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BEALE'S CONJECTURE REFUTATION USING PYTHAGOREAN THEOREM.

Sergey P. Klykov1,2

Fermentation expert, Poste Restante, Obolensk, Serpukhov district, Moscow region, Russia,

142279.

1Corresponding author.

2Alpha-Integrum, Ltd.

Email address: smlk03@mail.ru

ABSTRACT.

Based on the previously obtained proof of the Fermat's Last Theorem using the Pythagorean Theorem, it is shown, that the

equations of Beal's Conjecture are a kind of equations of the Pythagorean Theorem too: ap+bq =cr→ a(p/2)2+b(q/2)2

=c(r/2)2→A2+B2=C2, with integers a, b, c, p, q, r, and p,q,r >2, where are A= ap/2, B= bq/2 , C= cr/2. Since only Pythagorean Triples

are a necessary object for the smooth implementation of any integer solutions in the Pythagorean equations, the following

conclusion was made: only non-primitive Pythagorean Triples could serve as the main object for analysis in the Beal's Conjecture

case. However, this assumption did not give a positive result / confirmation of the hypothesis. Thus, the existence of at least one

irrational term, for example, A= ap/2, makes it possible to conclude that the numbers a, b, c are mutually prime numbers, which

excludes the possibility of Beal's Conjecture being true for the case 1/a+1/b+1/c<1. The case 1/a+1/b+1/c>1 is considered

separately, when conditions close to Beal's Conjecture make it possible to prove the assumption at a=1, b=1, c=√2.

Key words: Diophantine equations; Beal’s Conjecture; Fermat's Last theorem; Pythagorean

Theorem; Fermat's Last Theorem proof.

INTRODUCTION.

In [1], Fermat's Last Theorem, FLT, was proved using the Pythagorean Theorem, PT. This

article shows that the FLT equation, an+bn =cn and the equation derived from it, (bc)-n+(ac)-n

=(ab)-n, are showing different numbers of possible rational and irrational solutions. Taking into

account the proof of the PT and the previously published symmetry for the rational solutions for

“n” and “–n” values of the PT, the shortest solution for the FLT was shown. It was based on

point of view, that the FLT is the same thing as the PT by the way: an+bn =cn→

a(n/2)2+b(n/2)2=c(n/2)2→A2+B2=C2. Indeed, an elementary scheme for different “n” and “-n” values

can be presented to show symmetry and asymmetry between possible rational and irrational

solutions in various equations with degrees “n” and “–n”:

an+bn =cn→(1/a)-n+(1/b)-n=(1/c)-n→(abc/a)-n+(abc/b)-n=(abc/c)-n→(bc)-n+(ac)-n=(ab)-n, (1).

Based on Scheme (1), it is easy to determine that only for n = 2 and n = -2 there is symmetry in

the number of rational solutions for the first and last equations of this Scheme (1), i.e. all

solutions are rational.

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If there is at least one irrational solution in the first equation for n greater than 2, then the last

equation of Scheme (1) gives 2 irrational solutions. Two irrational solutions of the first equation

gives 3 irrational solutions in the last equation. That is, for n greater than 2, there is an

asymmetry in the number of rational and irrational solutions.

If consider Scheme (2):

an+bn =cn→ a(n/2)2+b(n/2)2 =c(n/2)2→A2+B2=C2, where A=an/2, B=bn/2, C=cn/2, (2).

The properties of Pythagorean triples, PTs, are known in an exhaustive way, and for this reason,

both of the above Schemes prove FLT in such a way that the FLT equations are part of the PT

equations and vice versa. That is,

FLT=PT, (3).

It follows from the above that all FLT equations are algebraic descriptions of such geometric

figures as rectangular triangles. Naturally, even for n = 2 (PT conditions), at least one or more

irrational solution(s) can be used. Fig.1. shows an example of describing a right-angled triangle

using PT and using FLT for an arbitrary sum of numbers 8+64=72:

PT equation (2√2)2+82=(3×23/2)2, (4),

FLT equation 23+43=(2×32/3)3, (5).

Figure 1. A right-angled triangle of light green color has a leg at the base with a length

expressed by an irrational number, which is numerically equal to two hypotenuses from two unit

right-angled triangles of dark green color with a=1, b=1, c=√2.

