11 A CONSTRUCTIVE PROOF OF BEALβS CONJECTURE
Nicholas J. Daras
Hellenic Military Academy, Department of Mathematics & Engineering Sciences, 16673 Vari Attikis, Greece
E-mail: [email protected]
Abstract.
We prove that there is no non-trivial integral positive solution to the generalized Fermat equation.
Keywords:
Generalized Fermat equation, greatest common divisor, Dirichletβs Theorem on arithmetic progressions, Bezoutβs Identity
Classification code MSC:
In the early twentieth century, David Hilbert presented twenty-three great mathematical problems which featured main directions of scientific research throughout the period that followed ([3, 4]).
By analogy, in 2016, John F. Nash and Michael Th. Rassias gave a list with seventeen, currently unsolved problems in modern mathematics, in the belief that these problems are expected to determine several of the main research directions at least during the beginning of the XXI century ([7]). The third problem in the series of this list refers to the exploration of the possibility of extending Fermat's Last Theorem. Even before Andrew Wiles announced his famous proof of this Theorem ([8]), various generalizations had already been considered, to equations of the shape π΄π₯ π + π΅π¦ π = πΆπ§ π , for fixed integers π΄ , π΅ and πΆ . In this direction, the Theorem of Henri Darmon and Andrew Granville states that if π΄, π΅, πΆ, π, π πππ π are fixed positive integers with π β1 + π β1 + π β1 < 1 , the equation π΄π₯ π +π΅π¦ π = πΆπ§ π has at most finitely many solutions in coprime non-zero integers x ; y and z ([2]). However, as is made clear through those mentioned by Michael A. Bennett, Imin Chen, Sander R. Dahmen and Soroosh Yazdani ([1, 7]), except the solutions identified by Preda MihΔilescu in the Catalan equation ([6]) and the solutions derived from some elementary numerical identities, in most cases, there is no non-trivial solution of this equation once we assume that
A = B = C = 1 . In this connection,
Bealβs Conjecture argues that if min{p, q, r} β₯ 3 , there are no non-trivial co-prime integral solutions to the generalized Fermat equation π₯ π + π¦ π = π§ π ([2, 7]). So far, many computational attempts have produced strong indications that this conjecture may be correct ([5]). The aim of this paper is to give a proof of this Conjecture. To this end, let us suppose that, on the contrary, Beal's Conjecture does not apply. This is equivalent to assuming that there are integer exponents π β₯ 3 , π β₯ 3 and π β₯ 3 and three coprime positive integers π₯ = ππ£ππ , π¦ = πππ and π§ = πππ satisfying a generalized Fermat equation of the form π₯ π Β± π¦ π = π§ π . Given any π , π , π in β€ + , π = πππ , and π β β€ + β {1} , put π π,π : = βπ¦ ππβ4 π(2π + π) + π , with π : = ππ¦ πβ2 , π = πππ . By adopting this definition, it is clear that π π,π is an integer and the following apply. Proposition 1 . There exist π β , π β β β€ + , for which the corresponding expression π§ β π¦ π π β ,π β represents a prime number. Proof . Since πππ(βπ¦ ππβ2 , β[π§ β π π¦ ]) = 1 , an application of Dirichletβs Theorem on arithmetic progressions, in its basic form, shows that there are infinitely many primes such that π β‘ β(π§ β π π¦ ) mod (βπ¦ ππβ2 ) . In particular, there exist infinitely many natural numbers π such that the integers βπ¦ ππβ2 π + (π§ β π π¦ ) are primes π . Since π , π¦ and π§ are odds, the numbers π are also odd and, therefore, they can be represented as differences of two squares ([π + 1] 2β ) and ([π β 1] 2β ) . Taking such an π and π β β€ + , π β β€ + , π β β€ + so that π + 1 = 2ππ§ π(πβ2) and π β 1 =2π, we see that π¦ ππβ2 {[ππ§ π(πβ2) ] β π } = Β±[π§ β π π¦ ] β π for a prime π and, therefore, if π β =ππ§ π(πβ2) , the integer π β = βππ§ π(πβ2) Β± π will be a root of the equation (π¦ ππβ2 )π β2 +(2π β π¦ ππβ2 )π β + [Β±(π§ β π π¦ ) β π] = 0 . We infer (Β±π¦ ππβ2 )π β2 + (Β±2π β π¦ ππβ2 )π β + (π§ βπ π¦ ) β π = π§ β π¦ π π β ,π β β π = 0 , and, thus, the proof of Proposition 1 is completed. Proposition 2 . We have π π β ,π β π§ πβ2 β π¦ πβ2 > 1 and πππ(π₯ πβ2 , π π β ,π β π§ πβ2 β π¦ πβ2 ) = 1 . Proof . Let π β and π β be as in Proposition 1. Because of this choice, it is easily verified that π π β ,π β π§ πβ2 β π¦ πβ2 > 1 . For the rest of the proof . To get a contradiction, suppose that πππ(π₯ πβ2 , π π β ,π β π§ πβ2 β π¦ πβ2 ) > 1 . This means that there were a natural number π > 1 such that π₯ πβ2 = ππ and π π β ,π β π§ πβ2 β π¦ πβ2 = ππ for some integers π and π . Multiplication of the first equation by π₯ , gives π₯ π = π(π π₯ ) , (1) while multiplication of the second one by π¦ and π§ gives π¦ π π β ,π β π§ πβ2 β π¦ π = π(π π¦ ) β Β±π¦ π = Β±π¦ π π β ,π β π§ πβ2 β π(π π¦ ) (2) and π§ π π β ,π β π§ πβ2 β π§ π¦ πβ2 = π(π π§ ) β π π β ,π β π§ π = π§ π¦ πβ2 + π(π π§ ) , (3) respectively. Adding (2) to (1), we obtain π₯ π Β± π¦ π = π(π π₯ β π π¦ ) Β± π¦ π π β ,π β π§ πβ2 or, by hypothesis, π§ π = π(π π₯ β π π¦ ) Β± π¦ π π β ,π β π§ πβ2 which can equivalently be written as π§ πβ2 (π§ β π¦ π π β ,π β ) = π(π π₯ β π π¦ ) . (4) Similarly, subtracting (3) from the multiple of (1) by π π β ,π β , we get π π β ,π β (π₯ π β π§ π ) =π(π π β ,π β π π₯ β π π§ ) β π§ π¦ πβ2 or, by hypothesis, β π π β ,π β π¦ π = π(π π β ,π β π π₯ β π π§ ) β π§ π¦ πβ2 which can equivalently be written as π¦ πβ2 (π§ βπ π β ,π β π¦ ) = π(π π β ,π β π π₯ β π π§ ) . (5) Having regard to the Proposition 1, from the relations (4) and (5), it follows that π > 1 divides both π§ πβ2 and π¦ πβ2 , which contradicts the statement that πππ(π₯, π¦, π§) = 1 under the condition that π₯ π Β± π¦ π = π§ π . So, it is impossible to hold πππ(π₯ πβ2 , π π β ,π β π§ πβ2 β π¦ πβ2 ) > 1 . Hence, πππ(π₯ πβ2 , π π β ,π β π§ πβ2 β π¦ πβ2 ) = 1 , and the proof of Proposition 2 is complete. We are now in position to give the main result of the paper. Theorem.
There is no non-trivial solution to the generalized Fermat equation π₯ π + π¦ π = π§ π , once we assume that πππ{π, π, π} β₯ 3 and πππ(π₯, π¦, π§) = 1 . Proof.
