Annual Research Symposium (ARS)
Permanent URI for this communityhttp://repository.kln.ac.lk/handle/123456789/154
Browse
16 results
Search Results
Item Roots of a cubic and simple proof of Fermat’s last theorem for n=3(University of Kelaniya, 2013) Ubeynarayana, C.U.; Piyadasa, R.A.D.; Munasinghe, J.; Ekanayake, E.M.P.Introduction: Fermat’s last theorem (FLT) which was written in 1637, became public in 1670, without proof. It has not only evoked the interest of mathematicians but baffled many for over three hundred and fifty years [1],[2]. It is well known that FLT despite its rather simple statement had been difficult to prove even for the small prime exponent .The main objective of this paper is to provide simple proof of the theorem for this special exponent using the Method of Taragalia and Cardom of solving a cubic which is much older than Fermat’s last theorem. Proof of Fermat’s last theorem for n=3 Fermat’s last theorem for 3 n can be stated as that the equation , ( , ) 1 3 3 3 z y x x y (1.1) is not satisfied by non-trivial integer triples x, y, z . Assume that the equation is satisfied by non-trivial integer triples x, y, z . If let x 3m1, y 3k 1, z 3s 1, then we have (2,0, 2)(mod3 ) 3 3 2 y x However 1(mod3 ) 3 2 z , therefore our assumption is wrong and we conclude that xyz 0(mod3) . Since we consider the equation (1.1) for positive and negative integer values, without loss of generality, we can assume that y 0(mod3) ,and let 0(mod3 ) m y . Then 3 1 3 3 3 z x 3 u , z y h , x y g m due to Barlow relations, where are respectively the factors of . Now, 3 2( ) 3 3 1 3 3 g u h x y z m (1.2) Since x y z x (z y) (x y) z y (z x) , x y z 0(3 ugh)It can be shown that [1], 3 3 3 3 3 x y z 3(x y)(z x)(z y) 3 h u g m (1.3) and therefore 3 2 3 0 3 3 3 1 3 g h u ugh m m (1.4) This necessary condition must be satisfied by the integer parameters , of respectively. We will first fix the parameters of and show that (1.4) is not satisfied by integer for any integer using the Method of Tartagalia and Cardoon[4] of finding a roots of a cubic. The equation (1.4) is of the form 3 0 3 3 3 g vwg v w (1.5) where 3 3 3 1 3 3 v w 3 u h m , uh vw m 1 3. 2 , and its roots can be written as 2 , w v w v and +w (1.6) with the cube root of unity. Now 3 3 , w v are the roots of the equation 0 2 3 H Gt t (1.7) where (3 ) 3 1 3 3 G u h m , . 3. 2 1uh H m Moreover real and distinct and this equation has only one real root, namely, g v w, since the discriminant of (1.5) is ,where ( 4 ) 2 3 G H is the discriminant of (1.7),and it is negative. Therefore, it has only one real root [4]. Note that 2 3 6 2 6 3 3 3 3 6 3 1 3 3 2 3 3 3 3 (G 4H ) 3 u 14.3 u h h (3 u h ) 4.3 u h m m m m which is positive when uh 0. On the other hand . 3 1 3 3 3 3 3 3 (3 u h ) 32.3 u h m m which is positive when If 1 2 1 2 1 1 1 1 v w ,v w ,v w (1.8) is another representation of the roots, we must have 2 2 1 1 1 1 v w v w,v w v w , v w v w 2 1 2 1 or v w v w v w v w 2 2 1 1 1 1 , , 2 1 2 1 v w v w In other words , ( ) ( ) 1 1 1 1 v v w w v v w w or ( ) ( ), 1 1 v w v w ( ) ( ) 1 1 v w v w This means that v v w w 1 1 , or , . 1 1 v w w v Therefore, roots must be unique. In particular, we must have a unique real root. From (1.4), it follows that the real root can be expressed in the form g h j m 1 3 or g j h m 1 3 where j is an integer satisfying (3, j) 1. This is due to the fact that ( )[( ) 3 ] 0(mod3 ). 3 3 2 m g h g h g h gh It is now clear that 3 h satisfies the equation 0 2 3 t Gt H , This means that (3 ) 8.3 0 6 3 1 3 3 3 3 3 3 3 h u h h u h m m implying that 0 u or , 0 h that is y 0 or x 0 in Fermat’s equation. Hence there is no non-trivial integral triple satisfying the Fermat equation (1.1). We have shown that Fermat’s Last Theorem for 3 n can be proved without depending on the method of infinite descent or complex analysis. The proof given above is short and simple, and simpler than proving [3] the theorem using the method of infinite descent.Item New Set of Primitive Pythagorean Triples and Proofs of Two Fermat’s Theorems(University of Kelaniya, 2012) Manike, K.R.C.J.; Ekanayake, E.M.P.; Piyadasa, R.A.D.Fermat