International Journal of Scientific & Engineering Research, Volume 4, Issue 8, August-2013

ISSN 2229-5518

Mystery of Beal Equation:

ax + by = cz

Author: DEBAJIT DAS

(Company: Indian Oil Corporation Ltd, Country: INDIA)

International Journal of Scientific & Engineering Research, Volume 4, Issue 8, August-2013

ISSN 2229-5518

1329

ABSTRACT

This paper contains the proof of Beal Equation Mystery i.e. without common factor among all the bases a Beal equation cannot exist and then the theory behind the formation of Beal Equation.

Keywords

Beal equation, INB-equation, Natural Equation, Nd operation, Ns operation, Zygote form

1. Introduction

ax + by = cz, where a, b, c, x, y, z all are of positive integers & x, y, z > 2 is known as Beal Equation .

If this Beal Equation exists there must be a common factor among all the bases a, b, c e.g. 23 + 23 = 24,

76 + 77 = 983, 194 + 383 = 573 etc.

The proof of this mystery is still unknown to all mathematical communities. I believe that I have been able to give a proof in favor of this mystery.

The proof is mainly based on the properties of Pythagorean equation a2 + b2 = c2 where a, b, c are of positive integers without any common factor among them.

The proof clearly shows that only one element of a, b, c can produce powers like 3/2, 2, 5/2, 3, ….. so as to receive three types of equations an + b2 = c2 or a2 + bn = c2 or a2 + b2 = cn barring two exceptional cases 1 + 23 = 32 and 72 + 25 = 34.

The LH odd element & RH odd element produces powers by Nd & Ns operations respectively and both cannot run simultaneously.

If these equations are converted to Pythagorean equation we may get the set of (a,b,c) as integers or irrationals depending upon n is even or odd.

These sets must be different from the sets when a Beal Equation is converted to Pythagorean equation after cancellation of common factor from Beal Equation.

Apart from the proof that without common factor among the bases a Beal Eq. cannot exist, it also shows how Beal equations are formed and there cannot be any common factor among the exponents of Beal Equation.

International Journal of Scientific & Engineering Research, Volume 4, Issue 8, August-2013

ISSN 2229-5518

1330

2. Natural Equation (or, N-Eq.)

For a< b <c the Pythagorean equation
a2 + b2 = c2 is said to be a Natural equation when the elements a,b,c all are of positive integers without any
common factor. (a,b,c) is said to be its prime set and (c – b) is called its Natural constant, denoted by k.
Now, from N-Eq. a2 + b2 = c2 or, (aµ)2 + (bµ)2 = (cµ)2 we can say that any prime set (a,b,c) can produce infinite number of composite sets (aµ,bµ,cµ) where µ is any positive integer.
(a,b,c) is said to be an irrational set when at least one element is an irrational number in the form of p + q√r
Now, for a N-Eq. all the elements cannot be even number as it mainly includes a prime set. Later on, it will be
shown that (odd)2 + (odd)2 = (even)2 also cannot exist. It exists only for (odd)2n-1 + (odd)2 = (even)2, where n =
1,2,3,4,……. Hence, only possibility for the existence of N-Equation is
(odd)2 + (even)2 = (odd)2
Now, as the left hand odd element of a N-Eq. can always be expressed as (α2 – β2), the said N-Eq. must be comparable with (α2 – β2)2 + (2αβ)2 = (α2 + β2)2, where α,β are the combination of odd & even numbers.
It indicates the existence of N-Eq. of following two kinds.
1st kind where k = α2 + β2 – 2αβ = (α – β)2 i.e. in the form of (2n – 1)2, where n = 1,2,3,4,…… Obviously, it is
(odd)2 + (even)2 = (odd)2
2nd kind where k is in the form of 2n2 for n = 1,2,3,4,….. Obviously, it is (even)2 + (odd)2 = (odd)2
(3rd element – 2nd element) will be always in the form of (2n – 1)2 and 2n2 for 1st kind & 2nd kind N-Eqs. respectively.
If any constant µ2, where µ is an odd integer be multiplied with all the elements of (α2 – β2)2 + (2αβ)2 = (α2 +

β2)2, the nature of k remains unchanged.

∴ we can say, the N-Eq also includes those composite sets where common factor is a square of an odd integer.

But in our concerned area as this common factor has got no significance we must assume that N-Eq is basically
for the equation of all prime sets. Apart from N-eq. all other sets of Pythagorean equation are definitely composite sets.

