Reasoning Regarding Beal's Conjecture

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Reasoning Regarding Beal's Conjecture Alexer Mitkovskiy mitkovskiy@gmail.com Befe considering Beal's conjecture it is necessary to resolve one mathematical question.

1 Reasoning Regarding Beal's Conjecture Alexer Mitkovskiy Befe considering Beal's conjecture it is necessary to resolve one mathematical question. I Let's compare two fmulas: = 2* = 2. From the point of view of algebra the right parts of both these fmulas are equal have a value of 2. But is this unequivocal from the point of view of geometry? We will show the geometrical analogues of both fmulas. In the left part of the first fmula we have two squares with the sides of each square equal to 1 the area of each square is also equal to 1 in the right part of the fmula we have a rectangle the length of one side equals 2 the length of the second side is 1. The rectangle's area equal s 2. The equality of the right left parts of the fmula is not in doubt = In the left part of the second fmula we also have two squares with the sides of each square equalling 1. The area of each square is also equal to 1. In the right part of the fmula we have a square whose side is equal to the diagonal of the square with the side of 1 the area of this square is equal to product of the sides. The length of the diagonal of a square with the side of 1 is equal to 2 - arithmetically an irrational number which to four decimals equals We will recollect the rule of multiplication of irrational numbers by themselves increasing to Also we will multiply with itself receiving a value of Thus the area of a square with the site of 2 is equal to the number located between to within four decimals. If greater accuracy is needed we can determine the area of the square to decimals but under no circumstances will the area of a square with sides of 2 precisely equal 2. It can only approach this value. Thus in nature there is no square with an area of 2. The area must also be an irrational number. Why in the first fmula did we get an exact value from our calculation in the second fmula only an approach to that value? In the first fmula we expressed the area using two numbers causing no difficulties in the second case we tried to express the value through one number we did not manage to make it absolutely

2 precise. Why do we consider both fmulas equivalent in algebro? This only a mathematical set assumption on which all mathematics is based. But it does not mean that we should not consider this assumption in our calculations reasoning. F example if we construct a direct rectangular parallelepiped on these squares (whose height can be any rational number) its volume will be an irrational number. No calculation can give this volume in rational numbers. This is valid f all squares whose side is an irrational number. Let's consider the equation II (1) Where - natural numbers Let's accept that have the general divider where is a natural number. Thus (1) it is possible to present the fmula as (2) Where m n are natural numbers The value in fmula (1) which is equivalent to value in fmula (2 is geometrically possible to present in the fm of a direct rectangular parallelepiped with basr height. The parallelepiped with base height which is numerically equal to in fmula (1) will be a part of this parallelepiped. The rest of the volume of parallelepiped will be equal to represents a parallelepiped with a base of a height of. Thus it is possible to assert that. It follows from this that if in equation (1) the sum one of the compnents have a general divider then the second component in the fmula also has the same general divider. Let's accept that have the general divider where is a natural number. Thus (1) it is possible to present the fmula as

3 Where m n are natural numbers It is possible to present sum геометрически in the fm of a direct rectangular parallelepiped with a base of both the height the volume of this parallelepiped will be numerically equal to. Thus It follows from this that if in equation (1) the components have a general divider the sum also has the same general divider. How in equation (1) can there not be other variants of two numbers with a general divider it is possible to conclude that If in an equation of the kind where are natural numbers any two numbers have a general divider also the third has the same general divider. This is a necessary sufficient condition. Let's consider the expression (3) Where are natural numbers it is possible to present equation (3) as It is easy to see that equation (3) is a special case of the equation (1) if two of the numbers the general divider it is possible to present it as have (4) Where m n k are natural numbers Fmula (4) geometrically represents a direct rectangular parallelepiped with a base of a height of which consists of two parallelepipeds; one with a base of height the second with a base of height. It follows from this that all values in fmula (4) have a general divider. It is necessary to notice that the general divider in this case is a necessary condition but insufficient. If we look attentively at a parallelepiped with a base of a height of we will see that this parallelepiped represents individual cubes built on one number with sides so that fmula (3) was carried out it is necessary that the parallelepiped height was a multiple of the sum of heights of the parallelepipeds with a base of height with a base of height was equal to the height of a parallelepiped with a base of a height of.