Let π β and π β be as in Proposition 1. Let also
π, π, πΜ, πΜ β β€ β {0} with
ππΜ β πΜπ β 0 . Then, the system β¦ (π)π + (βπ)π + (π π β ,π β π)π = Β±1(πΜ)π + (βπΜ)π + (π π β ,π β πΜ)π = Β±1(π₯ )π + (Β±π¦ )π + (βπ§ )π = 0 β§ with determinant (ππΜ β πΜπ)(π§ β π¦ π π β ,π β ) β 0 has unique solution π = β πβπΜππΜβπΜπ , π = β (πβπΜ)π§ +(πβπΜ)π πβ,πβ π₯ (ππΜβπΜπ)(π§ βπ¦ π πβ,πβ ) and π = β (πβπΜ)π¦ Β±(πβπΜ)π₯ (ππΜβπΜπ)(π§ βπ¦ π πβ,πβ ) . From Proposition 2, it holds true πππ(π₯ πβ2 , π π β ,π β π§ πβ2 β π¦ πβ2 ) = 1 , so an application of Bezoutβs Identity shows that the equation (π₯ πβ2 )π’ + (π π β ,π β π§ πβ2 )π£ + (βπ¦ πβ2 )π£ = Β±1 can be solved in β€ . Given any partial integer solution (π’, π£) β β€ of this equation and π, β β β€ , π β β , let us take the induced solutions π, π, πΜ, πΜ β β€ β {0} defined by
π: = π’ + π(π π β ,π β π§ πβ2 β π¦ πβ2 ) , π: = π£ β ππ₯ πβ2 , πΜ: = π’ + β(π π β ,π β π§ πβ2 β π¦ πβ2 ) and πΜ : = π£ β βπ₯ πβ2 . Observe that
π, π, πΜ, πΜ β 0 . Further, since (ππΜ β πΜπ) = Β±(π β β) β 0 and it is easily seen that the unique solution of the system is exactly (π, π, π) = (π₯ πβ2 , π¦ πβ2 , π§ πβ2 ) . But, on the other hand, considering the above-mentioned expression of the unique solution, we get π = π₯ πβ2 , π =(βπ§ )π· + (π₯ π π π β ,π β )πΈ and π = (βπ¦ )π· + (π₯ π )πΈπ₯ π , where we have used the notation π· β π πβ,πβ π§ πβ2 βπ¦ πβ2 π§ β π¦ π πβ,πβ and πΈ = β π¦ π πβ,πβ . Thus, especially in this case, with this formulation, we found β¦ π¦ πβ2 = (βπ§ )π· + (π₯ π π π β ,π β )πΈπ§ πβ2 = (βπ¦ )π· + (π₯ π )πΈ β§ . Since π₯ π Β± π¦ π = π§ π , this formulation guarantees that the system of equations β¦ (1 β π§ πΈ)π Β± (π¦ πΈ)π = βπ¦ π·(Β±π§ π π β ,π β πΈ)π + (βπ¦ π π β ,π β πΈ β 1)π = Β±π§ π· β§ would have a solution at (π, π) = (π§ πβ2 , π¦ πβ2 ) , which is impossible, since the determinant of the system equals zero and both lines representing it must in any case not be identical, otherwise we would have π π β ,π β = 0 and π§ = βπ¦ . We therefore conclude that there is no non-trivial solution to the generalized Fermat equation π₯ π Β± π¦ π = π§ π and, thus, the proof is complete. We finish by giving two immediate consequences. Corollary 1 (Fermatβs Last Theorem) T here is no non-trivial integer solution to the Fermat Equation π₯ π + π¦ π = π§ π provided that π β₯ 3 and πππ(π₯, π¦, π§) = 1 . Corollary 2 (Irrationality of Binomial Expansions) If π, π, π β₯ 3 , there is no (π₯, π¦) β β€ with πππ(π₯, π¦) = 1 and satisfying (π₯ π + π¦ π ) β β . Proof . Suppose there is a rational number (π πβ ) β β , with
π, π β β€ and πππ(π, π) =πππ(π₯, π¦, π) = 1 , satisfying π₯ π + π¦ π = (π πβ ) π . This implies that π π = π π (π₯ π + π¦ π ) . Since πππ(π, π) = 1 , we infer π π β (π₯ π + π¦ π ) , which means that there is a π β β€ satisfying (π₯ π + π¦ π ) =ππ π . It follows that π π = π π ππ π , which guarantees that π π π = 1 . But, this only applies if π = π =1 . In such a case, we would have π₯ π + π¦ π = π π and also πππ(π₯, π¦, π) = 1 , which, in view of the Theorem above, is impossible to hold true. References [
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