3. For a particular value of k of a N-Equation there exists a definite functional set of single variable [fk(x), φk(x), ψk(x)] to satisfy the N-Equation. Hence, [fk(x)]2 + [φk(x)]2 = [ψk(x)]2

Case I. When k = 1.
We have, a2 + b2 = (b + k)2 ∴ b = ½(a2/k – k).
∴general expression of all the sets is [a, ½(a2/k – k), ½(a2/k + k)]
For k = 1, the set becomes [a, ½(a2 – 1), ½(a2 + 1)] which accepts only odd numbers of a. Put a=2x + 1, to have
the prime set [2x+1, 2x2+2x, 2x2+2x+1]
Where, f1(x) = 2x+1, φ1(x) = 2x2+2x & ψ1(x) = 2x2+2x+1.
e.g. for x = 1,2,3,….. we have the prime sets (3,4,5), (5,12,13), (7,24,25),…..
Case II. When k = 2,
The set is [a, ½(a2/2 – 2), ½(a2/2 + 2)], which accepts only even numbers against a. It will produce two kinds of sets (i) all the composite sets for k=1, where µ = 2 against a = 6,10,14,18,22,………..
(ii) new prime sets against a = 8,12,16,20,………
∴put a = 4(x+1) & produced prime set is [4(x+1),{2(x+1)}2-1,{2(x+1)}2+1]
∴ f2(x) = 4(x+1), φ2(x) = 4x2+8x+3 & ψ2(x) = 4x2+8x+5.
e.g. for x = 1,2,3,4,……….we have the prime sets (8,15,17), (12,35,37), (16,63,65), (20,99,101),………
* For a = 4, the prime set is (4,3,5) which falls under Case-I due to a<b<c

4. General procedure to find out the functional set of a N-Equation for a particular value of k under 1st kind.

Say, k = 25 = 52. Now, if the leading set be (a,b,c) then c = b+25, a = c – 2.42 [ as 2.42 is just greater than 52] ∴ a = b – 7.
∴(b – 7)2 + b2 = (b + 25)2 or, b = 72. ∴ 1st set is (65,72,97).
Similarly, for the 2nd set a=c–2.52. [obviously, it will be a composite set]

International Journal of Scientific & Engineering Research, Volume 4, Issue 8, August-2013

ISSN 2229-5518

1331

∴ (b – 25)2 + b2 = (b + 25)2. or, b = 100. ∴ 2nd set is (75,100,125). Similarly, for 3rd set a = c – 2.62. ∴ (b – 47)2 + b2 = (b + 25)2.
∴b = 132. & 3rd set is (85,132,157).
Now, say a = Ax + B. ∴ for x = 1, A+B = 65 & for x = 2, 2A+B = 75.
Solving these two, A = 10 & B = 55. ∴ a = 10x + 55. Say, b = Ax2+Bx+C.

∴ for x = 1,2 &3; A + B + C = 72, 4A + 2B + C = 100 & 9A + 3B + C=132

Solving these three, A = 2, B = 22 & C = 48.
∴ b = 2x2 + 22x + 48. Obviously, c = 2x2 + 22x + 73.
∴[f25(x), φ25(x), ψ25(x)] = [(10x + 55), (2x2 + 22x + 48), (2x2 + 22x + 73].

5. General procedure to find out the functional set of a N-Equation for a particular value of k under 2nd kind.

Say, k = 18. ∴ k = 2.32.
Now, if the leading set be (a,b,c), then c = b + 18, a = c – 52 = b – 7.
∴ (b –7)2 + b2 = (b+18)2 or, b = 55 ∴ a = 48 & c = 73 & 1st set is (48,55,73)
similarly, for the 2nd set a = c – 72. ∴(b – 31)2 + b2 = (b + 18)2 or, b = 91
∴ a = 60 & c = 109 ∴ 2nd set is (60,91,109).
Similarly, for 3rd set a = c – 92. ∴ (b – 63)2 + b2 = (b + 18)2 or, b = 135.
∴ a = 72 & c =153 ∴ 3rd set is (72,135,153)
Now, say a = Ax + B ∴ for x = 1 & 2, A + B =48 & 2A + B =60
Solving, A = 12, B = 36. ∴ a = 12x + 36.
Now, say b = Ax2 + Bx + C

∴ for x = 1,2,3; A + B + C = 55, 4A + 2B + C = 91 & 9A + 3B + C = 135

Solving these three, A = 4, B = 24 & C = 27. ∴ b = 4x2 + 24x + 27
Obviously, c = b + 18 & c = 4x2 + 24x + 45.
∴[f18(x), φ18(x), ψ18(x)] = [(12x + 36), (4x2 + 24x + 27), (4x2 + 24x + 45].
In this way we can find out the functional set against any accepted value of k
Followings are the few examples of N-equation in functional form.
1st kind. For k = 1, (2x + 1)2 + (2x2 + 2x)2 = (2x2 + 2x + 1)2
For k = 9, (6x + 21)2 + (2x2 + 14x + 20)2 = (2x2 + 14x + 29)2
For k = 25, (10x + 55)2 + (2x2 + 22x + 48)2 = (2x2 + 22x + 73)2
For k = 49, (14x + 105)2 + (2x2 + 30x + 88)2 = (2x2 + 30x + 137)2
………………………………………………………………………………...

