Annual Research Symposium (ARS)
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Item On the systematic of anomalous absorption of partial waves by nuclear optical potential(Faculty of Graduate Studies, University of Kelaniya, 2008) Amarasinghe, D.; Munasinghe, J.M.; Piyadasa, R.A.D.An interesting phenomenon relating to the nuclear optical potential was discovered (Kawai M & Iseri Y,(1985)) [1] which is called the anomalous absorption of partial waves by the nuclear optical potential. They found, by extensive computer calculations, that, for a special combinations of the total angular momentum (j) ,angular momentum(/) ,energy (E) and the target nuclei(A), the elastic S-matrix elements corresponding to nucleon elastic scattering become zero. This phenomenon is universal for light ion elastic scattering on composite nuclei. [2]. It is very interesting that this phenomenon occurs for the realistic nuclear optical potential and it exhibits striking systematic in various parameter planes. For example, all nuclei which absorb a partial l waves of a definite node lie along a straight in the plane (Re, A 3 ) as shown in the figure , where Re is the closest approach and A is mass number of the target nucleus. Theoretical description of this systematic has been actually very difficult, though attempts have been made by the Kyushu group in Japan. In this contribution, we explain mathematically the most striking systematic of this phenomenon. Explanation of the systematic Partial wave· u 1 ( k, r) of angular momentum I and incident wave number k satisfies the Schrodinger equation d21 + [ k2 _ l (l : l ) _ 21 {V(r)+iW(r)}] u,(k,r)= 0 dr r 1i , where V (r) is the total real part and W (r) is the total imaginary part of the optical potential. Starting from this equation , one obtains (1) lu1(k,r)I2=2 XI du, 12 -g(ru1(r2Jdr (2) dr dr 0 where g(r)= [k2-�� V(r)-l(l r:1)J. If u1(k, r) is the anomalously absorbed partial wave, the corresponding S-matrix element is zero and hence in the asymptotic region I u1(k,r) I is almost constant. Therefore [1;1' -g(ru1(k,rt ] o (3) for large r. Now, from (1) and (3), it is not difficult to obtain[3] the equation _1 !!I 12 -- g'( r) wh (r) 'Jw ( )J 12d (4) 2 u1 - ( ) + 2 h r,., u, r lu,l dr 2g r g(r u,l 0 166 Proceedings of the Annual Research Symposium 2008- Faculty of Graduate Studies University of Kelaniya which is valid for large r , and has been numerically tested in case of an anomalously absorbed partial waves , where Wh(r) =- 2 W(r) . If W(r) decays much more rapidly n than V(r) in case of a partial wave under consideration lu,l2 =- g'( ( r) ) and by lu,l dr 2g r integrating this equation with respect to r, we obtain I iu1 (k, r)i2 (g(r) 2 = C (5) ,where C is a constant, and the equation (5) is valid for large values of r. In case of anomalous absorption of the partial wave, I u 1 ( k, r ) I is constant in the asymptotic region and therefore g(r) is also constant. We have found that for all partial waves corresponding to a straight line of definite node, g(r) is constant at the respective I [l(l + 1)]2 closest approach. For example, at Re = k , g(r) is constant for all partial waves lying on a straight line in case of anomalous absorption of neutron partial waves by the nuclear optical potential. Therefore, neglecting the spin-orbit potential , we get -I I 21tV0[1+exp[([l(/+l)F -1.17A3)]/arr1 =C0 n2 k where V0 is depth of the real potential and A is the target mass and the optical potential parameter ar = 0.75 and C0 is a constant. Therefore, in case of neutron, we get the linear relation [I (I+ 1)]2 = 1.11 A + C1 k . (6) where C1 is again a constant. This relation has found to be well satisfied in the cases we have tested numerically. The equation (6) well accounts for the anomalous absorption of neutron partial waves by the Nuclear Optical