z = m − λ:
iv) Finally, if we assume that x
z = m − λ and y
z = m resulting system:
x = (c − a · λ) · k
z = (a + b) · k with k ∈ Z, λ ∈ Q
y = (c + b · λ) · k
By the same method and the function f which is given by the general form
f(x, y, z) =





x = (C(z) − λ · A(x)) · k
z = (A(x) + B(y)) · k
y = (C(z) + λ · B(y)) · k





more generally with k ∈ Z+
, λ ∈ Q+
,
where λ = u/s, (u, s co-primes) or λ ∈ Z+ (6)
8

by sequence follows




x/k = (C(z) − λ · A(x))
z/k = (A(x) + B(y))
y/k = (C(z) + λ · B(y))



 ⇔




x/k = (C(z) · λ − A(x))
z/k = (A(x) + B(y))
y/k − x/k = (C(z) · λ + λ · A(x)) ⇔




⇔





x/k = (C(z) · λ − A(x)
z/k = (A(x) + B(y))
y/k − x/k = λ · z/k





⇔ x + λ · z = y (7)
We will therefore obtain table (9) which is directly related to the variables x, y, z







x + y · λ = z
y + x · λ = z
x = y + z · λ
y = x + z · λ







λ ∈ Q+
, where λ = u/s, (u, s co-primes ) or λ ∈ Z+
(8)
and also the simplified table (9) which is obtained by dividing each variable x, y, z by GCD[x, y, z] and
therefore the variables x′
, y′
, z′
will be obtained.







x′
+ y′
· λ = z′
y′
+ x′
· λ = z′
x′
= y′
+ z′
· λ
y′
= x′
+ z′
· λ







λ ∈ Q+
, where λ = u/s, (u, s co-primes ) or λ ∈ Z+
(9)
By simple extension of the theorem we can prove that these relations (8, 9) and apply the generalized case
Diophantine equation on the form a′
xq
+b′
yp
= c′
zw
which is equivalent to the form a′
x·xq−1
+b′
y ·yp−1
=
c′
z · zw−1
⇔ a · x + b · y = c · z, a = a′
xq−1
∧ b = b′
· yp−q
∧ c = c′
· zw−1
and the same applies for the
proof, since we accept the general solution of the equation a·x+b·y = c·z as originally defined, ie a, b, c ∈ Z+
.
Examples according to the theorem 3
1.6∧
3 + 5∧
4 = 29∧
2 ⇒ 4∗
6 + 5 = 29
2.7∧
2 + 2∧
5 = 3∧
4 ⇒ 3 + 2 ∗ 2 = 7
3.7∧
4 + 15∧
3 = 76∧
2 = 7 + (23/5)∗
15 = 76
4.2∧
7 + 17∧
3 = 71 ⇒ 2 ∗ 27 + 17 = 71
5.3∧
5 + 11∧
4 = 122∧
2 = 3 ∗ 37 + 11 = 122
6.1414∧
3 + 2213459∧
2 = 21063928∧
2 = 1414 + (442409/13)∗
65 = 2213459
7.17∧
7 + 726271∧
3 = 21063928 = 17 + (21063911/76271) ∗ 76271 = 21063928
Note that λ will be either an integer or explicit in the form mentioned above i.e. where λ = u/s (u, s
co-primes) or λ ∈ Z+
. This property generalizes everywhere to any triad that is a solution of the equation
xq
+ yp
= zr
where where x, y, z ∈ Z+
1 and final p, q, w ∈ Z+
=2.
9

4. The Generalized Fermat Equation [9]
We now return to the generalized Fermat equation xq
+ yp
= zr
(1) x, y, z, p, q, r ∈ Z+
∧ {q = 2, p =
2, r = 2} where x, y and z are integers, and the exponents p, q and r are (potentially distinct) positive
integers. We restrict our attention to primitive solutions, i.e. those with gcd(x, y, z) = 1, since, without such
a restriction, it is easy to concoct uninteresting solutions in a fairly trivial fashion. Indeed, if we assume,
say, that p, q and r are fixed positive integer, then we can choose integers u, v and w such that
uqr ≡ −1(modp), vpr ≡ −1(modq), wpq ≡ −1(modr)
In the case where they are given a, b, c with a + b = c multiplying this equation by auqr
bvpr
cwpq
we have

a(uqr+1)/p
bvr
cwq
p
+

a(ur)
b(vpr+1)/q
cwp
q
=

a(uq)
b(vp)
c(wpq+1)/r
r
We call (p, q, r) the signature of equation (1). The behaviour of primitive solutions depends fundamentally
upon the size of the quantity
σ(p, q, r) =
1
p
+
1
q
+
1
r
(2)
in particular, whether σ(p, q, r) 1, σ(p, q, r) = 1 or σ(p, q, r) 1. If we set χ = σ(p, q, r) − 1, then χ is
the Euler characteristic of a certain stack associated to equation (1). It is for this reason that the cases
σ(p, q, r) 1, σ(p, q, r) = 1 or σ(p, q, r) 1 are respectively termed spherical, parabolic and hyperbolic.
4.1 The spherical case σ(p, q, r) 1
In this case, we can assume that (p, q, r) is one of (2, 2, r), (2, q, 2), (2, 3, 3), (2, 3, 4), (2, 4, 3) or (2, 3, 5).
In each of these cases, the (infinite) relatively prime integer solutions of (1) belong to finite families of two
parameters; in the (more complicated) case (2, 3, 5), there are exactly 27 such families, as Johnny Edwards
[11] proved in 2004 via an elegant application of classical invariant theory. In the case (p, q, r) = (2, 4, 3),
for example, that the solutions x, y and z satisfy one of the following four parameterizations







x = 4ts s2
− 3t2

s4
+ 6t2
s2
+ 81t4

3s4
+ 2t2
s2
+ 3t4

y = ± s2
+ 3t2

s4
− 18t2
s2
+ 9t4

z = s4
− 2t2
s2
+ 9t4

s4
+ 30t2
s2
+ 9t4

s ̸= t · mod(2) and 3 s
where







x = ± 4s4
+ 3t4

16s8
− 408t4
s4
+ 9t8

,
y = 6ts 4s4
− 3t4

,
z = 16s8
+ 168t4
s4
+ 9t8
,
r is odd and 3s
where







x = ± s4
+ 12t4

s8
− 408t4
s4
+ 144t8

,
y = 6ts s4
− 12t4

,
z = s8
+ 168t4
s4
+ 144t8
,
s = ±1 mod (6) or
10

where



















x = 2 s4
+ 2ts3
+ 6t2
s2
+ 2t3
s + t4

23s8
− 16ts7
− 172t2
s6
− 112t3
s5
−22t4
s4
− 112t5
s3
− 172t6
s2
− 16t7
s + 23t8

