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Diophantine Geometry in Space And TEM Journal. Volume 8, Issue 1, Pages 78-81, ISSN 2217-8309, DOI: 10.18421/TEM81-10, February 2019. Diophantine Geometry in Space and 1 2 Viliam Ďuriš , Timotej Šumný 푬ퟐ 푬ퟑ 1Department of Mathematics, Constantine The Philosopher University in Nitra, Slovakia Abstract – In this paper, we are focusing on solving influenced the development of other parts of Diophantine equations with geometry. Integers that are mathematics. The fundamental link between algebra solutions of Diophantine equations are represented in and geometry occurred in the work “Geometry”, space by a lattice. We will use the fact that equations written by René Descartes. In the work, Descartes can be interpreted geometrically as curves or surfaces. introduces the coordinate system and thus builds the To determine the properties of the solutions, we will basics of analytical geometry. In addition to this, the use tangents (of the surface). mathematical field of “number theory” became Keywords – Geometry, Equations, Space, Surface, independent. The Diophantine equations had a major Solutions. influence on this area. However, the Diophantine equations were not distant from geometry, because 1. Introduction many special problems in geometry were directed towards the synthesis of the Diophantine equation, In this paper, we are focusing on using geometry and so these mathematical domains are to solve Diophantine equations. The use of geometry interconnected. It was from this connection that work in number theory tasks was among the primary was being done to unify Diophantine equations and solutions. The reason for this was that the geometric geometry. The connection between geometry and view often offered a clear view of the problem Diophantine equations was gradually crystallized solved, and the precision of geometric methods acted with the introduction of a discrete space. The discrete as an intuitive and logical procedure. Many processes space determined the basic properties of the solutions and types of problem solving from number theory via of Diophantine equations and thus provided the basic geometry are written in the Euclidean work conditions for the solution sought. This resulted in “Fundamentals”. However, many tasks turned out to one of the most elaborate works in the field of theory be difficult to solve using geometry, which prompted of Diophantine geometry, from the hands of the emergence of a new division of mathematics Hermann Minkowski. In his work, he used lattice called “Algebra”. This division of mathematics has points, which represented whole numbers, to move to evolved independently and has brought about a a discrete space. He characterizes the lattice as a number of discoveries in equation solutions and in discrete topology inserted into the Euclidean space. the research of mathematical structures that have Here we apply the idea of the lattice to퐿 the Diophantine equation in the form of ( , ) = in the space , so the Diophantine equation is DOI: 10.18421/TEM81-10 represented here as a curve. Similarly,푓 we푥 푦can apply푐 https://dx.doi.org/10.18421/TEM81-10 the idea to th퐸e2 equation in the form of ( , , ) = for , which can be reformulated to the equation Corresponding author: Viliam Ďuriš, Timotej Šumný, 푓 푥 푦( 푧 ) 푐 Department of Mathematics, Constantine The Philosopher using the assumed solution in the form of , = 퐸3 University in Nitra, Slovakia for . More about Minkowski’s number geometry theory can be found in the reference section푓 [1].푥 푦 We Email: [email protected] 2 use푑 the퐸 lattice topology in the article, determining the Received: 27 September 2018. properties of the solution for the Diophantine Accepted: 28 January 2019. equation [2], [3], [4], [5] by means of lattice points Published: 27 February 2019. and a tangent. We apply this method using a tangent and lattice points to familiar and demanding tasks, © 2019 Viliam Ďuriš, Timotej Šumný; such as Fermat’s theorem and Beal’s conjecture. published by UIKTEN. This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivs 3.0 License. The article is published with Open Access at www.temjournal.com 78 TEM Journal – Volume 8 / Number 1 / 2019. TEM Journal. Volume 8, Issue 1, Pages 78-81, ISSN 2217-8309, DOI: 10.18421/TEM81-10, February 2019. 