Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
John Wallis wikipedia , lookup
Wiles's proof of Fermat's Last Theorem wikipedia , lookup
Fundamental theorem of algebra wikipedia , lookup
Location arithmetic wikipedia , lookup
Line (geometry) wikipedia , lookup
Elementary mathematics wikipedia , lookup
Proof of the Ferma theorem Author: Valentin Alexeyevich Gorbunov Doctor of science, assistant professor of Moscow State University of Mines The author has graduated from Mechano-Mathematical Department of Moscow State University (1960-1966). He acquired his specialization at the Chair of Hydrodynamics. From 1967 till now he works at Moscow State University of Mines at first in the research laboratory and then at the Chair of Higher Mathematics at the post of assistant professor. During this period he published two books: * “Methods of mathematical physics in the problems of mining production” In particular in this book there is a part connected with a theory of explosion. The problems relating to this subject the author solved using numerical methods(difference schemes), where difference grids are used (the region is divided into cells). The author has occasionally come to proof of the Ferma theorem and apparently the usually used by the author methods of numerical methods (partition of the region) have been automatically applied by him. After proof the Ferma theorem the idea to use a grid for prime numbers in the coordinate system XOY for proof of the Goldbach theorem has come suddenly to him. Everything what the author has made in this field he expounded in his second book: * “Mathematical methods in the theory of information protection”, which was published quite recently. 1. Foreword In contrast to existing variant of proof of the Ferma theorem presented by U. Waylls the present proof is a direct proof and doesn’t use special cases of earlier proofs of this theorem. In our opinion a successful approach was found which allows to prove the Ferma theorem sufficiently simply and thus to destroy the existing opinion that it’s impossible to prove this theorem using elementary methods. 1 2. Proof of the Ferma theorem The Ferma theorem asserts that there exist no integer solutions of the following equation xn yn z n (1) under condition {n 2, x, y, z , n} N - set of integers, Proof Using the transformations u x z v y z (2) we can transform the equation (1) to the following form un vn 1 (3) It’s obvious that the variables u and v meet the following inequalities 0 u 1 0 v 1 (4) n n Let us designate as S n the sum of u v , and the graph of the equation S n 1 will be called a Ferma curve of the n -order. In the fig.1 the unit square is represented in which the unit circle S 2 1 and the Ferma curve S n 1 , n 3 are represented. By virtue of symmetry of the equation (3) relative to the variables u and v we shall later on examine only the upper part of the square relative to the collateral diagonal (matrix terminology). For any natural number z , z 1 we shall draw the uniform grid on the unit square with a rational step h 1/ z (fig. 1). The greater is z , the lesser is h , that is thicker is the grid in the unit square. Assertion of the Ferma theorem is equivalent to the fact that for any z number (and hence for any h ) the Ferma curves don’t pass through the node (rational) points in the unit square what should be proved. 1) 2 Fig1. Unit circle S 2 1 and the Ferma curve S n 1 , (n=3, 4, …) in the unit square covered by the uniform grid with the rational step h 1/ z . It’s known that there is an infinite set of rational points on the unit circle. If by fragmentation x y P0 0 ; 0 z0 z0 of the unit square for the natural number z 0 the rational point of the unit circle turns out to be a node point on the grid we shall call this point the Pythagorean point and the ray passing through it (going out from the origin if coordinates) will be called the Pythagorean ray 0 . x y M 0 0 ; 0 z1 z 2 with rational coordinates can become the node 1) Any point in the unit square point of the grid for the natural number z z1 z 2 . Thus hereinafter we shall imply speaking about a rational point that the denominators of the point coordinates are equal. If the Pythagorean number z 0 is a prime number there will be only one Pythagorean point on the unit circle in the upper part of the unit square on the grid for z 0 . For the composite numbers of the type z k p1 p2 ... pl , where p1 , p2 ,..., pl - are prime numbers of the type 4n 1 , and k - is a natural number in the factorization of which to prime factors there are no Pythagorean 3 numbers there will be l 2 Pythagorean points on the unit circle in the upper part of the unit square for the grid with the step h 1/ z . Let us denote as the ray forming the angle with the axis Ou . Along with this designation we give the numerical value tg to the parameter . Then the coordinates of points on this ray are connected with each other by the relationship v u . Substituting u instead of v into equation (3) we get the coordinates of the point