y = 3(s − t)(s + t) s4
+ 8ts3
+ 6t2
s2
+ 8t3
s + t4

z = 13s8
+ 16ts7
+ 28t2
s6
+ 112t3
s5
+ 238t4
s4
+112t5
s3
+ 28t6
s2
+ 16t7
s + 13t8
s ̸= tmod(2) and s ̸= tmod(3)
Here, s and t are relatively prime integers. Details on these parametrizations (and much more besides) can
be found in Cohen’s exhaustive work [10].
4.2 The parabolic case σ(p, q, r) = 1
If we have s(p, q, r) = 1, then, up to reordering, (p, q, r) = (2, 6, 3), (2, 4, 4), (4, 4, 2), (3, 3, 3) or (2, 3, 6).
As in Examples 1 and 2, each equation now corresponds to an elliptic curve of rank 0 over Q; the only
primitive non-trivial solution comes from the signature (p, q, r) = (2, 3, 6), corresponding to the Catalan
solution 32
− 23
= 1.
4.3 The hyperbolic case σ(p, q, r) 1
It is the hyperbolic case, with σ(p, q, r) 1, where most of our interest lies. Here, we are now once
again considering the equation and hypotheses [9]. As mentioned previously, it is expected that the only
solutions are with (x,y, z, p, q, r) corresponding to the identity 1p
+ 23
= 32
, for p = 6, or to
25
+ 72
= 34
, 73
+ 132
= 29
, 27
+ 173
= 712
, 35
+ 114
= 1222
,
177
+ 762713
= 210639282
, 14143
+ 22134592
= 657
, 92623
+ 153122832
= 1137
,
438
+ 962223
= 300429072
and 338
+ 15490342
= 156133
A less ambitious conjecture would be that (4) has at most finitely many solutions (where we agree to count
those coming from 1p
+23
= 32
only once). In the rest of this section, we will discuss our current knowledge
about this equation.
4.4 The Theorem of Darmon and Granville [1,4,9]
What we know for sure in the hyperbolic case, is that, for a fixed signature (p, q, r), the number of so-
lutions to equation (1) is at most finite:
Theorem 4. (Darmon and Granville [9]).
If A, B, C, p, q and r are fixed positive integers of equation Axq
+ Byp
= Czr
and p, q and w are
fixed positive integers numbers, with
1
p
+
1
q
+
1
r
1
then the equation has at most finitely many solutions in coprime non-zero integers x, y and z.
11

Proof.
The proof by Darmon and Granville is extremely elegant and we cannot resist giving a brief sketch. The
hypothesis 1/p + 1/q + 1/r 1 is used to show the existence of a cover ϕ : D → P1
that is ramified only
above 0, 1, ∞, where the curve D has genus ≥ 2. Moreover, this cover has the property that the ramification
degrees above 0 are all divisors of p, above 1 are all divisors of q, and above ∞ are all divisors of r. Now
let (x, y, z) be a non-trivial primitive solution to the equation Axq
+ Byp
= Czr
. The above properties of
the cover ϕ imply that the points belonging to the fiber ϕ−1
(Axp
/Czr
) are defined over a number field K
that is unramified away from the primes dividing 2ABCpqr. It follows from a classical theorem of Hermite
that there are only finitely many such number fields K. Moreover, by Faltings’ theorem, for each possible K
there are only finitely many K-points on D. It follows that the equation Axq
+ Byp
= Czr
has only finitely
many primitive solutions. It is worth noting that the argument used in the proof is ineffective, due to its
dependence upon Faltings’ theorem; it is not currently known whether or not there exists an algorithm for
finding all rational points on an arbitrary curve of genus ≥ 2.
4.5 Summary tables of what we know.
What we would really like to do is rather more distant than what Darmon’s theorem and Granville tells us.
And as we will see in the next chapters we will determine which cases are solvable. In the tables below, we
list all known (as of 2015) cases where Eq.(1) has been fully solved. For references to the original papers
we recommend the exhaustive search [10,11]. The 2 tables bring together all known infinite families treated
to date:
(p, q, r) reference(s)
(n, n, n) Wiles, Taylor-Wiles
(n, n, k), k ∈ {2, 3} Darmon-Merel, Poonen
(2n, 2n, 5) Bennett
(2, 4, n) Ellenberg, Bennett-Ellenberg-Ng, Bruin
(2, 6, n) Bennett-Chen, Bruin
(2, n, 4) Bennett-Skinner, Bruin
(2, n, 6) Bennett-Chen-Dahmen-Yazdani
(3j, 3k, n), j, k ≥ 2 immediate from Kraus
(3, 3, 2n) Bennett-Chen-Dahmen-Yazdani
(3, 6, n) Bennett-Chen-Dahmen-Yazdani
Bennett-Chen-Dahmen-Yazdani
(2, 2n, k), k ∈ {9, 10, 15} Bennett-Chen-Dahmen-Yazdani
(4, 2n, 3) Anni-Siksek
(2j, 2k, n), j, k ≥ 5 prime, n ∈ {3, 5, 7, 11, 13} And
Our second table lists ”sporadic” triples where the solutions to (1) have been determined, and infinite fam-
ilies of exponent triples where the (p, q, r) satisfy certain additional local conditions.
(p, q, r) reference(s)
(3, 3, n)∗
Chen-Siksek, Kraus, Bruin, Dahmen
(2, 2n, 3)∗
Chen, Dahmen, Siksek
(2, 2n, 5)∗
Chen
(2m, 2n, 3)∗
(2, 4n, 3)∗
(3, 3n, 2)∗
(2, 3, n), n ∈ {6, 7, 8, 9, 10, 15} Poonen-Schaefer-Stoll, Bruin, Zureick-Brown, Siksek, Siksek-Stoll
(3, 4, 5) Siksek-Stoll
(5, 5, 7), (7, 7, 5) Dahmen-Siksek
12

The asterisk here refers to conditional results. There are detailed tables for each case in total.
5. Formations of the diophantine equation
5.1 Lemma 1.
The number of forms of xq
+ yp
= zw
, x, y, z, p, q, w ∈ Z+
∧ {q 2, p 2, w 2} and after simplify-
ing the variables {x, y, z} with GCD[x, y, z] = 1 is limited to exactly 6 cases and for the exhibitors (p, q, w),
we accept from 4 only 3 cases of equality between them.
Proof
Depending on the ascending order of exponents {p, q, w} of original Diophantine xp
+ yq
= zw
, x, y, z, p, q,
w ∈ Z ∧ {q 1, p 1, w 1}. The key point of the equation is if analyzed each variable x, y and z must
have a common prime factor. After simplifying the terms of the number ε = GCD[x, y, z], where x = ε · λ,
y = ε · µ, z = ε · σ, {ε, λ, µ, σ ∈ Z+
} (2). Let’s analyze case and the other cases will be similar, according to
relations (1 and 2), the following equivalences are obtained:
xp
+ yq
= zw
⇔ λp
· εp
+ εq
· µq
= εw
· σw
⇔ λp
· εp−q
+ µq
= εw−q
· σw
, w p q 1 ∈ Z+ (3)
According to equation (3) the detailed order of all cases is as follows and as we can see there are only 6 :
1. λp
· εp−q
+ µq
= εw−q
· σw
, {w, p, q ∈ Z+
1}
2. λp
+ εq−p
µq
= εw−p
· σw
, {w, p, q ∈ Z+
1}
3. λp
· εp−q
+ µq
= εw−q
· σw
, {p, w, q ∈ Z+
1}
4. λp
· εp−w
+ εq−w
· µq
= σw
, {p, q, w ∈ Z+
1}
5. λp
· εp−w
+ εq−w
µq
= σw
, {q, p, w ∈ Z+
1}
6. λp
+ εq−p
µq
= εw−p
· σw
, {q, w, p ∈ Z+
1}
Let us now analyse the cases that the exponents p, q, w can take only then we can know which cases
we can accept that there are
I) p = q ∧ w = q ⇔ p = q = w
II) p = q ∧ w ̸= q
III) p ̸= q ∧ w = q
IV) p ̸= q ∧ w ̸= q
Relation (I) cannot be valid because leads to Fermat’s equation(If p, q, w 2 integers) and is therefore
impossible. All the others can be valid by assumption and as we will see below at least one of p, q and w
must be equal to 2 , but not a Pythagorean triad. In our analysis in Lemma 2 , however, we will accept
only IV. But the technique that will allow us to analyze each case can only be given by an algorithm and
this is given in Lemma 2. There we analyze 2 forms one is related to trying to find what the variables x, y, z
are for a range of values up to 100 but being the first among them and the second if is given the variables
x, y, z in which set of bases and exponents they are related to Lemma 1.
13