4 Differently a sufficient condition of the function of fmula (4) will be parity operation Или Или III Let's consider equation (3) provided that two of the numbers are mutually simple that is they have no general divider. If two numbers of are mutually simple the third has a general divider in the numbers then we have established earlier that all numbers should have a general divider. It follows from this that if two of numbers are mutually simple all the numbers in are mutually simple. Let's present fmula (3) as (5) Let's accept that The right part of fmula (5) geometrically represents a direct rectangular parallelepiped with a base square of height. The parallelepiped with volume a base squared of height lies In the volume of this parallelepiped. Taking into account it parallelepiped as it is possible to present the volume of the (6) Remaining from the above the volume equals we need to express through the third. F this purpose we will take advantage of the Pythagean Theem we will present in the fm of. We will substitute this value in fmula (6) to get (7) Fmula (7) geometrically shows that the volume of parallelepiped consists of volumes of two parallelepipeds one with a base of both with heights the second with a base of height. The area of the base of parallelepiped is a square with a side consists of the sum of two squares. F descriptive reasons we will present it in the fm of a schematic drawing:

5 Let's compare parallelepipeds (fmula 7) with (fmula 5) we will see that they differ in height. It is obvious that the height of these parallelepipeds cannot be identical because of the numbers regarding the problem of mutually simple values i.e.. It follows from this that can be me less than. We will consider a variant when. Our drawing then becomes The volume of the parallelepiped will represent the sum of volumes of three parallelepipeds; the first with a base of height volume the second with a base of height volume ; the third with a base of height volume. (8) From fmulas (5) (8) it follows that (9) Fmula (9) shows that the volume of parallelepiped consists of two volumes - parallelepiped parallelepiped which was fmed from the subtraction of the volume of parallelepiped of parallelepiped. In fmula (9) it is obvious that the height of parallelepiped is me than the height of parallelepiped as it is me than the volume of parallelepiped. We will present it in the fm of a schematic drawing.

6 Thus it is clear if from what follows that there is no necessity to consider a variant when. From change of places of the components the sum does not change. Otherwise if the value of one of component in degree x y a minus two is less than the value of the sum of degrees z a minus two the value of the second component in degree x y the minus two is me than value of the sum of degrees z a minus two. Using equation (5) the condition of fmula (9) should be satisfied the volume of parallelepiped should share We will transfm fmula (9) get The right member of equation from fmula (3) is equal to we will write it down in the fm of we will get After a reduction on we will get equation Equation (10) is a condition of equation (3) in the case when (10) are mutually simple natural numbers. We know that equation (10) has resolution as whole positive numbers only in the case where are Pythagean numbers. In this case from theem of Fermat Euler it follows that there is only one variant in the resolution of equation (10) f primitive Pythagean tripllets. If numbers are mutually simple it means that are primitive Pythagean triplets. If at least one of the numbers is not a primitive Pythagean triplet this means that the number will be irrational the parallelepiped volume on that base will also be an irrational number. We cannot express it in natural numbers. See Chapter 1 f an understing of this. So we have equation (3) equation (10) which is a condition of the solution of equation (3). To get the result in equation (10) from equation (3) we will increase both expressions of equation (10) by equation already found earlier get the (11) Let's compare equation (11) equation (3) in what follows After reduction on the right the left parts of the equations give (12)

7 Where solution. are mutually simple numbers equations (12) have no solution. Hence equation (3) also has no It would seem that it is possible to put an end to it but we have considered equation (3) f mutually simple numbers only in the case where However the variant of equation (3) is theetically possible by which. It is possible that if is a small number in large degree then there are great numbers in small degree. At the moment of writing of this article I cannot furnish unequivocal proofs of fmula (3). So equation has no solutions in natural numbers under the condition when represent simple numbers f any natural in the case It is ftunate that the variant considered includes the Last Theem of Fermat.

1.5. Inequalities. Equations and Inequalities. Box (cont.) Sections

1.5. Inequalities. Equations and Inequalities. Box (cont.) Sections 1 Equations and Inequalities Sections 1.5 1.8 2008 Pearson Addison-Wesley. All rights reserved 1 Equations and Inequalities 1.5 Applications and Modeling with Quadratic Equations 1.6 Other Types of Equations