Potential as shown in the figure below. : [:::::r�;;�����i����:::]::::::::::::::::::::::::] :::::::::--::::::::::::: 1 I I I I I I 7 L----------J------------L---------__ J _ ----------L----------- : ----- ------L----------- ::N :: 6 : ----------: ------------: -----------: _ ___________ :L _ -------i: --------..:---: ----------- ;::::" 5 ,1- ------ l l l : : ! - ---,------------r---- -------, ---- ------r--- --------,------------r----------- + : : : I : : : ::-::::: 4 :I- -----------;I ------ I I I I I ------:----- ---"t-------- ----:------------ 1------------:------------ 1....1 I I I I I I I 3 lI- ----------1I ------------I -----------1I ------------rI -----------1I ------------rI ----------- 2 ::- ----------1: ------------: -----------1: ------------: -----------1: ------------: ----------- I I I I I I I ,a , 2 3 4 5 6 All3 Gradient of straight line predicted by ( 6) is 1.1 7 and the actual value is 1.1828 . Very small discrepancy is due to the negligence of the spin-orbit potential.Item Singularities of the elastic S-matrix element,(Faculty of Graduate Studies, University of Kelaniya, 2008) Jayasinghe, W.J.M.L.P.; Piyadasa, R.A.D.It is well known that the standard conventional method of integral equations is not able to explain the analyticity of the elastic S-matrix element for the nuclear optical potential including the Coulomb potential. It has been shown[1],[2] that the cutting down of the potential at a large distance is essential to get rid of the redundant poles of the S-matrix element in case of an attractive exponentially decaying potential. This method has been found [3] to be quite general and it does not change the physics of the problem. Using this method , analiticity and the singularities of the S-matrix element is discussed. Singularities of the elastic S-matrix element Partial wave radial wave equation of angular momentum l corresponding to elastic scattering is given by, [ d2 - 2 /(/+ 1)] 2p [ . ] 2 + k - 2 u1(k,r)=-2 V(r)+ Vc(r)+ zW(r) u1(k,r) M r n (1) where V (r) is the real part of nuclear potential, W (r) is the imaginary part of the optical potential · (r) is the Coulomb potential, and k is the incident wave number. Energy dependence of the optical potential is usually through laboratory energy E1ah and hence it depend on k2 and therefore k2- 2 [V(r) + Vc(r) + i W(r)] is depending on k n through e . It is Well known that . (r) is independent Of k Jn Order tO make U 1 ( k, r) an entire function of k , we impose k independent boundary condition at the origin . Now, we can make use of a well known theorem of Poincare to deduce that the wave function is an entire function of k2 and hence it is an entire function of k as well. We cut off the exponential tails of the optical potential at sufficiently large Rm and use the relation -1 --d u1 =.:u ;<--l -(k,-r)--s--'-r (--k,R--m----') u-'-;(+l--- ( k,-r) u, dr u/-l(k ,r)-sr(k,Rm) uj+l(k,r) (2) to define St(k,Rm),where u,<-l(k,r) and u,<+l(k,r) stand for incoming and outgoing Coulomb wave functions respectively which are given by I (±l(k )-+· [r(/+ l+i 17) ]2 [J[2"+iU+nJ w (-2 'u1 ,r - _l . e 1 1k r ) r(Z+I-z17) +i,,/+2 (3) where Ware the Whittaker functions. In the limit Rm oo St (k,Rm) ,the nuclear part of the S-matrix element , becomes St { k) and the redundant poles removed[1 ],[2].Now, the nuclear S-matrix element , in terms of the Whittaker functions is given by 143 Proceedings of the Annual Research Symposium 2008- Faculty of Graduate Studies University of Kelaniya where w' 1 (2ikr)-(k,r) W I (2ikr) IIJ, 1+-2 in'" 1+-2 , ,r 2 Rm W. 1 (-2ikr)-(-k,r)W 1 (-2ikr) -IIJ, I+-2 -ill ' 1+-2 P1(k,r) = u;(k,r) ,and St (k) has an essential singularity at k = 0, which u1(k,r) , (4) is apparent from the Wister's definition of the Gamma function l(z) smce z= l+ 1 ±i lJ .However, this singularity has no any