5.2 Lemma 2.
Prove that there is a relation connecting the bases and exponents to form a simplified algorithm to get
the fastest calculation of equation xp
+ yq
= zw
.
Proof
I) Dividing each variable {x, y, z} with GCD[x, y, z] and simplifying exhibitors, removing the minimum
of exhibitors from every exhibitor shows a simplified form as final variables, according to the shape.
x1 = (GCD [x, y, z])
∧
(p − Min[p, q, w])∗
(x/GCD [x, y, z])
∧
p;
x2 = (GCD [x, y, z])
∧
(q − Min [p, q, w])
∗
(y/GCD [x, y, z])
∧
q;
x3 = (GCD[x, y, z])∧
(w − Min[p, q, w])∗
(z/ GCD[x, y, z])∧
w;
Therefore, we can write a simple but quick program with mathematica, which gives us the value of let’s
say up to 100 for the variables, with additional conditions for each variable with the others being prime
numbers.Surely this could be feasible if we get many instances by constructing the corresponding program,
which would be grouped by basis and exponents.
Clear[x, y, z, p, q, w]
Do [If [(x)∧
p + (y)∧
q == (z)∧
wGCD[x, y, z] = 1,
x1 = (GCD[x, y, z])∧
(p − Min[p, q, w])∗
(x/GCD[x, y, z])∧
p;
x2 = (GCD[x, y, z])∧
(q − Min[p, q, w])∗
(y/GCD[x, y, z])∧
q;
x3 = (GCD[x, y, z])∧
(w − Min[p, q, w])∗
(z/GCD[x, y, z])∧
w
ff = GCD[x1, x2, x3];
d1 = FactorInteger [x1/ff]
d2 = FactorInteger [x2/ff];
Print[”(”,x,”∧”,p,”,”,y,”∧”,q,”,”,z,”∧”,w,”)”,”,”,x1,”,”,x2,”,”,x3,”,”,d1,”,”,,d2,”,”,d3]],
{x, 1, 100}, {y, 1, 100}, {z, 1, 100}, {p, 3, 10}, {q, 2, 10}, {w, 3, 10}]
Additionally the function FactorInteger [ ] automatically configures all analyzes powers of primes of each
variable. In the whole process we accept p 2 and q = 2 and w 2 and furthermore (xmax =
ymax = zmax = 100) and then we see that at least one exponent is equal to 2 . We also call σ(p, q, w) =
1/p + 1/q + 1/w and compare each case, a condition we will discuss in chapter 4.
Results:
(25
, 72
, 34
), 32, 49, 81, {{2, 5}}, {{7, 2}}, {{3, 4}}, σ(p, q, w) 1
(35
, 102
, 73
), 243, 100, 343, {{3, 5}}, {{2, 2}, {5, 2}}, {{7, 3}}, σ(p, q, w) 1
(34
, 462
, 133
), 81, 2116, 2197, {{3, 4}}, {{2, 2}, {23, 2}}, {{13, 3}}, σ(p, q, w) 1
(73
, 132
, 29
), 343, 169, 512, {{7, 3}}, {{13, 2}}, {{2, 9}}, σ(p, q, w) 1
14

If we choose to have p 2 and q 2 and w = 2 then we will get the Printout according to the command
{x, 1, 100}, {y, 1, 100}, {z, 1, 100}, {p, 3, 10}, {q, 3, 10}, {w, 2, 10}]
Results:
27
, 173
, 712

, 128, 4913, 5041, {{2, 7}}, {{17, 3}}, {{71, 2}}, σ(p, q, w) 1
54
, 63
, 292

, 625, 216, 841, {{5, 4}}, {{2, 3}, {3, 3}}, {{29, 2}}, σ(p, q, w) 1
63
, 54
, 292

, 216, 625, 841, {{2, 3}, {3, 3}}, {{5, 4}}, {{29, 2}}, σ(p, q, w) 1
(74
, 153
, 762
), 2401, 3375, 5776, {{7, 4}}, {{3, 3}, {5, 3}}, {{2, 4}, {19, 2}} σ(p, q, w) 1
In total we have 4 more cases than the already known ones we will see in chapter 4. As we observe,
however, we also have cases with σ 1 which result from the computational analysis.
II. To do in the given case especially this analysis knowing the values of x, y, z we can construct the
corresponding program
Clear[x, y, z]
x := 34; y := 51; z := 85;
Do [If [(x)∧
p + (y)∧
q == (z)∧
wGCD[x, y, z] = 1,
x1 = (GCD[x, y, z])∧
(p − Min[p, q, w])∗
(x/GCD[x, y, z])∧
p;
x2 = (GCD[x, y, z])∧
(q − Min[p, q, w])∗
(y/GCD[x, y, z])∧
q;
x3 = (GCD[x, y, z])∧
(w − Min[p, q, w])∗
(z/GCD[x, y, z])∧
w
ff = GCD[x1, x2, x3];
d1 = FactorInteger [x1/ff]
{p, 2, 10}, {q, 2, 10}, {w, 2, 10}]
Results
(34∧
5, 51∧
4, 85∧
4) , 544, 81, 625, ({{2, 5}, {17, 1}}, {3, 4}, {5, 4})
To give a typical example, we get the equation 345
+ 514
= 854
. Notice that it is analyzed as {2∧
5∗
17∧
1
+3∧
4 = 5∧
4}. Of course this case is not acceptable as it is easy to see that the truth of Theorem 3 is valid.
So the results of the first programme are the triads we are mainly interested in. In chapter 7 we have some
examples that we examine to see if they agree with these 2 Lemmas.
6. Theorems that prove the Proposal 1(Of Beal’s)
Proposal 1. The equation xp
+ yq
= zw
has no solution in positive integers x, y, z, p, q, w, when ap-
ply (p, q, w 2) .
15