2nd kind. For k = 2, (4x + 4)2 + (4x2 + 8x + 3)2 = (4x2 + 8x + 5)2
For k = 8, (8x + 12)2 + (4x2 + 12x + 5)2 = (4x2 + 12x + 13)2
For k = 18, (12x + 36)2 + (4x2 + 24x + 27)2 = (4x2 + 24x + 45)2
For k = 32, (16x + 72)2 + (4x2 + 36x + 65)2 = (4x2 + 36x + 97)2
……………………………………………………………………………….
If the element against linear part is in the form of p√q i.e. αx + β = p√q or,
x = (p√q – β)/α, then satisfying it to the quadratic part ax2 + bx + c on LHS we get (Integer) + p√q/α(b – 2aβ/α). If it is to be integer b – 2aβ/α = 0 & it is really = 0 for both kinds of equations.
→ bα = 2aβ or, bn = 2β where n = √k for 1st kind & n = √(k/2) for 2nd kind.
→ one of the LH elements can be in the form of pn.√p & it will produce odd power & this element has no further rational zygote elements (discussed later)
For k=1&2, it will produce only prime sets, but in other cases some composite sets will appear intermittently as explained reason earlier in two kinds of N-eqs. In general, For k = 2n where n = 0,1,3,5,……., N-Eq produces only prime sets.
Coefficients of x2 for 1st kind & 2nd kind are respectively 2&4. If it is taken into consideration then with the help of first two leading sets we can find out the functional sets of all leading sets.

6. General expression of all the leading sets for different accepted values of k under 1st kind

Natural equation.

International Journal of Scientific & Engineering Research, Volume 4, Issue 8, August-2013

ISSN 2229-5518

1332

Here, [b + (2x – 1)2 – 2y2]2 + b2 = [b + (2x – 1)2]2
where, 2y2 is just greater than (2x – 1)2 by an integer value.
or, b2 – b.4y2 + 4y4 – 4y2(2x – 1)2 = 0 or, b = 2y2 ± 2y(2x – 1)
or, b = 2y2 + 4xy – 2y, neglecting (-) sign.
∴ a = 4x2 + 4xy – 4x –2y + 1 & c = 4x2 + 2y2 + 4xy – 4x – 2y + 1
∴ leading functional set (a,b,c) = [Lfk(x), Lφk(x), Lψk(x)] =
[4x2 + 4xy – 4x - 2y + 1, 2y2 + 4xy –2y, 4x2 + 2y2 + 4xy – 4x – 2y + 1]

∴for k = 1, put x = 1 & y = 1, to get the leading prime set (3,4,5)

for k = 9, put x = 2 & y = 3, to get the leading set (27,36,45)
for k = 25, put x = 3 & y = 4, to get the leading set (65,72,97)
for k = 49, put x = 4 & y = 5, to get the leading set (119,120,169) & so on

7. General expression of all the leading sets for different accepted values of k under 2nd kind

Natural equation.

Here, k = 2x2,∴ (b + 2x2 – y2)2 + b2 = (b + 2x2)2 where, y is an odd number so that y2 is just greater than 2x2 by an integer value.
∴ b2 – 2by2 + (y2 – 4x2)y2 = 0 or, b = y2 ± y√(y2 – y2 + 4x2). Neglect (-)sign.
∴ b = y2 + 2xy, a = y2 + 2xy + 2x2 – y2 = 2x2 + 2xy & c = y2 + 2xy + 2x2
∴ leading functional set (a,b,c) = [Lfk(x), Lφk(x), Lψk(x)] = [2x2 + 2xy, y2 + 2xy, y2 + 2xy + 2x2]

∴ for k = 2, put x = 1, y = 3 to get the leading prime set (8,15,17)

for k = 8, put x = 2, y = 3 to get the leading set (20,21,29)
for k = 18, put x = 3, y = 5 to get the leading set (48,55,73)
for k = 32, put x = 4, y = 7 to get the leading set (88,105,137) & so on.

8. Natural equation in Zygote form.

In Zygote form, a N-equation can be written as,
1st kind
{(y+2x-1)2 – (y)2}2 + {2.y.(y+2x-1)}2 = {(y+2x-1)2 + (y)2}2 [ as leading set]
or, [{f(x)}2 – {φ(x)}2]2 + [2.f(x).φ(x)]2 = [{f(x)}2 + {φ(x)}2]2
where, for a particular value of k, {f(x) - φ(x)} is constant.
2nd kind
{2.x.(x+y)}2 + {(x+y)2 – (x)2}2 ={(x+y)2 + (x)2}2 . [as leading set] or, [2.f(x).φ(x)]2 + [{f(x)}2 – {φ(x)}2]2 = [{f(x)}2 + {φ(x)}2]2 . where, for a particular value of k, {φ(x)} is constant.