n lying at the intersection of the ray with the Ferma curve S n 1 : un 1 n 1 n vn n 1 n (5) From this point on we shall be interested only in the rays passing through the node (rational) points of the grid in the unit square. On these rays the parameter takes on rational values. We shall distinguish the rays 0 passing through the rational points on the unit circle and the rays passing through the rational points in the unit square and not lying on the unit square. Let us proof the following lemma which will be useful for us below. Lemma. x y ; z 0 z 0 lies on the ray 0 the length of radius-vector R of this 1). If the rational point point is a rational number ( x and y - are natural numbers); x y ; z z lies on the ray the length of radius-vector R of this 2). If the rational point point is an irrational number ( x , y and z - are natural numbers). x y P0 0 ; 0 z 0 z 0 - the point with rational coordinates Let us prove the first part of the lemma. Let – belongs to the unit circle P0 S 2 1 . Then the coordinates of this point comply with the equation x02 y 02 z 02 (6) where x 0 , y 0 and z 0 - are natural numbers. 4 x y ; z 0 z 0 with rational coordinates lie on the ray 0 passing through the Let the point point P0 , S 2 1 . Then 0 y0 y x0 x and the length of radius-vector of the point will be equal to 2 2 2 x y y x x x R 1 20 1 0 z0 z0 x0 z0 z0 x0 - rational number. ----------------------2). Proof of this formula is given in the appendix 2. x y ; z z with the rational Let us prove the second part of the lemma. Let the point coordinates lies on the ray , which intersects the unit circle in the irrational point P u ; v , u y m n v k ; k . Since x - is a rational number, the numerators of the fractions u and v - are rational numbers: n m . It follows from here that the denominator k m 2 n 2 - is an irrational number. Length of radius-vector of the point will be equal to 2 2 2 x y y x x x k n R 1 1 z z z m m z z x - irrational number. 2 Hence the length of radius-vector of the rational point on the ray 0 (passing through the rational point of the unit circle) is a rational number and the length of radius-vector of the rational point lying on the ray (intersecting the unit circle in the irrational point) is an irrational number. Let x0 , y 0 , z 0 - be a Pythagorean triple of numbers with pairwise co-prime numbers. Such Pythagorean triples of numbers are called primitive. Then on the grid in the unit circle for the natural number z 0 there is a Pythagorean point P0 u0 ,v0 , where u 0 x0 / z 0 ; v0 y 0 / z 0 on the 5 unit circle. The ray 0 passing through the point P0 doesn’t contain other node points of the grid for z 0 . Let us prove it. x y ; z 0 z 0 of the grid lies on the ray 0 fir z 0 . Then Let the node point y 0 x y0 x x0 (7) Since x 0 and y 0 are co-prime numbers y will be an integer if x is a multiple of x 0 that is x m x 0 , where m - is a natural number, m 1 . Then y m y 0 and the coordinates of the point will be u m x0 m y0 v z0 , z 0 . It follows from here that the value of radius- vector length of the point will be R m , where m 1 is an integer. Since the greatest value of the radius-vector length of any point inside the unit square is less than 2 the point will be situated outside the unit circle. Thus if the Pythagorean triple of numbers x0 , y 0 , z 0 - is primitive , then there are other node points of the grid on the Pythagorean ray 0 besides the Pythagorean point P0 on the unit circle. So on the grid in the unit square for the natural number z 0 all Ferma curves intersect the ray 0 not in the node points and hence there will be no integer-valued solutions in these points according to the equation (1). Then the transformed equation (3) will have irrational solutions for all n , n 2 : un vn 1 n 1 0 n 0 n 1 0 n x0 n x0 y 0 n n y0 n x0 y 0 n n (8) It should be noted that according to the formulas (8) the coordinates of the points of the Ferma curves on the ray 0 are formed from the coordinates of the Pythagorean point x y P0 0 ; 0 z 0 z 0 and hence they don’t depend on other fragmentations of the unit square (it’s obvious that the point coordinates doesn’t depend on fragmentation of the unit square at all). 6 Since the numerators of coordinates of the point n u n ; vn are integers solutions of the n equations (8) will be irrational if we prove that x0 y 0 n n is an irrational number (by n 2 ). Let us prove it. Let for some natural number n , n 2 , n x0 y 0 n n z 0 k - be a rational number, where z 0 k x0 k y 0 n ; z z 0 0 lie at the intersection of and k are co-prime natural numbers and let the point the Ferma curve S n 1 with the Pythagorean ray 0 . z 0 z0 It’s obvious that k since the length of radius-vector of the point n is greater than one (see fig.1) 2 2 k x0 k y 0 k z0 Rn 1 z 0 z 0 z 0 (9) It follows from the assumption that the point n with rational coordinates lies on the Ferma curve S n 1 that the equation (1) should have an integer-valued solution k x0 n k y0 n z0 n (10) It follows from the previous equation that z 0 should be divisible