physical meaning and is an outcome of treating 21Jk as well defined quantity for all k including k = 0 in the corresponding r Schrodinger equation .The infinite number of zeros and poles of S- matrix element due to the Gamma functions associated with S - matrix element have to be interpreted 1 carefully. S;'(k)=O at the zeros of ---- f(l+1+i1J ) and then the total wave function reduces to [ . ] I J( '7 +i(/+I)ZZ"j uj-l(k,r)=-i f(l+1-'7) e l 2 2 W (2ikr) f(l+1+zlJ) i11,1+l2 which is also zero. Even though the corresponding energies of these states are negative since the corresponding wave number is given by ? k= z · z,z2 e- 2 n= 0' 1' 2 , ... 11 (n+l+1) they are not physically meaningful bound states as found in[1],[2] long ago. These states are unphysical since poles are redundant poles. This fact is clearly understood by the fact that all these poles are absent in the physically meaningful total S - matrix element. For large 1k1, Sin' (k) (- ) { e-21k r S(k), where S(k) = [-ik+(k)] • +2k [ik-(-k)]. smce W = e- " for large k. Therefore the S-matrix element has an essential singularity at infinity, which is on the imaginary axis. It is clear that there are no redundant poles in the total S-matrix element is free from redundant poles sinceSJ (k) =SeS t , where Se = f(l + 1 + i7J) f(l+1-i1]) .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 Simple theorem on the integral roots of special class of prime degree polynomial equations(University of Kelaniya, 2008) Piyadasa, R.A.D.Even in case of a simple polynomial x3 + l5xb + 28 = 0 , where (3, b) = 1 , it may be extremely difficult to discard the integral solutions without knowing the number b exactly .In this case, one can make use of the method of Tartaglia and Cardan [Archbold J.W.1961] and its solutions can be written as u + v,Ul:o + vw2 ,uw2 + vw, where u3, v3 are the roots of the equation x2 + 28x -125b3 = 0, and {J) is the cube root I . . [-28 ± .J282 + 500b3 J 3 of umty .Also, u or v can be wntten as 2 and this expression is obviously zero only when b = 0. Therefore if b :f. 0 , it is very difficult to determine that I k [-28±.J282+500b3J3. . = Th h 'lib I' d · h 2 IS an mteger or not . e t eorem w1 e exp rune m t e . following , is Capable of discarding all integral solutions of this equation using only one condition (3, b) = 1 . The theorem in its naive form discards all integral solutions of the polynomial . xP + pbx-cP = 0, where p is a prime and (p,b) = (p,c) = 1 Theorem xP + pbx-cP = 0 has no integral solutions if (p,b) = (p,c) = 1 , where b,c are any integers and p is any prime . Proof Proof of the theorem is based on the following lemma Lemma If (a,p) = (b,p) = 1, and ifs= aP-bP is divisible by p , then p2 dividess. This is true even when s = a P + bP and p is odd. Proof of the Lemma s=aP-a-(bP-b)+a-b and since s is divisible by p and aP-a ,bP-b are divisible by p due to Fermat's little theorem, it follows that a-b is divisible by p. aP -bP= (a-b)[(ap-l -bP-1)+b(aP-2 -bP-2)+ .. ·+bP-3(a-b)+ pbP-1)] (1) From ( 1 ), it follows that s is divisible by p2 • Proof of the lemma for a P +bP is almost the above. It is well known that the equation xP + pbx-cP = 0 (2) has either integral or irrational roots. If this equation has an integral root l, let x = l and (p,l) = 1. Then , lP -cP+ pbl= 0 . From the Lemma, it follows that p2 I CF - cP) .Therefore pI b , and this is a contradiction. Therefore equation has no integral roots which are not divisible by p .If it has an integral solution which is divisible by p , then let x=pf3k,(p, k)=l. Then we have, (p/3 k)P + pbp/3 k- cP = 0 ,and hence pI c ,which is again a contradiction since(p, c) = 1 which completes the proof. As an special case of the theorem , consider the equation x3 + 15xb + 28 = 0 (3) which can be written as x3 +1+15xb+33 =0 (4) and it is clear that