9. Example Chart of N-equation in Zygote form.

1st kind for k = 1, (22 – 12)2 + (2.2.1)2 = (22 + 12)2. (32 – 22)2 + (2.3.2)2 = (32 + 22)2. (42 – 32)2 + (2.4.3)2 = (42 + 32)2.
………………………………… for k = 9, (62 – 32)2 + (2.6.3)2 = (62 + 32)2. (72 – 42)2 + (2.7.4)2 = (72 + 42)2.
(82 – 52)2 + (2.8.5)2 = (82 + 52)2.
…………………………………
for k = 25, (92 – 42)2 + (2.9.4)2 = (92 + 42)2.
(102 – 52)2 + (2.10.5)2 = (102 + 52)2. (112 – 62)2 + (2.11.6)2 = (112 + 62)2.
…………………………………
2nd kind for k = 2, (2.4.1)2 + (42 - 12)2 = (42 + 12)2. (2.6.1)2 + (62 - 12)2 = (62 + 12)2. (2.8.1)2 + (82 - 12)2 = (82 + 12)2.
…………………………………

International Journal of Scientific & Engineering Research, Volume 4, Issue 8, August-2013

ISSN 2229-5518

1333

for k = 8, (2.5.2)2 + (52 - 22)2 = (52 + 22)2. (2.7.2)2 + (72 - 22)2 = (72 + 22)2. (2.9.2)2 + (92 - 22)2 = (92 + 22)2.
…………………………………
for k = 18, (2.8.3)2 + (82 - 32)2 = (82 + 32)2. (2.10.3)2 + (102 - 32)2 = (102 + 32)2. (2.12.3)2 + (122 - 32)2 = (122 + 32)2.
…………………………………
For common factor between f(xr) & φ(xr), it will be a composite set. So far prime set is concerned the number in
the form of 2(odd no.) cannot be the element of a N-eq. it comes as a composite set when both the zygote
elements are odd. After cancellation of c.f. it will follow a N-eq. of prime set

2 2 2

2 2 2 2 st

In zygote form of N-equation(a0 – b0 )
1) it is of 2nd kind.
+ (2.a0.b0)
= (a0 + b0 ) . if (a0/b0) < (√2 + 1) it is of 1
kind & if > (√2 +

10. Power Characteristics of three elements of a N-equation a2 + b2 = c2.

It is quite understood that all the three elements are individually capable of producing power and how they produce we can discuss one by one.
3rd Element
The prime numbers cp, p=1,2,3,….. obtained from c of a N-Eq. a2 + b2 = c2. are responsible to produce any composite values of c having the prime factors taken from cp only.
Let us consider two N-Equations,
(a1)2 + (b1)2 = (cp)2 = {(ap)2 + (bp)2}2 & (a2)2 + (b2)2 = (cq)2 = {(aq)2 + (bq)2}2 where cp & cq are prime numbers &
a,b stand for odd number & even number respectively.

Now let us take the identity {(ap)2 + (bp)2}{(aq)2 + (bq)2} = (apaq ± bpbq)2 + (apbq + aqbp)2
Say, = (α)2 + (β)2, [ This process can be said as Ns-operation.]
Now, with the help of (α)2 + (β)2, we can generate a new N-equation like
{α2 - β2}2 + (2αβ)2 ={α2 + β2}2 = (cpcq)2.
We will get two kinds of N-equation like

2 + β 2

= (cpcq)2
= α 2
+ β 2
i.e. of Double wings. For a composite no. of c it may have several wings
If ap = aq & bp = bq then we will get c = (cp)2 of single wing. Following the same treatment/Ns-operation with
(cp)2 versus (cp)2 we can generate the power of cp like 2n, n = 1,2,3,…… of single wing.
If aq = a1 & bq = b1 we will get c = (cp)3 of double wings, one of prime set & other of composite set having common factor (cp)2. For double wings of prime sets there must be Ns-operation in between cp & cq, obviously cp

≠ cq

∴ for aq = a1 & bq = b1 we have the N-equation by applying N-operation in between {(b2 - a2)2 + (2ab)2} & a2 +
b2 [i.e mother versus zygote for the N-equation {(b2 - a2)2 + (2ab)2 = (a2 + b2)2 or say, p2 + q2 = (a2 + b2)2 ]
[{b(3a2-b2)}2 – {a(3b2-a2)}2]2 + {2ab(3a2-b2)(3b2-a2)}2
= [{b(3a2-b2)}2 + {a(3b2-a2)}2]2 = [(a2 + b2)3]2 = (c3)2 ……… (1)no. The LHS can be expressed in terms of p,q also as (b2 - a2) = p & 2ab = q. It comes like
{p(3q2-p2)}2 + {q(3p2-q2)}2 = (c3)2 ……… (2)no.