6.1 Theorem 5
Any equation form xp
+ yq
= zw
with positive integers x, y, z, p, q, w where p, q, w 1 , is transformed
into a final Diophantine equation with GCD(x, y, z) = 1 then and only then, when at least one exponent
equals 2. This equation will belong to a class of equations with exponents that be consistent with the
criteria σ(p, q, r) 1, σ(p, q, r) = 1 or σ(p, q, r) 1 with a limited number equations, in accordance with
chapter 4.
Proof
The number of the forms of xq
+ yp
= zw
, x, y, z, p, q, w ∈ Z+
∧ {q = 2, p = 2, w = 2} after simplifying
the terms of the GCD[x, y, z], Lemma 1, Lemma 2 limited to 6. Depending on the ascending order of expo-
nents {p, q, w} of original Diophantine equation xp
+yq
= zw
, x, y, z, p, q, w ∈ Z+
∧{q = 2, p = 2, w = 2}
and after simplifying the terms with the number ε = GCD[x, y, z], we receive a total of 6 cases where any
stemming detail has as follows
1. λp
· εp−q
+ µq
= εw−q
· σw
, w p q = 2 ∈ Z+
2. λp
+ εq−p
µq
= εw−p
· σw
, w q p = 2 ∈ Z+
3. λp
· εp−q
+ µq
= εw−q
· σw
, p w q 2 ∈ Z+
4. λp
· εp−w
+ εq−w
· µq
= σw
, p q w = 2 ∈ Z+
5. λp
· εp−w
+ εq−w
µq
= σw
, q p w = 2 ∈ Z+
6. λp
+ εq−p
µq
= εw−p
· σw
, q w p = 2 ∈ Z+
But these exhibitors must comply with the Fermat-Catalan criteria, but here we will analyse them in
general terms, distinguishing 3 general cases:
if we accept that p,q and w are fixed positive integers and that these exponents must satisfy the crite-
ria of chapter 4, and after first accepting p, q, w = 2, we will prove that at least one exponent equals 2
using these criteria alone. So according to this logic the following 3 cases will apply:
Case 1rd
0 1/p + 1/q + 1/w 1
In order to we calculate the exhibitors present in the open interval (0, 1) solve the inequality as z and
we get
1/w 1 −
p + q
p · q
⇒ w
p · q
p(q − 1) − q
The inequality has integer solutions which arise only in accordance with the 3 equations:
(1). p · (q − 1) − q = 1
(2). q = φ · (p · (q − 1) − q)
(3). p = ε · (p · (q − 1) − q)
ε, φ ∈ Z
16

1.From the first equation it follows that
p · (q − 1) = q + 1 ⇒ p = 1+q
q−1 = 1 + 2
q−1 which implies 2 prerequisites:
i) q − 1 = 1 ⇒ q = 2 ∧ p = 3
ii) q − 1 = 2 ⇒ q = 3 ∧ p = 2
because should the (q − 1) must divide 2
And for 2 exhibitor cases we get w 6 ⇒ w ≥ 7
Therefore Thus arise the two triads p = 3, q = 2, w ≥ 7 and p = 2, q = 3, w ≥ 7
2. Similarly from the second equation q = ϕ · (p · (q − 1) − q) we get:
q = ϕ · p · (q − 1) − q · φ ⇒ p =
q · (1 + ϕ)
ϕ · (q − 1)
i)ϕ(q − 1) = 1 ⇒ ϕ =
1
q − 1
= 1 ∧ q − 1 = 1 ⇒ q = 2
p =
q · (1 + ϕ)
ϕ · (q − 1)
=
2 · 2
1
= 4
w
p · q
p(q − 1) − q
=
4 · 2
4 · 1 − 2
= 4, w ≥ 5
Hence the triad
p = 4, q = 2, w ≥ 5
ii)q = σ(q − 1) ∧ (1 + ϕ) = λ · ϕ
a)ϕ =
1
λ − 1
⇒ λ − 1 = 1 ⇒ (λ = 2 ∧ ϕ = 1)
β)q · (σ − 1) = σ ⇒ q =
σ
σ − 1
= 1 +
1
σ − 1
== 2 ∧ σ − 1 = 1 ⇒ (σ = 2 ∧ q = 2)
p =
q · (1 + ϕ)
ϕ · (q − 1)
=
2 · 2
1 · 1
= 4, w
p · q
p(q − 1) − q
=
4 · 2
4 · (2 − 1) − 2
= 4, w ≥ 5
Therefore resulting triad
|p = 4, q = 2, w ≥ 5|
17

iii)q = σ · ϕ ∧ (1 + ϕ) = λ · (y − 1)
a)λ =
1 + ϕ
y − 1
=
1
q − 1
+
ϕ
q − 1
∧ q − 1 = 1 ⇒ (q = 2 ∧ λ = 3)
σ · ϕ = 2 ⇒ (σ = 1 ∧ ϕ = 2), (σ = 2 ∧ ϕ = 1)
q = 2 ∧ ϕ = 1 ⇒ p =
q · (1 + ϕ)
ϕ · (q − 1)
=
2
1
2
1
= 4, w
p · q
p(q − 1) − q
=
4 · 2
4 · 1 − 2
= 4
q = 2 ∧ ϕ = 2 ⇒ p =
q · (1 + ϕ)
ϕ · (q − 1)
=
2
2
3
1
= 3, w
p · q
p(q − 1) − q
=
3 · 2
3 · 1 − 2
= 6
Thus arise the two triads
p = 4, q = 2, w ≥ 5 and p = 3, q = 2, w ≥ 7
3.Similarly from equation p = ε · (p · (q − 1) − q) take that:
p = ε · p · (q − 1) − q · ε ⇒ q =
p · (1 + ε)
ε · (p − 1)
i)ε(p − 1) = 1 ⇒ ε =
1
p − 1
= 1 ∧ p − 1 = 1 ⇒ p = 2
q =
p · (1 + ε)
ε · (p − 1)
=
2 · 2
1
= 4
w
p · q
p(q − 1) − q
=
4 · 2
2 · 3 − 4
= 4, w ≥ 5
Therefore shows the triad
q = 4, p = 2, w ≥ 5
ii)p = ε(p − 1) ∧ (1 + ε) = λ · ε
a)ε =
1
λ − 1
⇒ λ − 1 = 1 ⇒ (λ = 2 ∧ ε = 1)
b)p · (ε − 1) = ε ⇒ p =
ε
ε − 1
= 1 +
1
ε − 1
== 2 ∧ ε − 1 = 1 ⇒ (ε = 2 ∧ p = 2)
q =
p · (1 + ε)
ε · (p − 1)
=
2 · 2
1 · 1
= 4, w
p · q
p(q − 1) − q
=
4 · 2
2 · (4 − 1) − 4
= 4, w ≥ 5
Hence the triad
p = 2, q = 4, w ≥ 5
18