used the well known primitive Pythagorean set [1], [2], (1.1) ,where > 0 and are of opposite parity to prove the two theorems (1) Fermat’s last theorem for (2) the area of a Pythagorean triangle cannot be a square of an integer. Historically the above set was well known long ago and no other primitive whole set was available in the literature. However, the following complete primitive set can be obtained from the Pythagoras’ equation easily. The set (1.2) where > 0 and are both odd and can be obtained from the Pythagoras equation , (1.3) To obtain this set, assume that is odd and is even. Obviously, is odd. Then , , and ( ) are co-prime. Hence , , where , and are odd. Now, , , , where a > b > 0 and both a, b are odd and coprime, give the complete primitive set of Pythagorean triples. 2. Proof of Two Theorems (1) Fermat’s last theorem for can be stated as there are no non-trivial integral triples satisfying the equation + , where is odd. This can be stated as there is no non-trivial integral triples satisfying the equation: = . (2) The second theorem can be stated as there are no integers satisfying the equation ,where = , . (2.2) In terms of the new Pythagorean triples this can be stated as = , where are odd. In other words, there are integers satisfying the equation = (2.3) where we have used the fact that are squares. Now, we will show that there are no non-trivial integer triples satisfying the equation (2.4) using the new set of Pythagorean triples, in order to prove the two theorems. In the following, we prove the two theorems using the method available in the literature [1], [2] but using the new set of Pythagorean triples: , ab, and - If , , , Assume that is the smallest integer satisfying the equation (2.4). Now, we have - = ,where . Therefore, we deduce by the method of infinite descent that there are no non-trivial integer triples satisfying the equations (2.4) or (2.3). This completes the proof of the two theorems.Item On the Integer Roots of Polynomial Equation(University of Kelaniya, 2012) Dharmasiri, K.G.E.U.; Dahanayaka, S.D.; Piyadasa, R.A.D.Integer solutions of polynomial equations are very important in number theory. However, solutions of general polynomial equations of degree five or higher than five cannot be solved in radicals due to Abel-Ruffini theorem. Even in case of a quadratic equation, it is not easy at all to discard all integer solutions without knowing the coefficients of the independent variable. For example, the simple cubic 5 12 0 3 2 q bx x x , where 1) , 2( q , has no integer roots. But, even in this case, it is not easy to justify this fact right away. The simple theorem is given in this paper which is capable of discarding integer roots of this equation at once. This theorem and the lemma [1] which was used in a previous paper are capable of discarding all integer roots of a special class of polynomial equations of any degree. Solution of polynomial equations First of all we consider simple statements with respect to the quadratic equation 0 2 ax bx c (1) The equation (1) has no integer roots if c ba, , all are odd or c is odd and ba, are of the same parity. In general, the polynomial equation ... 0, 1 1 0 1 n n n n a x a x a x a has no integer roots if na is odd and 0 1 1 , ,..., n a a a are even. Let us call this simple theorem, the theorem of odd parity. This follows at once since zero is even and integer roots of the equation are odd factor of n a . From (1) and the theorem of odd parity, we can deduce that b 4ac 2 cannot be an integer when a,b,c all are odd. Now consider the polynomial equation of the form 0 p p x pbx c (2) It can be shown that this polynomial equation (2) has no integer roots for any odd prime p , where (p,b) (p,c) 1. By the theorem of odd parity, the equation has no integer roots when c and b are odd. If a is an integer root of the equation, then we must have a c pba 0 It is clear that 0(mod ) 2 a c p p p . This condition can’t be satisfied since (ba, p) 1 and hence the equation (2) cannot have integer roots. The equation of the type 0, p m p p pm x p bx c d p (3) where 1m , 1) , ( pcd , has no