by k . But according to the assumption z 0 and k are co-prime numbers. Hence n x0 y 0 n n can’t be a fractional rational number. n Now let x0 y 0 n n - be a natural number and z 0 z 0 . Then on the grid in the unit square x y n 0 ; 0 z 0 z 0 is a node point and lies on the Ferma curve and for a natural number z 0 the point S n 1 (according to the assumption). It’s obvious that for the Pythagorean triples of numbers x0 , y 0 , z 0 for which z 0 y0 1 x0 y 0 n n (e.g. (3, 4, 5), (5, 12, 13) and so on), n can’t be an integer because in this case the following inequalities should be true y 0 n x0 y 0 y 0 1 n n (11) In the other cases when z 0 y0 1 (e.g. (20, 21, 29)) it follows from the assumption that 7 x y n 0 ; 0 z 0 z 0 lies on the Ferma curve S n 1 that this point will be a node point on the the point grid in the unit square for the natural number z 0 which isn’t a multiple of z 0 , ( z 0 z 0 and z 0 is a prime number). According to the formulas (8) by n 2 solution of the equation (3) should be rational and solution of the equation (1) should be integer–valued. But on the grid in the unit square the point P0 won’t be a node point for z 0 and hence by n 2 the equation (1) will not have an integervalued solution in this point. Thus the assumption that by some n , (n 2) , the fact that by n 2 , n x0 y 0 n - is a natural number z 0 leads to n x0 y 0 z 0 isn’t a natural number that is we come to a 2 2 contradiction. n It follows from irrationality of x0 y 0 n n lying at the intersection of the Ferma curve that the length of radius-vector of the point n S n 1 with the Pythagorean ray 0 is also an irrational number Rn u n vn 2 2 x0 n n n x0 y 0 2 y0 n n n x0 y 0 2 z0 n x0 y 0 n n (12) The node points on the Pythagorean ray 0 can appear if we consider the grids in the unit square for the Pythagorean triple of numbers similar to the triple x0 , y 0 , z 0 that is if we consider the grids in the unit square for natural numbers which are multiples of z 0 . But for such numbers the solution of the equation (1) doesn’t change. It’s obvious that there are infinitely many rational points on the ray 0 and every of them can become a node point on the grid for the natural number z by appropriate selection of the fragmentation step of the unit square h 1/ z . According to the lemma the length of radius vector of the rational point on the ray 0 - is a rational number. Hence for any fragmentations of the unit square there is the following inequality on the Pythagorean ray 0 R Rn (13) where Rn - is the length of radius-vector of the point n on the Ferma curve S n 1 and R - is the length of radius-vector of any node (rational) point on the ray 0 . 8 The coordinates of the point n on the Pythagorean ray 0 can be written in the following form x0 y0 n ; n n n n n n x0 y 0 x0 y 0 R hn n z0 or n x0 hn ; y0 hn , where 1 n x0 y 0 n n (14) The coordinates of the Pythagorean point P0 can be written in the form P0 x0 h0 ; y0 h0 (15) where h0 1 / z 0 - is the fragmentation step of the unit square. It’s obvious that hn h0 since Rn 1 , hn h0 Rn . If we plot the square with the side equal to the length of radius-vector of the point n and cover this square by the uniform grid with the step hn Rn / z 0 the point n in this square will be a node point (Pythagorean). In this point analogously to the Pythagorean point P0 S 2 1 the Pythagorean theorem holds true integrally for cells (see fig.2). Fig.2. a)Uniform grid in the unit square for the Pythagorean number z 0 , h0 1 / z 0 , x y P0 0 ; 0 z 0 z 0 - is a Pythagorean point; x0 y0 n ; n n n n n n x0 y 0 x0 y 0 - is a point lying at the intersection of the Ferma curve S n 1 with the Pythagorean ray 0 . 9 b)Uniform grid in the square plotted on the radius-vector of the point n of the Ferma curve S n 1 (in the point n the Pythagorean theorem holds true integrally for the cells as well as for the point P0 in the unit square). Finally let us consider the rays passing through the node (rational) points of various fragmentations of the unit square, not being Pythagorean rays for any fragmentations. These rays intersect the unit circle in irrational points. For example 3 / 2 or 6 / 5 and so on. x y ; z z of the grid in the unit Let us suppose that on one of such rays the node point square lies on one of the Ferma curves for some natural number z . It follows from here: 1)The fragmentation step of the square plotted on the radius-vector of the point will be the same as in the unit square h 1/ z ; 2)The number of elements (cells) of the square plotted on the radius-vector of the point will be greater than the number of fragmentation elements of the unit square since R 1 ; 3)The point should be a node (Pythagorean) point in the square with the side R on the ray which isn’t a Pythagorean ray. It’s obvious that the square with the side R can’t be covered by the uniform grid with the rational step h 1/ z since in this case R is an irrational number (see the lemma). In the node points of the grid of the unit square the following condition is fulfilled: the