iii)p = ε · ϕ ∧ (1 + ε) = λ · (p − 1)
a)λ =
1 + ε
p − 1
=
1
p − 1
+
ε
p − 1
∧ p − 1 = 1 ⇒ (p = 2)
ε · ϕ = 2 ⇒ (ε = 1 ∧ ϕ = 2), (ε = 2 ∧ ϕ = 1)
p = 2 ∧ ϕ = 1 ⇒ q =
p · (1 + ϕ)
ϕ · (p − 1)
=
2
1
2
1
= 4, w
p · q
p(q − 1) − q
=
4 · 2
2 · 3 − 4
= 4
p = 2 ∧ ϕ = 2 ⇒ q =
p · (1 + ϕ)
ϕ · (p − 1)
=
2
2
3
1
= 3, w
p · q
p(q − 1) − q
=
3 · 2
2 · 1 − 2
= 6
Thus arise the two triads
q = 4, p = 2, w ≥ 5 and q = 3, p = 2, w ≥ 7
Total we have 12 cases for exhibitors and cyclically we will have
(i)
p = 3, q = 2, w ≥ 7 ∧ p = 2, q = 3, w ≥ 7
w = 3, p = 2, q ≥ 7 ∧ w = 2, p = 3, q ≥ 7
w = 3, q = 2, p ≥ 7 ∧ w = 2, q = 3, p ≥ 7
q = 4, p = 2, w ≥ 5 ∧ q = 2, p = 4, w ≥ 5
w = 4, p = 2, q ≥ 5 ∧ w = 2, p = 4, q ≥ 5
w = 4, q = 2, p ≥ 5 ∧ w = 2, q = 4, p ≥ 5
Which in relation to equations take the form
(ii)
x3
+ y2
= zw
, w ≥ 7
x2
+ y3
= zw
, w ≥ 7
x2
+ yq
= z3
, q ≥ 7
x3
+ yq
= z2
, q ≥ 7
xp
+ y2
= z3
, p ≥ 7
xp
+ y3
= z2
, p ≥ 7
x2
+ yq
= z4
, q ≥ 5
x4
+ yq
= z2
, q ≥ 5
x2
+ y4
= zw
, w ≥ 5
x4
+ y2
= zw
, w ≥ 5
xp
+ y2
= z4
, p ≥ 5
xp
+ y4
= z2
, p ≥ 5
Characteristics mention the work of Jamel Ghanouchi ”A new approach of Fermat-Catalan conjecture” that
achieves the same result.
The generalized Fermat conjecture (Darmon and Granville, 1995; Darmon, 1997), also known as the
Tijdeman-Zagier conjecture and as the Beal conjecture (Beukers, 2012), is concerned with the case χ
1.
19

It states that the only non-trivial primitive solutions to xq
+ yp
= zw
with σ(p, g, r) 1 are
25
+ 72
= 34
, 73
+ 132
= 29
, 27
+ 173
= 712
, 35
+ 114
= 1222
,
177
+ 762713
= 210639282
, 14143
+ 22134592
= 657
, 92623
+ 153122832
= 1137
,
438
+ 962223
= 300429072
and 338
+ 15490342
= 156133
.
The generalized Fermat conjecture has been documented for many signatures (p, q, r), including many in-
finite families of signatures, starting with Fermat’s last theorem (p, p, p) by Wiles (1995). The remaining
cases are reported in Chapter 4.
Case 2rd. 1/p + 1/q + 1/w = 1
i) From case 1 shows that overall we have 12 cases for exhibitors and and we roundly take:
p = 3, q = 2, w ≥ 7 ∧ p = 2, q = 3, w ≥ 7
w = 3, p = 2, q ≥ 7 ∧ w = 2, p = 3, q ≥ 7
w = 3, q = 2, p ≥ 7 ∧ w = 2, q = 3, p ≥ 7
q = 4, p = 2, w ≥ 5 ∧ q = 2, p = 4, w ≥ 5
w = 4, p = 2, q ≥ 5 ∧ w = 2, p = 4, q ≥ 5
w = 4, q = 2, p ≥ 5 ∧ w = 2, q = 4, p ≥ 5
⇔
p = 3, q = 2, w 6 ∧ p = 2, q = 3, w 6
w = 3, p = 2, q 6 ∧ w = 2, p = 3, q 6
w = 3, q = 2, p 6 ∧ w = 2, q = 3, p 6
q = 4, p = 2, w 4 ∧ q = 2, p = 4, w 4
w = 4, p = 2, q 4 ∧ w = 2, p = 4, q 4
w = 4, q = 2, p 4 ∧ w = 2, q = 4, p 4
(i) (ii)
But the inequality (ii), for example, p = 3, q = 2, w 6 as well as the inequality q = 4, p = 2, w 4 which
is characteristic of the group of exhibitors according to the criterion 0 1/p + 1/q + 1/w 1, so for the
exponent group to have equality, 12 relations will apply cyclically as follows:
(iii)
p = 3, q = 2, w = 6 ∧ p = 2, q = 3, w = 6
w = 3, p = 2, q = 6 ∧ w = 2, p = 3, q = 6
w = 3, q = 2, p = 6 ∧ w = 2, q = 3, p = 6
q = 4, p = 2, w = 4 ∧ q = 2, p = 4, w = 4
w = 4, p = 2, q = 4 ∧ w = 2, p = 4, q = 4
w = 4, q = 2, p = 4 ∧ w = 2, q = 4, p = 4
ii) Pending from only the case 3/p = 1 ⇒ p = 3 which implies p = q = w = 3. But this case according to
the proof of Fermat’s theorem does not accept solutions with exponents greater than 2.
Case 3rd
. 1/p + 1/q + 1/w 1
Originally accept that p = 2, q = 2 and w = 2. We examine three cases:
i) p = q = w = 2 which is true
20

ii) p = q = 2 ⇒ w 2 which is true we cyclically for the other exhibitors that p = w = 2 ⇒ q 2
and q = w = 2 = p 2.
iii) For all other cases will apply in accordance with the relation (iii) the second case, because now would
force the inequality 6, i.e total of 12 relations for all exhibitors.
p = 3, q = 2, {2 = w = 5} ∧ p = 2, q = 3, {2 = w = 5}
w = 3, p = 2, {2 = q = 5} ∧ w = 2, p = 3, {2 = q = 5}
w = 3, q = 2, {2 = p = 5} ∧ w = 2, q = 3, {2 = p = 5}
q = 4, p = 2, {2 = w = 3} ∧ q = 2, p = 4, {2 = w = 3}
w = 4, p = 2, {2 = q = 3} ∧ w = 2, p = 4, {2 = q = 3}
w = 4, q = 2, {2 = p = 3} ∧ w = 2, q = 4, {2 = p = 3}
(iv)
6.2 Theorem 6
The equation xp
+ yq
= zw
with positive integers x, y, z and extra (p, q, w = 2) and p, q and w are
fixed positive integers is solved if and only if apply the conditions of Theorem 5, (1, 2, 3) cases for exponents
p, q, w with extra (x, y, z) = 1, and at least one of them equal 2. Therefore Beal’s Conjecture is true with
the above conditions, because accepts that there is no solution under the condition that all values of the
exponents greater of 2.
Proof
For the equation xp
+ yq
= zw
with positive integers x, y, z, (p, q, w = 2) demonstrated that solved if
and only if apply the conditions of Theorem 5 (i, ii, iii) for the exponents p, q, w with extra (x, y, z) = 1, so
we have analytical
i) 1/p + 1/q + 1/w 1
According to Theorem 5, and 1 case, there is a solution to obtain values for the group of exhibitors {p, q, w}
as follows:
p = 3, q = 2, w ≥ 7 ∧ p = 2, q = 3, w ≥ 7
w = 3, p = 2, q ≥ 7 ∧ w = 2, p = 3, q ≥ 7
w = 3, q = 2, p ≥ 7 ∧ w = 2, q = 3, p ≥ 7
q = 4, p = 2, w ≥ 5 ∧ q = 2, p = 4, w ≥ 5
w = 4, p = 2, q ≥ 5 ∧ w = 2, p = 4, q ≥ 5
w = 4, q = 2, p ≥ 5 ∧ w = 2, q = 4, p ≥ 5
Which clearly shows that p = 2 or q = 2 or w = 2. Therefore least one exponent = 2.
ii) 1/p + 1/q + 1/w = 1
It happens the second case, Theorem 5, for exist solution will arrive at values for the group of exhibitors
21