integer roots. If x is an integer root, ) (mod 0 p p m x c p and therefore 0(mod ) 1 m x c p and we can write j pc x m1 , where j is an integer co-prime to .p Now, we can write m p m m p p pm c p j p b c p j c d p [( ) ]( ) 1 1 1 and hence we must have 0 pm p p p pm p j d p which implies p divides j . This is a contradiction and we deduce that the equation (3) has no integer roots.Item Integer roots of two polynomial equations and a simple proof of Fermat‟s last theorem(University of Kelaniya, 2011) Piyadasa, R.A.D.; Perera, B.B.U.P.Fermat‟s last theorem (FLT),possibly written in 1637,despite its rather simple statement, is very difficult to prove for general exponent n [1]. In fact, formal complete proof of FLT remained illusive until 1995 when Andrew Wiles and Taylor[1],[2] put forward one based on elliptic curves[3]. It is well known that their proof is lengthy and difficult to understand. Main objective of this paper is to provide a simpler and shorter proof for FLT. It is shown that FLT can be proved by showing that two polynomial equations have no integer roots when the independent variable satisfies certain conditions. Theorem: The polynomial equations in x 2. ( ) 0 1 p m p pm p x p uhdx h p u 2 ( ) 0 p m p p x uhdp x h u where u,h, p,d are integers co-prime to one another , p is an odd prime and m 2 , have no integer roots co-prime to h for any integer values of it when u,h are both odd or of opposite parity[4],[5]. Lemma If ( , ) 0(mod ) p p m F a b a b p and (a, p) (b, p) 1, then 0(mod ) 1 m a b p and m 2 Proof of the theorem: We first consider the equation 2. ( ) 0 1 p m p pm p x p uhdx h p u The integer roots of this equation are the integer factors of p pm p h p u 1 and let us assume that it has an integer root. This integer root obviously must be co-prime to u,h, p since they are co-prime. If an integer satisfies the equation , then ( ) 2. ) 0 1 p p m pm p g h p uhd p u and 0(mod ) 1 m g h p . Therefore, we can write g h p j m1 , where the integer j is co-prime to d,h, p .Now, our equation can be written as m m p m p pm p h p j h p j uhdp h p u 1 1 1 1 ( )[( ) 2 ] and we use the remainder theorem to check weather the linear factor h p j m1 in h a factor of the polynomial p pm p h p u 1 in h . If so, 0 1 pm p p pm p p j p u This is impossible since ( j, p) 1, and we conclude that (1) has no integer roots we need. If g satisfies the equation 2 ( ) 0 p m p p g uhdp g h u and g must be a factor of p p h u . We also assume that h,u are both odd or of opposite parity which is relevant to Fermat‟s last theorem. First of all, we will show that g h u using the relation i i p i p i i p p p i p i C u h h u i p h u h u 2 2 1 1 1 1 ( ) .( 1) . ( ) Our equation takes the form g h u ughdp If g h u 0 , then we must have ( ) - (-1) ] 0 2 - 2 - [ .( ) 2 -3 2 -3 2 -1 -5 1 -3 -3 p p p m p p p C uh h u p u h p dp p h u If both u and h are odd, or , of opposite parity ,then the term ( ) - (-1) ] 2 -[ .( ) 2 -3 2 -3 2 -1 -4 1 -3 -3 p p p p p p C uh h u p u h p p h u is odd since p( 3) is an odd prime and therefore the equation ( ) - (-1) ] 0 2 - 2 - [ .( ) 2 -3 2 -3 2 -1 -5 1 -3 -3 p p p m p p p C uh h u p u h p dp p h u will never be satisfied since m 2dp is even. Hence, g h u . From the equation ( ) 2 .( 1) . . ( ) 0 2 1 1 2 1 1 i i p i i i p i p i p p m C u h h u i p g h u ughdp we conclude that g (h u) 0(mod p) , which follows from the lemma Therefore , we can write g h (u p j) k , where k 1 , j 0 and ((u p j),h) 1 k is since (g,h) 1. Since g h (u p j) k is an integer root of the equation we can write [ ( )][( ) 2 ] p p k k p 1 m h u h u p j h u p j uhdp As before , using the remainder theorem, we get ( ) 0 k p p u p j u But this equation will never be satisfied since j 0 .Item Possible quark confinement by a non-relativistic model(University of Kelaniya, 2011) Karunatathne, S.; Piyadasa, R.A.D.Confinement of quarks by an infinitely deep potential well is well [1] known. We are interested in confinement of quarks by the singular potential 2 1 r when the effective potential , 2 2 ( 1) r r l l is negative, where 2 1 2 . However, we have found that the corresponding