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The above method of proof for the FLT has also been successfully tested for proving similar

Diophantine equations with a large number of terms, than the FLT. The results are being

prepared for publication.

Beal’s Conjecture,[2], postulates that: If ap+bq=cr , where a, b, c, p, q, and r are non-zero

integers with p, q, r ≥ 3, then a, b, and c have a common prime factor.

The Fermat-Catalan Conjecture and related problems, [3]: There are only finitely many triples of

coprime integer powers xp; yq; zr for which xp+yq=zr with 1/x+1/y+1/z< 1. The article contains a

set of examples that show the absence of common factors. Many other works are devoted to

these problems, for example, [4]. The article [5] also can be considered separately, because all

examples from this article can be reduced to the sum of 1+1=2 and 1/x+1/y+1/z>1.

For example: 327+1285=646=236=235+235; 10247+12810=271=270+270, etc.

All of the works noted above and many not mentioned cannot be considered as a proof/refutation

of Beal’s Conjecture.

PROOF (REFUTATION): The equation ap+bq=cr, which was presented in the form

a(p/2)2+b(q/2)2=c(r/2)2→A2+B2=C2, with integers a, b, c, p, q, r and p,q,r >2, where are A= ap/2,

B= bq/2 , C= cr/2, have no PTs as integer solutions. This would be possible, for example, in the

case of p = q = r = 2, but this contradicts the Beal’s Conjecture conditions. Therefore, at least one

number, A or B, or C, may be irrational. In general, all three numbers can be irrational. For this

reason, the symmetry of rational solutions in the equations A2+B2 =C2 against

(BC)-2+(AC)-2=(AB)-2 will be broken, which excludes the presence of "a common prime

factor". Example 1: Let's the C value is an irrational number. Then the equation (BC)-2+(AC)-2=(AB)-2 will

have two irrational terms inside the brackets on the left, (…), and one rational on the right, and it is impossible.

Example2: Let’s The B (or A) value is an irrational number. Then the equation (BC) -2+(AC)-2=(AB) -2 will have

one irrational term and one rational inside the brackets on the left, (...), and one irrational on the right, and it is

impossible. And so on.

CONCLUSION: Thus, when n = 2, symmetry ensures the presence of integers only for PTs,

and other cases provide only irrational solutions in the absence of symmetry, which prohibits

"a common prime factor" and refutes Beal's Conjecture, so Beal’s Conjecture=PT.

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REFERENCE.

1. Sergey P. Klykov (2021) On the issue of the relationship between Fermat’s Last

Theorem, fl t, and the Pythagorean Theorem. DOI: 10.13140/RG.2.2.36466.02243 Link:

https://bit.ly/3wxriRW

2. "Beal Conjecture". American Mathematical Society, Link:

https://www.ams.org/profession/prizes-awards/ams-supported/beal-conjecture and

https://www.ams.org/profession/prizes-awards/ams-supported/beal-conjecture-mauldin-

letter-to-editor.pdf

3. R. Daniel Mauldin. A Generalization of Fermat’s Last Theorem: The Beal Conjecture

and Prize Problem (Engl.)//Notices of the AMS. — 1985. — Vol. 44, no. 11. — P.

1436—1437. Link: http://www.ams.org/notices/199711/beal.pdf

4. Poonen, Bjorn; Schaefer, Edward F.; Stoll, Michael (2005). "Twists of X(7) and

primitive solutions to x2 + y3 = z7". Duke Mathematical Journal. 137: 103–158. Link:

https://arxiv.org/pdf/math/0508174.pdf

5. ГИПОТЕЗА БИЛА. ДОКАЗАТЕЛЬСТВО. ЯКОВЛЕВА О.Н., журнал

УСПЕХИ СОВРЕМЕННОЙ НАУКИ И ОБРАЗОВАНИЯ, Том: 2,

Номер: 9, 2016, Страницы: 159-160, (Beal’s Conjecture. Proof., O.N.

Jakovleva , SUCCESSES OF MODERN SCIENCE AND EDUCATION

journal, V.2, №9, 2016, P.159-160). (The article was found through the web

resource: https://elibrary.ru/contents.asp?id=34267157 )

June, 17, 2021.