{p, q, w} as follows:
p = 3, q = 2, w = 6 ∧ p = 2, q = 3, w = 6
w = 3, p = 2, q = 6 ∧ w = 2, p = 3, q = 6
w = 3, q = 2, p = 6 ∧ w = 2, q = 3, p = 6
q = 4, p = 2, w = 4 ∧ q = 2, p = 4, w = 4
w = 4, p = 2, q = 4 ∧ w = 2, p = 4, q = 4
w = 4, q = 2, p = 4 ∧ w = 2, q = 4, p = 4
Which also seems that p = 2 or q = 2 or w = 2. Therefore least one exponent equal 2.
iii) 1/p + 1/q + 1/w 1
For the third case, the Theorem 5, to obtain a solution we will arrive at values for the group of exhibitors
{p, q, w} as follows:
p = 3, q = 2, {2 = w = 5} ∧ p = 2, q = 3, {2 = w = 5}
w = 3, p = 2, {2 = q = 5} ∧ w = 2, p = 3, {2 = q = 5}
w = 3, q = 2, {2 = p = 5} ∧ w = 2, q = 3, {2 = p = 5}
q = 4, p = 2, {2 = w = 3} ∧ q = 2, p = 4, {2 = w = 3}
w = 4, p = 2, {2 = q = 3} ∧ w = 2, p = 4, {2 = q = 3}
w = 4, q = 2, {2 = p = 3} ∧ w = 2, q = 4, {2 = p = 3}
in which at least appear that one of the p = 2 or q = 2 or w = 2.
Therefore at least one exponent equals 2 to have a solution and hence play Beal’s Conjecture is true,
because it recognizes that there is no solution if all values of the exponents greater 2.
6.3 Lemma 3
According to the known identity that satisfies the raising integer power will apply:
aϕ
=

1.ϕ = 0
2.α∗
m, (α, m, ϕ ∈ Z+
)

Proof
If a, ϕ ∈ Z+
then will apply to the exhibitor of a:
i) a0
= 1 applicable
ii) aϕ
= α · αϕ−1
= α · m, m ∈ Z+
∧ m = aϕ−1
which this applies.
22

6.4 Formulations of the equation so that the exponents can be calculated by algorithm
According to Theorem 2 that the Diophantine equation xp
+ yq
= zw
where we apply that q, p, w ∈ Z+
we
define the F(x, y, z) function variables, x, y, z in particular integers as follows
F(x, y, z) =

∃x, y, z ∈ Z3
: A(x) = xq−1
, B(y) = yp−1
, C(z) = zw−1
∧ A(x)x + B(y)y = C(z)z

with additional constraints q, p, w ∈ Z ∧ {q 1, p 1, w 1}.
For (kerf= 0) which means to solve the system of equations F(x, y, z) = 0 we must be true kerF =

∃(x, y, z) ∈ Z3
: f(x, y, z) = 0

⊆ Z3
and finally after replacing the f becomes
i) If we first divide initially with x and x ̸= 0 the variables {y, z}, all the terms of the equation xd
+yp
= zw
,
we get with the existing conditions and we have:
A(x) = xq−1
, B(y) = yp−1
, C(z) = zw−1
. ∧ A(x)x + B(y)y = C(z) · z.
Thus according to Theorem 1 will apply to:
f(x, y, z) =




x = (C(z) · λ − B(y)) · k
y = (A(x) − C(z)) · k
z = (A(x) · λ − B(y)) · k



 (1) with more generally k ∈ Z+
, λ ∈ Q+
.
But in accordance with Lemma 3 will have the equivalences:
A(x)x−1
= A(x) · m1 ∧ B(y)y−1
= B(y) · m2 ∧ C(z)z−1
= C(z) · m3(2)
From (1) and (2) implies that
f(x, y, z) =







A(x) · m1 = (C(z) · λ − B(y)) · k
k =
B(y)y−1
A(x) − C(z)
C(z) · m3 = (A(x) · λ − B(y)) · k







⇔







(A(x))(A(x) − C(z)) · A(x) · m1 = A(x) · (C(z) · λ − B(y)) · B(y)y−1
k =
B(y)y−1
A(x) − C(z)
(−C(z)) · (A(x) − C(z)) · C(z) · m3 = (−C(z)) · (A(x) · λ − B(y)) · B(y)







and after replacing the relations and simplifying the system we will arrive at the final
(A(x) − C(z)) ·

A(x)2
· m1 − C(z)2
· m3

= (C(x) − A(x)) · B(y)y
⇔
relationship:
C(z)2
· m3 − A(x)2
· m1 = B(y)y
(3)
23

ii) Also if we divide by y if y ̸= 0 the variables {x, z}, we will certainly force the same method on the
function F, given will by the general form:
f(x, y, z) =




y = (C(z) · λ − A(y)) · k
x = (B(x) − C(z)) · k
z = (B(x) · λ − A(y)) · k



 (4)
With extra generally κ, λ ∈ Q+
. But according to Lemma 3 and relation (3), we have the equivalences:
f(x, y, z) =





k =
A(y)x−1
B(y) · C(z)
B(y) · m2 = (C(z) · λ − A(x)) · k
C(z) · m3 = (B(y) · λ − A(x)) · k





⇔





k =
A(y)x−1
B(y) − C(z)
(B(y))(B(y) − C(z)) · B(y) · m1 = B(y) · (C(z) · λ − A(x)) · A(x)x−1
(−C(z)) · (B(y) − C(z)) · C(z) · m3 = (−C(z)) · (B(y) · λ − A(x)) · A(x)x−1





and after replacing the relations and simplifying the system we arrive at the final relationship
(B(y) − C(z)) ·

B(y)2
· m2 − C(z)2
· m3

= (C(x) − B(y)) · A(x)x
⇔ C(z)2
· m3 − B(y)2
· m2 = A(x)x
(5)
iii) Finally, in the same way and dividing by the variable z the {x, y}, where z ̸= 0 will occur in accordance
with the basic relation and in such a use of Theorem 1, the function F which is given by the general form:
F(x, y, z) =




x = (C(z) + λ · B(y)) · k
z = (A(x) + B(y)) · k
y = (C(z) − λ · A(x)) · k



 with more generally k ∈ Z+
, λ ∈ Q+
(6)
But in accordance with Lemma 3 and relation (3), we have the equivalences
f(x, y, z) =





A(x) · m1 = (C(z) + λ · B(y)) · k
C(z) · m3 = (C(z) − λ · A(x)) · k
k =
C(z)z−1
A(x) + B(z)





⇔





(A(x)) · (A(x) + B(y)) · A(x) · m1 = A(x) · (C(z) + λ · B(y)) · C(z)z−1
(B(y)) · (A(x) + B(y)) · B(y) · m2 = (B(y)) · (C(y) − λ · A(x)) · C(z)z
− 1
k =
C(z)z−1
A(x) + C(z)





and after operations arrive and clear the system in the final relationship
(A(x) + B(y)) ·