series solution is not a bound wave function. Now, we assume that quarks are localized to a small region and obtain the bound states in the following way. Consider the Schrödinger equation in the form 0 ( 1) 2 2 2 2 2 u r r l l k dr d u (1.1) and let us choose such l(l 1) . Then (1.1) reduces to 0 2 2 2 k u dr d u and the wave function kr u r e ( ) and the total radial wave function is given by R(r) = r e r u r kr ( ) (1.2) which is normalizable and the normalization constant 2 1 (2k) . We conclude that non relativistic quarks having nonzero angular momentum can be bound by the inverse square potential and the quark wave function can be made highly localized acquiring sufficient energy 2 2 2 k . We use the experimental value of the size(diameter) of the nucleon of 1.6 fm to determine the value of k . We can attribute this value to the mean square radius given by 2 | | 0.64 2 2 2 0 2 r kr e dr r kr (1.3) The equation (1.3) gives 2 (0.64) 1 2 k (1.4) We have assumed that the quark mass is and therefore the quark binding energy E is given by 2 2 2 k , and if we use to be one third of the nucleon mass , then E 62 48.437 2 938 197. 197 3 2 2 0.556 fm 2k 1 r Strength of the potential can be found in this case by using l(l 1) . If l 1, 2 2 2 1 . Therefore 60MeV 938 . 3.200.200 2 1 Another important point to be mentioned here is that attractive potential can be bound to potential centre of a circular orbit by an inverse square potential only if the total energy of the particle is zero in case of classical mechanics. Therefore, our quark bound states might be stable if they are confined to a very small region and they are undisturbed. This conclusion is actually based on classical mechanics but plausible since speeds of quarks should be big.Item A simple analytical proof of Fermat’s last theorem for n = 7(Research Symposium 2010 - Faculty of Graduate Studies, University of Kelaniya, 2010) Piyadasa, R.A.D.It is well known that proof of Fermat’s last theorem for any odd prime is difficult and first proof for n 7 was given by Lame [1] ,and Kumar also has given a proof for a special class of primes (Regular primes)which includes the case n 7 .However, these proofs are lengthy and difficult and may not easily be extended for all odd primes. The prime n 7 differs from n 5 since 2.7.115 is not a prime, whereas 2.5111 is a prime. Then it follows from the famous theorem of Germain Sophie that the corresponding Fermat’s equation , ( , ) 1 7 7 7 z x y x y may have two classes of integer solutions, xyz 0(mod7) and xyz 0(mod7) if we assume that the Fermat equation has non-trivial integer solutions for x, y, z . This fact is proved using the simple argument [3] of Oosterhuis. The main objective of this paper is to give a simple analytical proof for the Fermat’s last theorem n 7 using general respective parametric solutions corresponding two classes of solutions of the Fermat equation ,which has already been extended for all odd primes.Item Method of Infinite Descent and proof of Fermat's last theorem for n = 3(Research Symposium 2010 - Faculty of Graduate Studies, University of Kelaniya, 2010) Piyadasa, R.A.D.The first proof of Fermat’s last theorem for the exponent n 3 was given by Leonard Euler using the famous mathematical tool of Fermat called the method of infinite decent. However, Euler did not establish in full the key lemma required in the proof. Since then, several authors have published proofs for the cubic exponent but Euler's proof may have been supposed to be the simplest. Paulo Ribenboim [1] claims that he has patched up Euler’s proof and Edwards [2] also has given a proof of the critical and key lemma of Euler’s proof using the ring of complex numbers. Recently, Macys in his recent article [3, Eng.Transl.] claims that he may have reconstructed Euler’s proof for the key lemma. However, none of these proofs is short nor easy to understand compared to the simplicity of the theorem and the method of infinite decent The main objective of this paper is to provide a simple, short and independent proof for the theorem using the method of infinite decent. It is assumed that the