2. Lattice points Then we will use In geometry, a Diophantine equation can be [ , 0] , 1 = (1,0) interpreted as a curve or surface. For surface 푥 푥 푥 equations, we determine the projection into the plane ( 푆 푠 ) 1 = grad( [ ]) 1 . Consequently, we can easily determine the −1 푥 푥 푥 lattice points in the space , by specifying the 푂 − 푆 ∙= 1푏 ∙ grad(퐹[퐴]) ∙ 푖 푗 푥method푥 for the tangent from the textbook [6]. −1 We will use the coordinate퐸 2system , , by For the inverse−푠푥 vector푏 ∙ is 푥valid∙ 퐹 퐴 specifying the integers by means of a scale on the axes , . Then, every point that is important〈푂 푥 푦 to〉 us grad( [ ]) grad( [ ]) = has integer coordinates [ , ], , , and grad( [ ]) grad푇( [ ]) therefore푥 푦 the coordinates are also lattice points. Next, −1 퐹 퐴 퐴 푥0 푦0 푥0 푦0 ∈ 푍 퐹 퐴 we can use a line to connect the origin of the For Diophantine equations,� it퐹 is퐴 useful∙ to 퐹know퐴 that [0,0] [ , ] coordinate system and . , while this is not a necessary condition. Other For the line passing through the origin of the 0 0 useful information is 푂 퐴 푥 푦 푥 coordinate system, we can write the equation in the 푠 ∈ 푍 form of ( ) grad( [ ]) = ; = −1 푦0 By way푂 − of퐴 the∙ tangent퐹 퐴 above the훼 lattice,훼 ∈ 푄 we can 푦 ∙ 푥 divide the Diophantine equation into the sets that 푥0 Let ( , ) = 1, then = determine the conditions for the solution above the 0 푦 set of integers. Furthermore, if we assume that there 푥0 푦0 푥0 푞 = is a solution, the equation can be parameterized so as to solve tasks in and similarly in by creating For the tangent we can푦 find푞 the∙ 푥 points which intersect a surface cut in the level to obtain the curve 3 2 the axes , . To determine such a line, we will use f( , ) = . 퐸 퐸 affine transformation. Then, we can determine 푧the0 relationship between 푥 푦 the푥 tangent푦 푐 in the assumed solution and the line that Let originated from the beginning of the coordinate system with the expected solution: ( , ) = c We can use the following퐹 푥 푦equation for the tangent = 푥 − 푥0 푥0 + ∙ 훼 훼 ∈ 푄 ( ) grad( [ ]) = 0 0 2.1 Fermat’s푦 last− 푦theorem푦 Another condition푂 − is퐴 ∙ 퐹 퐴 푏 One of the well-known problems in mathematics is Fermat’s last theorem [7], [8], [9]. This is aimed at grad( [ ]) = 0 finding a solution to the Diophantine equation in the form of + = , > 2 . We were motivated By joining the above푠 ∙ equations퐹 퐴 we get by the well푛-known푛 Pythagorean푛 triads in the form of 3 + 4 =푥 5 . 푦Fermat푧 pointed푛 out that there are no grad( [ ]) + ( ) grad( [ ]) = integer2 2 solutions2 for the equation in the form of + = and also said that there is no solution Then− 푠we∙ get 퐹 퐴 퐴 − 푂 ∙ 퐹 퐴 푏 between3 3 integers3 for any equation in the form of 푥 + 푦 = 푧 , > 2 . Despite the convincing claim + ( ) grad( [ ]) = that푛 he 푛found푛 evidence, no evidence was found. After nearly푥 푦 300 푧years,푛 it was the British mathematician �푠 =퐴 − 푂 � ∙ =퐹 퐴, 푏 We will use , and an inverse Andrew Wiles who was able to prove this claim in vector grad( [ ]) in the space E 1994, improving upon what Fermat did not know. 푠 �푆푗 − 퐴� 푗 푥 푦 −1 Nonetheless, we will use the theory of lattice points 2 +퐹 퐴 grad( [ ]) grad( [ ]) for Fermat’s last theorem. Let us consider = grad( [ ]) −1 the Diophantine equation in the form of −�퐴 − 푆푗 푂 − 퐴� ∙ 퐹 퐴 ∙ 퐹 퐴 −1 + 푏 ∙ 1 =퐹 퐴 grad( [ ]) + = , > 2 −1 푛 푛 푛 −�퐴 − 푆푗 푂 − 퐴� ∙ 푏 ∙ 퐹 퐴 푥 푦 푧 푛 TEM Journal – Volume 8 / Number 1 / 2019. 79 TEM Journal. Volume 8, Issue 1, Pages 78-81, ISSN 2217-8309, DOI: 10.18421/TEM81-10, February 2019. grad( [ ]) = ( , , ) The known solutions are: ( ) 푛grad−1 (푛−[1 ]) =푛−01 0 0 0 1 + 2 = 3 퐹 퐴 푥 푦 −푧 푎 3 2 Let us assume푂 the− 푋existence∙ 퐹of퐴 the solution and 2 + 7 = 3 therefore determine the conditions in the way that we 135 + 72 = 24 fix one coordinate and, for the remaining 2 3 9 coordinates, determine the conditions resulting from 2 + 17 = 71 the projection into the planes and 7 3 2 3 + 11 = 122 푥푦 푥푧 5 4 2 33 + 1549034 = 15613 = 푛−1 8 2 3 푦 − 푦0 푦0 1414 + 2213459 = 65 푛−1 0 0 3 2 7 푥 − 푥 푥 9262 + 15312283 = 113 = 3 2 7 푛−1 17 + 76271 = 21063928 푧 − 푧0 푧0 푛−1 0 − 7 3 2 Let us use the 푥projection− 푥 in푥 0the plane and fix 43 + 96222 = 30042907 8 3 2 푥푦 Fermat-Catalan’s conjecture says that the equation 푧0 = 푛−1 ± = with the condition + + < 1 has 푦 − 푦0 푦0 Let , 푛−1 푎 푏 푐 1 1 1 0 a final number of solutions for , , , , , . The 푥 − 푥 0 푎 푏 푐 =푥 푥equation푦 solutions푧 are known in the publication [10]. 훼 훽 ∈ 푄 = 푎 푏 푐 푥 푦 푧 0 0 푦 − 푦 훼푦 On the other hand, Beal’s conjecture uses the =0 0 푥 − 푥 푛−훽1푥 equation: 0 훼 푦 푛−1 훽 푥0 + = , , < 2 = 1 푎 푏 푐 푛−1 푥 푦 푧 푎 푏 푐 = 0 This conjecture is intended to address the solutions 훼 1 푦 ( ) 푛−1 with the condition , , = , , for 훽 푥0 , , > 2. This conjecture is linked by the basic = 0 0 0 푛 equation to Fermat-Catalan’s푥 푦 푧conjecture.훿 훿휖푁 Both of 훼푦0 푦0 these푎 푏 푐 conjectures are divided by the option, in which ( , ) =푛1 0 0 at least one of the exponents gains the value 2.
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