Examples:

1st prime element of c is 5 (= 22 + 12) ∴ 32 + 42 = 52.
For c = 52, (42 – 32)2 + (2.4.3)2 = (42 + 32)2 or, 72 + 242 = (52)2.
For c = 53, 53 = 52.5 = (42 + 32)(22 + 12)
By N-operation 53 = (4.2+3.1)2 + (4.1-3.2)2 or, 53 = 112 + 22.
∴ required N-equation, (112 – 22)2 + (2.11.2)2 = (112 + 22)2 or, 1172 + 442 = (53)2.

Or, we can directly get the equation from (1) no where a=1, b=2 or from (2)no where p=3, q=4. For c = 5.13 i.e c = (12 + 22)(32 + 42) ∴ c = (2.2 ± 1.3)2 + (2.3 + 2.1)2.
or, c = (72 + 42) & (12 + 82) ∴ required N-equations are:
i) (82 – 12)2 + (2.8.1)2 = (82 + 12)2 or 162 + 632 = (65)2.
ii) (72 – 42)2 + (2.7.4)2 = (72 + 42)2 or 332 + 562 = (65)2.

International Journal of Scientific & Engineering Research, Volume 4, Issue 8, August-2013

ISSN 2229-5518

1334

∴ the generalized equation for cn against the N-eq. a2 + b2 = c2 = {a 2
following the same tedious successive N-operations as given below,
+ b 2}2
can be easily established by
{ ⎢an – nc2 an-2.b2 + nc4 an-4.b4 - nc6 an-6.b6 +……..⎪}2 + { ⎢nc1 an-1.b1 - nc3 an-3b3+ nc5 an-5.b5 - ……..⎪ }2 = (cn)2 =

n n n-2 2 n

n-4 4 n

n-6 6

2 n n-1 1 n

n-3 3 n

n-5 5 2 2

[{⎢a0
– c2 a0
.b0
+ c4a0
.b0
– c6 a0
.b0
+…….⎪}
+ { ⎢ c1 a0
.b0
- c3 a0
b0 + c5 a0
.b0
- ……..⎪} ]
However, we can prove the above mentioned relation by mathematical induction as we have proved Binomial
Theorem.
To express this relation fully in terms of a0 & b0, put a = a0 & b = b0 and replace n by 2n on LHS only and we get the equation like,

2n 2n

2n-2

2 2n

2n-4

4 2n

2n-6 6 2

{ ⎢a0 –
c2 a0
.b0 +
c4 a0
.b0 –
c6 a0
.b0
+………⎪} +

2n-1

1 2n

2n-3 3 2n

2n-5 5

2 n 2

n n n-2 2 n

n-4 4 n

n-6 6

{ ⎢2nc1 a0
.b0 -
c3 a0
b + c5 a0
.b0
- ……..⎪ }
= (c )
= [{⎢a0 –
c2 a0
.b0
+ c4 a0
.b0
– c6 a0
.b0

n-1 1 n

n-3 3 n

n-5 5 2 2

+…….⎪}2 + { ⎢nc1 a0
.b0
- c3 a0
b0 + c5 a0
.b0
- ……..⎪} ]
LH odd element
Let us consider product of two LH odd elements under zygote expression i.e. apaq = {(a1)2 - (b1)2}{(a2)2 - (b2)2} = (a1a2 ± b1b2)2 - (a1b2 ± a2b1)2
This operation can be said as Nd-operation.
∴ for a N-eq. a2 + b2 = c2, whose zygote form is (a 2
– b 2)2 + (2a b )2 =
(a 2
+ b 2)2
applying repeated Nd-operation over (a 2
– b 2) we get an equation where LH odd element produces
power.

n n

n-2 2 n

n-4 4 n

n-6 6 2

i.e. [{a0
+ c2 a0
.b0
+ c4 a0
.b0
+ c6 a0
.b0
+…….} –

n-1 1 n

n-3 3 n

n-5 5 2 2

{ nc1 a0
.b0
+ c3 a0
b0 + c5 a0
.b0
+ ……..} ] ±

2n-1

1 2n

2n-3

3 2n

2n-5 5 2

{ 2nc1 a0
.b0 +
c3 a0
b0 +
c5 a0
.b0
+ …….. } =

2n 2n

2n-2

2 2n

2n-4

4 2n

2n-6 6 2

± {a0

+ c2 a0
.b0 +
c4 a0
.b0 +
c6 a0
.b0
+…….}
As Nd & Ns operations over two zygote elements cannot run simultaneously, the odd elements on either side of a
N-equation cannot produce power more than two simultaneously.
If possible, say there exists an equation like an + b2 = cm where a & c are the odd elements & b is even.