this equation has no integral root l = O(mod 3) since 1 is not divisible by 3. If this equation has an integral root k which is not divisible by 3 , then k3 + 1 + 15kb + 32 = 0 from which it follows that 31 b due to the Lemma(in case of negative c ) and is a contradiction . Therefore the equation has no integral roots. In case of p = 2 , it follows from the theorem that the equation x2 - 2 bx- q2 =0 where (2, q) = 1 = (2, b) 'has no integral roots. Again from the theorem it follows that x P - pcx - p f3p a P - bP = 0 , where (p, c) = 1 = ( b, p) and p is a prime, has no integral solutions .. In particular here, p f3p a 3, bP are two components o( F ermat triples. It is easy to deduce that this equation has no integral roots. This theorem may hold for some other useful forms of polynomial equations.Item Analytical proof of Fermat's last theorem for n=4(University of Kelaniya, 2008) Piyadasa, R.A.D.Fermat's last theorem for n = 4 is usually proved [1] using the famous mathematical tool of the method of infinite descent of F ermat. In this contribution, it will be shown that the parametric solution of the polynomial equation d4 = e4 + g4, (e, g) = 1 can be obtained using a simple mathematical technique and thereby the proof of the theorem can be done, without depending on the sophisticated structure of primitive Pythagorean triples of Fermat[1] given by X= 2lm, y =!2-m2 , z = !2 + m2 ' where l >m> 0 and l, m are of opposite parity. The main objective of this contribution is to introduce a new simple mathematical technique which may be very useful in some other problems as well. If the equation (1) has a non-trivial integral solution for (x, y, z) , then one of e, g is even and we can assume that d, e, g are positive. If gis even, (d2-g2)(d2+g2)=e4 and terms in the brackets are eo-prime and hence , one writes d2 +g2 =x4 d2-g2 =y4 (2a) (2b) From these two equations, we get 2d2 = x4 + y4. (2c) Therefore (x2 -d)(x2+ d)=( d-y2 )(d + y2) and it is easy to deduce that terms in the brackets on the left-hand side or on the right-hand side of this equation may have only factor 2 in common since all numbers are odd and x, d, y are eo-prime to one another. In the following, a new simple mathematical technique is used to obtain the parametric solution for x, y ,d, g from this single equation. If x2 -d = d-y2, 2d = x2 + y2 and therefore 4d2 = x4 + y4 + 2x2 y2 , which means d2 = x2y2, and it leads to a contradiction since (d, e) = 1. Similarly we can easily show thatx2-d:t=d+y2. Now, let (d-y2)= a (x2-d) ,to obtain x2-d = ba-1(d-y2), b where(a, b) = l.Then k (x2 +d )= (d+ y2), x2+ d = ab-1( d + y2) .Now, let us form the a following two simultaneous equations, to obtain, x2-d =ba-1(d-y2) (a) x2 + d = ab-1(d + y2) (b) (3) Proceedings of the Annual Research Symposium 2008- Faculty of Graduate Studies University of Kelaniya (4) Since ( d, y) = 1 , b 2 + a 2 = dk , where k has to be determined. Then , one easily obtains 2 2ab+b2-a2 a2+b2 . 2ab = a2 -b2 + y2 k , y = , d = .Now , from(3), It follows that k k 2abx2 = (a2 +b2)d +(a2 -b2)y2 = (a2 +b2)2 +(a2 -b2)(2ab+b 2 -a2) k (5) It is clear from (5) that a and b cannot be of opposite parity since then k2 x2 y2 cannot be either odd or even. Hence a and b are both odd. and therefore k2 = 4 or 41 k2. a2 +b2 Thereforex 2y2 =(4a2b2 -(a2 -b2)2)14 =e2, d = , ab (a2 -b2)=g2 as given 2 below , which is the parametric solution of the equation (1 ), where a, bare parameters. 