area of the square plotted on the radius-vector of the node point is equal to the integer number of elements (cells) of fragmentation of the unit square, but it’s impossible to cover this square by such elements except for squares plotted for Pythagorean (node) points of the unit circle. We can see from the fig.2 that the grid for points of the Ferma curves stretches out along the Pythagorean rays 0 that is the fragmentation step increases (with conservation of the number of elements): hn Rn / z 0 , where Rn - is an irrational number (by n 2 ). The number of Pythagorean points on the unit circle for the chosen number z 0 is equal to the number of Pythagorean (node) points on all Ferma curves on their local grids (with the hn irrational step of fragmentation Rn z0 1 n x0 y 0 n n ). 10 Thus the cause for what the equation (1) doesn’t have integer-valued solutions by n 2 is quite evident: on the Pythagorean rays 0 in the squares plotted on radius-vectors of the points of Ferma curves the Pythagorean theorem holds true integrally for the cells (with conservation of the number of fragmentation elements), see fig.2. The theorem was proved. Appendix 1 The chosen approach for proof of the Ferma theorem allows to give a geometric interpretation of this theorem. A geometric image of the degree a by n 1 is a segment with the length a ; and by n 2 - a square with the side a ; by n 3 - a cube with the edge a . By n 2 three squares can be set up in correspondence to the Pythagorean point x y P0 0 ; 0 z 0 z 0 : the square plotted on the radius-vector of the point P0 and two squares plotted on the coordinates of this point, see fig.3. 11 Fig.3. Number of fragmentation elements of the unit square is equal to the sum of elements of the squares plotted on the coordinates of the Pythagorean point P0 x0 h0 ; y 0 h0 , h0 1 / z 0 . By n 3 three cubes can be set up in correspondence to the Pythagorean point P0 : the cube plotted of the radius –vector of the point P0 , (the unit cube) and two cubes with the edges equal to coordinates of this point, see fig.4. Fig.4. x y P0 0 ; 0 z 0 z 0 - the Pythagorean point; 1 – the unit cube plotted on the radius-vector of the point P0 ; 2;3 – the cubes plotted on the coordinates of the Pythagorean point; h0 1 / z 0 - fragmentation step of the unit square; Pythagorean theorem is integrally fulfilled for the cells on the edges of the cubes 1, 2, 3. 12 Three cubes plotted on the radius-vector of the point 3 and on the coordinates of this point x0 h3 ; y0 h3 . Fig.4. x0 y0 3 ; 3 3 3 3 3 3 x0 y 0 x0 y 0 - the point lying at the intersection of the Ferma curve S 3 1 with the Pythagorean ray 0 ; R3 z0 3 x0 y 0 3 3 - radius-vector of the point 3 ; x0 h3 and y 0 h3 -values of length of the edges of the cubes 1, 2, 3 correspondingly; On the edges of these cubes 1,2,3 the Pythagorean theorem is also fulfilled integrally for the cells. If we introduce a designation : V1 - the volume of the cube plotted on the radius-vector of the point being considered, and V2 and V3 - the volumes of the cubes plotted on the coordinates of this point, then the following relationships hold true in the point P0 : 13 V1 1 V2 V3 1 (1) and in the point 3 the following relationships are fulfilled: V1 1 V2 V3 1 (2) In both cases the Pythagorean theorem is fulfilled integrally (for the cells) on the faces of the cubes. It follows from here the following conclusion: on the Pythagorean rays 0 the unit cube V1 1 of the specified fragmentation h0 1 / z0 is divided into two cubes V2 ;V3 consisting of equal elements but belonging to another cube (not the unit cube), and namely to the cube plotted on the radius-vector of the point 3 of the Ferma curve S 3 1 (situated on the same ray 0 ) with analogous fragmentation and the irrational step h3 h0 R3 . n By analogy we can consider a geometric image of the degree a , where n - is any natural number. The unit cube with the base z 0 h0 z 0 h0 and the height z 0 h0 n2 the fig.4. In this case prisms with the base h0 h0 and the height h0 is represented in n2 will be elements of fragmentation of the unit cube. In particular by n 2 we have a prism with the height equal to one; by n 3 the height of prism is equal to h0 and the element of fragmentation is an elementary cube. The greater is n the less is the prism height: h0 n2 0 by n , since 0 h0 1 . The following geometric elements are represented on the fig.5: the cube plotted on the radiusvectors of the point P0 and the point n and the parallelepipeds plotted on the coordinates of these points. On the bases of these parallelepipeds the Pythagorean theorem is also fulfilled integrally for the cells. 14 Fig.5. x y P0 0 ; 0 z 0 z 0 - the Pythagorean point; x0 y0 n ; n n n n n n x0 y 0 x0 y 0 - the point lying at the intersection of the Ferma curve and the Pythagorean ray 0 ; 1 – the unit cube plotted on the radius-vector of the point P0 (or the cube plotted on the radiusvector of the point n of the Ferma curve S n 1 ); 2,3 – the parallelepipeds plotted on the coordinates of the point P0 or n . 15