2 2 2 2

So far Nd operation over ‘a’ is concerned the previous relation is an/2 + b1
= c1
i.e. b = 2.b1 c1

2 2 m/2 2 2

So far Ns operation over ‘c’ is concerned the previous relation is a1
+ b1 = c
i.e. b = 2.a1 b2

2 2 2 2

But 2.b1 c1
≠ 2.a1 b2
Or, by forward movement, consider the eq. a2 + b2 = c2 where a & c are in place of odd & b is in place of even integer. When ‘a’ will produce power first even element will be 2b2c2 & when c will produce power first even element will be 2a2b2. If these two elements are equal then b = 0 & a = c & the equation does not stand.
Hence the existence of two odd elements in power form is absurd.
LH odd element produces even power by Nd operation in between two main elements & produces odd power by Nd operation in between main element & its zygote elements. Same thing is true for RH element under Ns operation.
*zygote odd element is the powerless element of a N-eq. which can be expressed as a 2

– b 2
(for LHS) or a 2 +
b0 (for RHS) where a0,bo are integers
Even Element of N-Eq.
When the 3rd element of a N-eq. produces power the even element also cannot produce power.
We have observed that when 3rd element produces power n, Even element = 2.{LH odd element of Ns operation for power n/2}.{Even element of Ns operation for power n/2} = 2. A.B (say) where the odd element A is always
powerless and there is no common factor in between A & B. Hence, 2AB cannot be in power form.
When the LH odd element produces power then also even element cannot produce power,
We have, Even element of Nd(for power n) = 2.{ RH odd element of Nd(for power n/2)}. {Even element of Nd(for power n/2)} = 2CB (say), where C is powerless and there is no common factor in between C & B. Hence, 2CB cannot produce power.

International Journal of Scientific & Engineering Research, Volume 4, Issue 8, August-2013

ISSN 2229-5518

1335

11. Two exceptional cases.

The property of unity whose power can be assumed as any positive integer, produces two exceptional relations.

2 2 2

2 2 2 2 3 2

For the N-Equation in zygote form, (a0
– b0 ) + (2a0b0)
= (a0
+ b0 ) , if a0 = √2 & b0 = 1 it is 1 + 2
= 3 . Again
by Ns operation where a0 = 23/2 & b0 = 1, we get 72 + 25 = 34.
Barring the above two exceptional cases, we cannot have any equation like
ax + by = cz for no common factor among a, b, c where more than one element of bases can produce power more than two.
It implies the fact that Beal equation (ax + by = cz) cannot exist if (a,b,c) is a prime set i.e. there is no common factor among a, b, c.

12. How Beal Equation forms:

The N-equation a2 + b2 = c2 where a, c are odd elements & b is even elements can be redefined in power form as an + b2 = c2 where a, b, c are powerless & n = 2, 3, 4, ……
a2 + b2 = cn where a, b, c are powerless & n = 2, 3, 4, ……
a2 + bn = c2 where a, b, c are powerless & n = 2, 3, 4, …… Exceptional cases, 1n + 23 = 32 & 72 + 25 = 34.
Three important properties of N-Eq. a2 + b2 = c2
1. If least element is in the form of p√q other elements must be integers, difference being followed by same k-rule ( either (2n – 1)2 or 2n2 form). If the greatest element is in the form of p√q, other elements must be integer.
2. 1 cannot be the element of a N-equation.
3. Only one term of N-eq. is in power form greater than 2.
Let us say the BN-equation which is obtained by modifying the Beal equation as (ax/2)2 + (by/2)2 = (cz/2)2 and after cancellation of common factor the modified equation can be said as BN-eq(irrational) or IBN-equation. IBN-equation must differ by the above three properties of N-equation.
This implies that INB-equation can be of following six types.
1. a2 + b2 = (p√q)2
2. a2 + (α√β)2 = (p√q)2
3. (γ√δ)2 + (α√β)2 = (p√q)2
4. 1 + (α√β)2 = p2
5. a2 + (α√β)2 = p2 where for a ≠ 1, (p – a) is not in the form of 2n2 or
(2n – 1)2, n = 1,2,3,…….
6. (γ√δ)2 + (α√β)2 = p2
Type-1 can be differed from N-eq. if any one of a & b or both are equal to 1. If both are one, obviously 1 + 1 = (√2)2 & choosing the c.f. as 2n it is 2n + 2n = 2n+1. Here the INB prime set is fixed i.e. (1, 1, √2).
1 + b2 = (p√q)2 cannot exist. Because (p√q) is always in the form of √{2(odd no.)} & cannot be expressed as
p√p so that common multiplier can be chosen for the INB set (1, b, p√q) to produce Beal equation.
For the INB set (1, α√β, p) no common multiplier can be chosen unless α = β
Say, the INB prime set is in the form of (1, α√α, β).

choose the common multiplier (βm+n√β), where m,n are (+) integers.