2 2ab+a2-b2-a2-b2 2a (a-b) 2 2 Now, x -d = = and k is a factor of a +b and if k k it is a factor of a-b , one deduces that k is 2 or a factor of a or b. Since (a, b)= 1, we conclude that k = 2 . Therefore x2 y2 = a2 b2 -( a2 -b2 )2 /4 Since (x2-y2)(x2 + y2) = x4-y4 = 2g2, which follows from(2a),(2b), it IS easy to deduce (6) Therefore a, b, (a2-b2) should be perfect squares. Now, if a=r2, b =s2, then r4 -s4 = t2 for some integers r, s, t. The famous and the only theorem that Fermat has proved is that there are no integers r, s, t satisfying r4-s4 = t2. Hence the Fermat's last theorem for n = 4 can be deduced. It is quite interesting that applying the mathematical technique used in this contribution ,we have shown[2] very easily that the equation r4-s4 = t2 has no non- trivial integral solution for r,s,t ,and then the Fermat's last theorem for n = 4 follows at once[1].Item Simple and analytical proof of Fermat's last theorem for n=3(University of Kelaniya, 2008) Piyadasa, R.A.D.It is well known that Fermat's last Theorem, in general, is extremely difficult to prove although the meaning of the theorem is very simple. It is surprising that the proof of theorem for n = 3 ,the smallest, corresponding number, given by Leonard Euler, which has been recommended for amateurs[!], is not only difficult but also has a gap in the proof. Paulo Rebenboim claims[1] that he has patched up Euler's proof, which is very difficult to understand, however. It was shown[2] that the parametric solution for the x,y,z in the equation z3 = y3 +x3,(x,y) = 1 could be obtained easily with one necessary condition that must be satisfied by the parameters. Fairly simple analytical proof of the Fermat's last theorem for n = 3 was given [2] using this necessary condition. In this contribution very much simpler proof is given ,which is very suitable for amateurs. Fermat's last theorem for n = 3 can be stated as the equation z3 = y3 + x3, (x,y) ·= 1. (1) has no non-zero integral solution for (x,y,z). If we assume a non-zero( xyz * 0) solution for (x,y,z) , then one of (x,y,z) is divisible by 3 . Since we can assume for (1) for negative integers, without loss of generality one can assume that y is divisible by 3 . Then if y = 3fJ ay, the parametric solution of (1), can be expressed as X = 3fJ a()o + 03 y = 3fJ a()o + 33fJ-1a 3 z = 33fJ-1 a3 + 3fJ a()o + 83 and a necessary condition satisfied by the parameters is ()3 -83 -2.3fJ aoB -33[3-1a 3 =0 In this equation ,() is a factor of z, 8 is a factor of x and r = B5 + 32P-I a2 Proof of the Fermat's last theorem for n = 3 (a) (b) (c) (d) Expressing 33/3-I a3 as 33/3-3 a3 + 8.33/3-3 and substituting e = 3fJ-1 g + o m (d),one gets (g- 2a)(o2 +3fl-1go+32fl-3(g2 +2ag+4a2 ))=32fl-3a3 (2) the condition,()= 3/3-l g + 8 is due to a simple lemma used in [2] from which it follows that f3 > 1 .It is easy to deduce that g-2a is divisible by 3 2/3-3 since (3, o) = 1 and fJ > 1 . If a= ± 1 , then ,g = ±2+32P-3>0. Now, (82 + 3/3-l g8 + 32f3-\g2 +. 2ag + 4a2 ) = ±1 is never satisfied since (2) can be expressed as 3/3-1 2 (8 +-- g )2 +32P-3(L+2g +4)= ±1 2 4 (3) (4) If a* ±1, we deduces from (2) that g-2a= 3 213-3s 3 where s* 1 and s is a factor of a . This is because factor of g cannot be a factor of both 8 and a since B = 3/3-l g + 8 and a is a factor of y . Hence, (5) where a= sq, (s, q) = 1. Now, this quadratic equation ing must be satisfied by 2a +3 213-3s 3. If the roots of this quadratic are a1 and a 2 =2a + 32/3-3 s3, then 82 + 4a2 .32/3-3 _ q3 2a.32f3-3 + 3p-t8 ala 2 = 3213_3 , a1 + a 2 =- 3213_3 It is easy to obtain from these two relations that (6) Hence, 2a + 3213-3s 3 is a factor of 82 + 4.a2 .32/3-3- q3 . In other words, 2a +3 2/3-3s 3 is a factor of the expression (2a)3 -(2a)2 .8s33 2/3-3 -882s3 or, -32/3-3s 3 is an integral root of the equation x3 -8s33213-3x2 -882s3= 0 (7) (8) This means that 32/3-3s 3 is a factor of 882. s3 which contradicts (3, 8)= 1 , that is (3,x) = 1, and the proof of theorem is complete. It should be emphasized that the necessary condition needed for our proof can be obtained without obtaining the parametric solution of (1 ), making the proof given here much shorter.
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