A(x)2
· m1 + B(y)2
· m2

= (A(x) + B(y)) · C(z)z
24

⇔ A(x)2
· m1 + B(y)2
· m2 = C(z)z
(7)
For the case of Conjecture Beal, as we form Diophantine equation ax
+ by
= cz
.
They will apply to replacement equivalents
A(x) = a, B(y) = b, C(z) = c
And therefore arise 3 Diophantine equations of the first degree where calculated variables m1, m2, m3 and
are resolved by known each category
c2
· m3 − a2
· m1 = by
(3)
c2
· m3 − b2
· m2 = ax
(5)
a2
· m1 + b2
· m2 = cz
(7)
with m1, m2, m3 ∈ Z+
But according to Theorem 5 6, two of the variables {x, y} or {x, z} or {y, z} will have values through
the set {3, 5, 7}. If therefore the system of primary Diophantos equations {i, ii, iii} import these values we
calculate the variables m1, m2, m3 per category and therefore the remaining variables.
1st
c2
· m3 − α2
· m1 = by
(i1)
ax−1
= α · m1 (ii1)
cz−1
= c · m3 (iii1)
2nd
c2
· m3 − b2
· m2 = αx
(i2)
by−1
= b · m2 (ii2)
cz−1
= c · m3 (iii2)
3rd
α2
· m1 + b2
· m2 = cz
(i3)
by−1
= b · m2 (ii3)
αx−1
= c · m1 (iii3)
with a, b, c, m1, m2, m3 ∈ Z+
According to these cases we can use a program to calculate exponents and variables cyclically so that their
values are fully calculated. In the following we will see examples of how they are calculated with a program
in mathematica.
7. Finding exhibitors theoretically-by Lemma 1 and Theorems (5 6).
7.1. Calculates of x, y, z on Diophantine equations
18x
+ 9y
= 9z
, 32x
+ 32y
= 4z
, 13x
+ 7y
= 2z
, 7x
+ 7y
= 98z
19x
+ 38y
= 57z
, 34x
+ 51y
= 85z
, 33x
+ 66y
= 33z
7.2. Diophantine solutions
7.2.1. Example 18x
+ 9y
= 9z
By analysis under Lemma1 takes the form:
18x
+ 9y
= 9z
⇔ (2 · 9)x
+ 9y
= 9z
⇔ 2x
+ 9y−x
= 9z−x
⇔ 2x
+ 32(y−x)
= 32(z−x)
25

Investigation:
1. Must x = 3, (y − x) = 0 as well as 2(z − x) = 2, from which implies x = 3, y = 3, z = 4, which is
accepted.
2. If x = 4, y = 4 and z = 5 is not acceptable.
3. If x = 2 then we have 2 cases
i) 2(z − x) = 4, and z = 4 impossible also because 32(y−x)
= 34
− 32
= 77.
ii) 2(z − x) = 3, and z = 7/2 also impossible not intact.
7.2.2. Example 32x
+ 32y
= 4z
By analysis under Lemma1 takes the form 1 + 25(y−x)
= 22z−5x
Investigation:
1. Must x − y = 0 and then x = y and also 2z − 5x = 1 ⇔ z = (1 + 5x)/2.If x = 1 then z = 3 which is
accepted. In General It holds for x = 2k + 1, k in Z i.e z = 5k + 3, x = y = 2k + 1, k in Z.
7.2.3. Example 13x
+ 7y
= 2z
By analysis is the case 1/p + 1/q + 1/w 1 with p = 3q = 2w 7. Therefore we have x = 3, y = 2 and
z = 9 and takes the form 133
+ 72
= 29
.
7.2.4. Example 7x
+ 7y
= 98z
By analysis under Lemma1 takes the form 1 + 7y−x
= (2z
) · 7(2z−x)
Investigation:
1. If x = 3, (y − x) = 0 as well 2z − x = 0 implies z = 3/2 impossible.
2. If x = 4, (y − x) = 0 as well 2z − x = 0, from which entails z = 4/2 = 2, which is not accepted
3. If z = 3, (y − x) = 1 and 2z − x = 0 it follows ó τx = 6k and y = 7 which is accepted
4. If z = 1, (y − x) = 0 and 2z − x = 0 it follows ót x = 2 and y = 2, which is accepted
7.2.5. Examples 19x
+ 38y
= 57z
By analysis under Lemma1 takes the form 19x−y
+ 2y
19y−z
= 3z
Investigation:
1. If x − y = 0, z − x = 0, y = 1 from where follows x = y = z = 1 which is accepted
2. If x − y = 1, y = 3, y − z = 0, z = 3 = x = 4 and y = z = 3 which is accepted
26

7.2.6. Example 34x
+ 51y
= 85z
By analysis under Lemma 1 takes the form 2x
17x
+ 3y
17y−z
= 3z
Investigation:
1. If x − y = 1, y − z = 0, z = 4 comes from where x = 5, y = z = 4 which is accepted
2. If x − y = 0, y − z = 0, z = 1 = x = 1, y = z = 1 which is accepted
7.2.7. Example 33x
+ 66y
= 33z
By analysis under Lemma1 takes the form 3x−y
11x−y
+ 2y
= 3z−y
11z−y
Investigation:
1. If x − y = 0, z − y = 1, y = 5 from where follows z = 6, y = x = 5 which is accepted
8. Calculation of Exhibitors. (Programs for Mathematica)
8.1. The first case
According to the second method of dividing initially with x where x ̸= 0 the variables {y, z}, where terms
in the equation xq
+ yp
= zw
⇒
f(x, y, z) =


y = (C(z) · λ − A(y)) · k
x = (B(x) − C(z)) · k
z = (B(x) · λ − A(y)) · k

 extra generality k ∈ Z+
, λ ∈ Q+
. (1)
And according to the analysis of the second method results in the
c2
· m3 − b2
· m2 = αx
(i)
by−1
= b · m2(ii)
cz−1
= c · m3(iii)
(1)
from where with a suitably choice of y according to the first method, we calculate the the m2, m3 ∈ Z+
.That
transformed as in program format mathematica accept as the basis {a, b, c} of the equation a∧
m+b∧
n = c∧
v
with a = 7; b = 7; c = 98;
Program 8.1.1
Clear [m2, m3, y]
a = 7; b = 7; c = 98
Table[Reduce[m3∗
c∧
2 − m2∗
b∧
2 = a∧
(x)m3 0m2 0, {m2, m3}, Integers], {y, 1, 6, 1}]
Results:
{False, C[1][Element] Integers C[1] = 0 m2 == 195 + 196 C[1] m3 == 1 + C[1]
27