------------

∴ α√α.βm+n√β = αβm.√αβ2n+1 ∴ αβm = αβ2n+1 or, m = 2n + 1

∴ for m = 2n+1, Beal equation follows, (1)2 + (α√α)2 = (β)2.
or, (βm+n√β)2 + {α√α.βmβ(2n+1)/2}2 = (βm+n√β.β)2.

or, (βm+n+1/2)2 + (βm+n+1/2α3/2)2 = (βm+n+3/2)2. or, β6n+3 + (αβ2n+1)3 = β6n+5

International Journal of Scientific & Engineering Research, Volume 4, Issue 8, August-2013

ISSN 2229-5518

1336

e.g. for the irrational prime set (1, 2√2, 3), put n = 1,2,3,……..
∴ 39 + 543 = 311; 315 + 4863 = 317 & so on.
Here also the INB-prime set is fixed i.e. (1, 2√2, 3)
When the INB prime set is in the form of (1,√α, β√β), choosing the common factor αm+n we have (αm+n)2 + (αm+n√α)2 = (αm+n β√β)2.

Now, αm+n β√β = αm β√(α2nβ) or, αm β = α2n β ∴ m = 2n.

∴ (α3n)2 + (α3n+1/2)2 = (α3nβ3/2)2. or, (α)6n + (α)6n+1 = (α2nβ)3.
e.g. we have the irrational prime set (1, √7, 2√2)
∴ for n = 1, 76 + 77 = 983. for n = 2, 712 + 713 = 48023. & so on.
Here the existence of INB prime set is infinite.
For INB prime set (√α, β√β, δ√δ), choose the common factor α√α
∴ α4 + (αβ)3 = (αδ)3. e.g. we have the prime set (√19, 2√2, 3√3), choose the common factor 19√19.
∴ 194 + 383 = 573.
Here the existence of INB prime set is infinite.

For the INB prime set like {αp, βn√(βδ), θq}, where p,q > 1, choose the common factor δn where n is divisible by p,q.

∴ (αpδn)2 + {(βδ)n. √(βδ)}2 = (θqδn)2, & it can produce a solution of Beal eq. e.g. we have the prime set (32, 22√(2.17), 52)
∴ (32.172)2 + {22.172√(2.17)}2 = (52.172)2. or, 514 + 345 = 854.
Here the existence of INB prime set is infinite.

13. Theory behind the formation of Beal Equation.

A prime set of three numbers (a, b, c) where a + b = c can produce a INB-prime set to produce a Beal equation after choosing a common multiplier provided one of the numbers of a, b & the number c are in power form or all the three numbers are in power form which is only for an exceptional case
1 + 8 = 9.
If a or b ≠ 1, then the powerless term must have at least one factor with power greater than 3, say b = βnγ =
βn – r.βrγ where n > 3, n – r > 3. For the corresponding other factor α = βrγ, αp – 1 will be common multiplier to produce Beal equation where GCF of (p – 1) with power of a, c separately & that of p with (n – r) will be greater
than 3.
If a or b = 1, say b = 1, then for the powerless term, say b, bn will be CM where GCF of n with power of c ≥ 3 i.e.
β = 1. It is true for a = 1 = b also.
In more simplified way we can say, for any two numbers A & B ( > 1) which are prime to each other if it is found Am ± Bn = γα .β where α > 3, then select a number p from 4,5,6 ….. , α such that GCF of p & α, (p – 1) & m, (p – 1) & n all are ≥ 3. If p exists then Beal eq. will exist with common multiplier βp – 1
Obviously, to produce Beal equation m, n, α cannot be all prime.
For m ≠ n; m, n cannot be prime.
If α is prime; m or n cannot be prime.
As the consecutive nos. p & (p – 1) do not have any common factor in between them, hence for any Beal equation Ax + By = Cz, (x, y, z) is a prime set i.e. no common factor lies among x, y, z. In between two there can be a common factor. This implies all the powers cannot be even.

14. Few examples for the existence of Am ± Bn = γα .β where α > 3, γ ≥ 1

Any even number can be expressed as N = 2n.p where p is an odd integer & n is an integer known as degree of intensity of N i.e. D(N) = n.
Say, N = Am – Bn where A & B are two odd nos. ≥ 3 & prime to each other and m, n both are odd.
⇒ N = (1 + e1)m – (1 + e2)n
After Binomial expansion, N = e1(an odd integer) – e2(an odd integer)
Say, e1 = 2p.o1 & e2 = 2q.o2 where o1 & o2 are odd integers.