equation 3 3 3 z y x , (x, y) 1 has non trivial integer solutions for (x, y, z) and their parametric representation [5] is obtained with one necessary condition that must be satisfied by the parameters. Using this necessary condition, an analytical proof of the theorem is given using the method contradiction. The proof is based on the method of finding roots of a cubic formulated by Tartagalia and Cardan [4], which is very much older than Fermat’s last theorem.Item Useful identities in finding a simple proof for Fermat’s last theorem(Research Symposium 2010 - Faculty of Graduate Studies, University of Kelaniya, 2010) Piyadasa, R.A.D.; Shadini, A.M.D.M.; Perera, B.B.U.P.Fermat’s last theorem, very famous and difficult theorem in mathematics, has been proved by Andrew Wiles and Taylor in 1995 after 358 years later the theorem was stated However, their proof is extremely difficult and lengthy. Possibility of finding s simple proof, first indicated by Fermat himself in a margin of his notes , has been still baffled and main objective of this paper is, however, to point out important identities which will certainly be useful to find a simple proof for the theorem.Item Effect of a long-ranged part of potential on elastic S-matrix element(Research Symposium 2010 - Faculty of Graduate Studies, University of Kelaniya, 2010) Shadini, A.M.D.M.; Piyadasa, R.A.D.; Munasinghe, J.It has been found that quantum mechanical three-body Schrödinger equation can be reduced to a set of coupled differential equations when the projectile can be easily breakable into two fragments when it is scattering on a heavy stable nucleus [1]. This coupled set of differential equations is solved under appropriate boundary conditions, and this method, called CDCC, has been found to be a very successful model in high energy quantum mechanical three body calculations [2]. It can be shown, however, that the coupling potentials in the coupled differential equations are actually long-range [3],[4] and asymptotic out going boundary condition, which is used to obtain elastic and breakup S-matrix elements is not mathematically justifiable. It has been found that the diagonal coupling potentials in this model takes the inverse square form at sufficiently large radial distances [3]and non-diagonal part of coupling potentials can be treated as sufficiently short-range to guarantee numeral calculations are feasible. Therefore one has to justify that the long range part of diagonal potential has a very small effect on elastic and breakup S-matrix elements to show that CDCC is mathematically sound .Although the CDCC method has been successful in many cases, recent numerical calculations[5],[6]indicate its unsatisfactory features as well. Therefore inclusion of the long range part in the calculation is also essential. The main objective of this contribution is to show that the effect of the long range part of the potentials on S-matrix elements is small.Item Simple proof of Fermat’s last theorem for n =11(Research Symposium 2010 - Faculty of Graduate Studies, University of Kelaniya, 2010) Shadini, A.M.D.M.; Piyadasa, R.A.D.Proof of Fermat’s last theorem for any odd prime is difficult. It may be extremely difficult to generalize any available Proof of Fermat’s last theorem for small prime such as n 3,5,7 to n 11[1]. The prime n 11 is different from n 13,17,19 in the sense that 2n 1 23 is also a prime and hence the corresponding Fermat equation may have only one type (Class.2) of solutions due to a theorem of Germaine Sophie[1],[2]. In this contribution, we will give a simple proof for the exponent n 11 based on elementary mathematics. The Darbrusow identity[1] that we will use in the proof can be obtained as Darbrusow did using the multinomial theorem on three components[1]. In our proof, it is assumed that the Fermat equation 11 11 11 z y x , (x, y) 1 has non-trivial integer solutions for (x, y, z) and the parametric solution of the equation is obtained using elementary mathematics. The proof of the theorem is done by showing that the necessary condition that must be satisfied by the parameters is never satisfied.