C[1][Element] Integers C[1] = 0 m2 == 189 + 196 C[1] m3 == 1 + C[1],
C[1]Element] Integers C[1] = 0 m2 == 147 + 196 C[1] m3 == 1 + C[1]
C[1][Element] Integers C[1] = 0 m2 == 49 + 196 C[1] m3 == 2 + C[1]
C[1][Element] Integers C[1] = 0 m2 == 147 + 196 C[1] m3 == 13 + C[1]}
Using a program re-try one of the cases the result of the programming 1 and calculate the exhibitors
to agree data the initial equation a∧
m + b∧
n = c∧
v.
Continue... Program 8.1.2.
Clear[x,y,x1,y1,z1]
a := 7; b := 7; c := 98;
Reduce[x == 13+ky = 147+196ky1 == 1+Log [b∗
y] /Log[b]z1 = 1+Log [c∗
x] /Log[c]x1 ==
Log [c∧
2 ∗ x − b∧
2 ∗ y] /Log[a]0 = k = 100, {x1, y1, z1}, Integers]
Results:
(k = 85x = 98y = 16807)x1 = 6y1 = 7z1 = 3
8.2. The Second Case.
According to the second method of dividing a crack where yby ̸= 0 variables {x, z}, where terms in the
equation xq
+ yp
= zw
⇒
f(x, y, z) =


x = (C(z) · λ − B(y)) · k
y = (A(x) − C(z)) · k
z = (A(x) · λ − B(y)) · k

 Extra generally k ∈ Z+
, λ ∈ Q+
c2
· m3 − α2
· m1 = by
(i)
ax−1
= α · m1(ii)
cz−1
= c · m3(iii)
(2)
from where with a suitably choice of y according to the first method, we calculate the m1, m3 ∈ Z+
That
transformed as in program format mathematica accept as the basis {a, b, c} of the a∧
m + b∧
n = c∧
v with
a = 7; b = 7; c = 98;
Program 8.2.1
Clear [m1, m3]
a := 7; b := 7; c := 98;
Table [Reduce [m3∗
c∧
2 − m1∗
a∧
2 = b∧
(y) m1 0 m3 0, {m1, m3}, Integers], {y, 1, 6, 1}]
28

Results:
{False,
C[1] [Element] Integers C[1] = 0 m1 == 195 + 196 C[1] m3 == 1 + C[1]
C[1][Element] lntegers C[1] = 0 m1 == 189 + 196 C[1] m3 == 1 + C[1]
C[1][Element] Integers C[1] = 0 m1 == 147 + 196 C[1] m3 == 13 + C[1]}
Using a program re-try one of the cases the result of the programming? 1 and calculate the exhibitors
to agree dedomana the initial equation a∧
m + b∧
n = c∧
v with a = 7; b = 7; c = 98;
Continue... Program 8.2.2
Clear[x, y, x1, y1, z1]
a := 7; b := 7; c := 98;
Reduce [z == 13 + kx == 147 + 196kx1 == 1 + Log [a∗
x] /Log[a]z1 == 1 + Log [c∗
z] /Log[c]y1
== Log [c∧
2∗
z − a∧
2 ∗ x] /Log[b]0 = k = 100, {x1, y1, z1}, Integers]
Results:
(k = 85z = 98x = 16807)x1 = 7y1 = 6z1 = 3
8.3. The third case.
According to the second method of dividing a initial where zz ̸= 0 variables {x, y}, where terms in the
equation xq
+ yp
= zw
⇒
f(x, y, z) =


x = (C(z) + λ · B(y)) · k
z = (A(x) + B(y)) · k
y = (C(z) − λ · A(x)) · k

 Extra generally with k ∈ Z+
, λ ∈ Q+
.
α2
· m1 + b2
· m2 = cz
(i)
by−1
= b · m2 (ii)
αx−1
= α · m1 (iii)
(3)
from where with a suitably choice of y according to the first method, we calculate the m1, m2 ∈ Z+
. That
transformed as in program format mathematica accept as the basis {a, b, c} of the equation a∧
m+b∧
n = c∧
v
with a = 7; b = 7; c = 98;
29

Program 8.3.1.
Clear[m1,m3]
a := 7; b := 7; c := 98;
Table[Reduce[m1∗
a∧
2 + m2∗
b∧
2 == c∧
(z)m1 0m2 0, {m1, m2}, Integers], {z, 1, 6, 1}]
Results:
{m1 == 1 m2 == 1,
C[1] Integers 1 = C[1] = 195 m1 == C[1] m2 == 196 − C[1],
C[1] Integers 1 = C[1] = 19207 m1 == C[1] m2 == 19208 − C[1],
C[1] Integers 1 = C[1] = 1882383 m1 == C[1] m2 == 1882384 − C[1],
C[1] Integers 1 = C[1] = 184473631 m1 == C[1] m2 == 184473632 − C[1],
C[1] Integers 1 = C[1] = 18078415935 m1 == C[1] m2 == 18078415936 − C[1]}
Using a program re-try one of the cases the result of the programming 1 and calculate the exhibitors
to a = 7; b = 7; c = 98;
Continue Program 8.3.2.
Clear [x, y, x1, y1, z1]
a := 7; b := 7; c := 98
Reduce [x == 1y == 1y1 == 1 + Log [b∗
y] /Log[b]x1 == 1 + Log [a∗
x] /Log[a]z1 ==
Log [a∧
x1 + b∧
y1] /Log[c]0 = k = 100, {x1, y1, z1}, Integers]
Results:
(k ∈ Zx = 1y = 1)x1 = 2y1 = 2z1 = 1
Therefore there are 3 solutions (6, 7, 3), (7, 6, 3) and (2, 2, 1).
According to these results it is obvious that we can calculate the exponents in each case if we know the
bases.If there is no correspondence then we will not find an integer solution in the second program analysis
in each case separately.
8.4.Theorem 7.(F.L.T) For any integer n 2, the equation xn
+ yn
= zn
has no positive
integer solutions
An equation of the form xa
+ yb
= zc
(Beals’) to have a solution, according to theorems {5, 6}, must
have at least one exponent equal to 2. And since in Fermat’s last theorem we have a = b = c = n, it follows
directly that the only solution that Fermat’s equation xn
+ yn
= zn
can have is when n = 2. So for n 2
there is no solution.
30

Epilogue
According to the logic for Beals’ equation to hold, we obtain a global proof in Theorems 5,6 after considering
all cases σ(x, y, z) 1, σ(x, y, z) 1 and σ(x, y, z) = 1 with co-prime bases. So as a final conclusion, at least
one exponent of the equation, must be equal to 2 to be solved and not always. This helps in many cases
to solve three-variables diophantine equations when considering exponents greater than 3. By this method,
Fermat’s Last Theorem (σ (x, y, z) = 1) is very easily and understandably proved as we have seen.
31

References
[1] The Darmon–Granville Equation with Algebraic Exponents, John A. Zuehlke.
[2] Mchel Waldschmidt Moscow, Open Diophantine Problems Mathematical.
[3] Michael A. Bennett and Chris M. Skinner, Ternary Diophantine Equations via Galoi Reresentations
and Modular Forms.
[4] H. Darmon and A. Granville, On the equations zm
= F(x, y) and Axp
+ Byq
= cZr
, London Math.
Soc. 27(1995), 513-543.
[5] Fermat’s Last Theorem, Mantzakouras Nikos, https://www.researchgate.net/publication/
342349669_Proof_of_Fermat’s_Last_Theorem
[6] The Diophantine Equation ax
+ by
= cz
by Nobuhiro Terai.
[7] A new approach of Fermat-Catalan conjecture Jamel Ghanouchi.
[8] Mathematica ”Main Theorem of Gröbner”, www.wolfram.com.
[9] M. A. Bennett, I. Chen, S. R. Dahmen and S. Yazdani, Generalized Fermat equations: a miscellany,
Int. J. Number Theory 11 (2015), 1–28.
[10] H. Cohen, Number theory. Vol. II. Analytic and modern tools, Graduate Texts in Mathematics, 240,
Springer, New York, 2007.
[11] J. Edwards, A complete solution to x2
+ y3
= z5
. J. Reine Angew. Math. 571 (2004), 213–236.
32

Proof of Beal's conjecture

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