International Journal of Scientific & Engineering Research, Volume 4, Issue 8, August-2013

ISSN 2229-5518

1337

⇒ if p ≠ q, N = 2α.(an odd no.) where α = Min (p, q) & to create Beal eq. α > 3
For p = q = α,

2 n 2 α m 3 n

3 2α

N = 2α[(mo1 – no2) + (mc2.o1 –
c2.o2 )2
+ ( c3.o1 –
c3.o2 ).2
+ …….. ]
= 2α + λ.(an odd integer) where λmin = 1. So α > 2
If D(mo1 – no2) ≥ 2 then α > 1.
For m, n both are even,

2 n 2

N = (me1 – ne2) + (mc2e1 –
c2e2 ) + ………

⇒ D(N) > Min (p, q) + 1 ⇒ to create Beal eq. Min( p, q ) > 2

Say m, n are combination of even & odd.
For p = q = α, N = 2α.(an odd integer). Hence, to create B eq. α > 3
For p ≠ q, Min(D(me1), D(ne2)) > 3
Say, A & B are even & for any integer values of m, n (>3),
Say, N = (1 + aαg)m – (1 + aαh)n = aα(an even no.), after binomial expansion. To create Beal eq. α > 3
Say, A is even & B is odd & m, n are any integers > 3
Say, N = (1 + aαg)m – (1+ 2k aαh)n = aα(an odd no), after binomial expansion. To create Beal eq. α > 3
For a particular acceptable value of α there can be infinite no. of sets (m, n).
For the INB-eq. a + b = c, any two number taken from each side must be in power form. Say, a & c (both > 1) are in extreme power form with bases a1 & c1. Then there must be a common factor among (a1 – 1), b, (c1 – 1) in the form of θk where θ = 1, 2, 3, 4,.…. & k > 3 [θ = 1 for γ = 1]. If not, it is to be understood that a, c are not
prime to each other. Already there exists a common factor γ.
Let consider the case N = Am + Bn where A & B are any two odd integers ( > 1) & m, n are any two integers ≥ 3
Say, A = 1 + γα.I1 & B = 1 + γα.I2
⇒ N = (1 + γα.I1 )m + (1 + γα.I2)n

2 n 2 2α

= 2[ 1 + (mI1 + nI2)γα + (mc2I1
= 2[ 1 + γα.P]
+ c2I2 )γ
+ …….. ]
Now γα.P & (1 + γα.p) two consecutive nos. cannot have a common factor γα.

⇒ N = 2(an odd no.) & D(N) = 1

In spite of considering common factor γα result is free from γα. Hence, it is not capable of producing Beal eq. for
γ ≠ 1.

p1 p2

Here the result 2(odd no.) i.e. 2.o1
(* subject to Proof)
.o2
…..has the limitation p1, p2, p3,…≤ 3
Say, A is odd & B is even.
N = (1 + e1)m + (1 + o1)n where e1 = γα.e2 & o1 = γα.o2, obviously γ is odd.

2 n 2 2α

= 2 + (mc1.e2 + nc1.o2)γα + (mc2.e2
+ c2.o2 )γ
+ ……..
= 2 + γα.P ⇒ N does not have a c.f. γα as two consecutive odd nos. cannot have a common factor.
Hence, it is also not capable of producing any Beal eq. for γ ≠ 1.
Let us analyze the case Am – Bn = γαβ = 1where A is odd & B is even integer
Here γ = 1 = β & α can be any integer & obviously for m = n it is not possible i.e. m ≠ n.
⇒ common multiplier is βp – 1 where p – 1 can be any integer which on multiplication on both sides there will be
GCF of (p – 1) with m & n separately ≥ 3 It is not possible for m, n ≥ 3, so far Beal eq. formation is concerned
at γ = 1.
Now consider the case, either m or n = 2 & other ≥ 3 where there exists a unique solution i.e. 32–23 = 1
In the above all we have discussed considering the fact that A, B are prime to each other. If there is a common factor in between A & B, then it is always possible to have a relation like Am ± Bn = γαβ.
[ it is of the type of INB-Eq. (γ√δ)2 + (α√β)2 = (p√q)2 ]

International Journal of Scientific & Engineering Research, Volume 4, Issue 8, August-2013

ISSN 2229-5518

1338

Conclusion

Whether my proof is correct or wrong it needs to be examined by an expert number-theorist and after all it should be accepted by all mathematical communities.
The theory by which Beal Equation is proved as I believe, also reveals the existences of two type of prime numbers. Some prime numbers can satisfy the RH odd element of N-equation & the rest cannot. First one can produce a relation a2 + b2 but second one cannot.
I personally believe that from this theory there is a possibility of finding out so many properties of prime numbers.

References

Books

[1] Academic text books of Algebra.