ARS 2008
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Item Structure of primitive Pythagorean triples and the proof of a Fermat's theorem(University of Kelaniya, 2008) Jayasinghe, W.J.M.L.P.; Piyadasa, R.A.D.In a short survey of survey of primitive Pythagorean triples (x,y,z) 0 < x < y < z , we have found that one of x, y, z is divisible by 5 and z is not divisible by 3, there are Pythagorean triples whose corresponding element are equal , but there cannot be two Pythagorean triples such that (x10 y��" z1 ), (x1, zP z 2 ) , where z1 and z 2 hypotenuses of the corresponding Pythagorean triples. This is due to a Fermat's theorem [1] that the area of a Pythagorean triangle cannot be a perfect square of an integer, which can directly be used to prove Fermat's last theorem for n = 4. Therefore the preceding theorem is proved using elementary mathematics, which is the one of the main objectives of this contribution. All results in this contribution are summarized as a theorem. Theorem If (x, y, z) is a primitive Pythagorean triangle, where z is the hypotenuse, then z is never divisible by 3, andJ? = O(mod3) , xyz = O(mod5) ,and there are Pythagorean triangles whose corresponding one side is the same. But there are no two Pythagorean triangles such that (x1,y"z1) , (x"z"z 2 ) ), where z"z 2 are hypotenuses. Proof of the theorem Pythagoras' equation can be written as z2 =y2+x2 ,(x,y)= 1 (1) and if z = O(mod 3) ,then since J? is not divisible by 3, z2= y2 -1+ x2 -1+ 2 .Now, it follows at once from Fermat's little theorem that z cannot be divisible by 3. If xyz is not divisible by 5, squaring (1), one obtains z4 = y4 + x4 + 2x2 y2 and hence z4 -1 = y4 -1+ x 4 -1+ 2(x2 y2 ± 1)+ t, where t =- 1 or 3. Therefore xyz = O(mod5). It is easy to obtain two Pythagorean triples whose corresponding two elements are equal, from the pair-wise disjoint sets which have recently been obtained in Ref.2. For example 3652 =3642 +2 2 3652 =3642 +2 2 3652 =3572 +�� 2 .Now, assume that there exists ' ' two primitive Pythagorean triples of the form a z = bz + cz dz =az +cz (1) (2) It is clear that a is odd and c = O(mod 3) .F rom these two equations, one obtains immediately d2 - b2 = 2c2, d2 + b2 = 2a2, and therefore d4 -b4 = 4c2a2 = w�� (3) It has been proved by Fermat, after obtaining the representation of the primitive Pythagorean triples as x = 2rs,y = r2 -s2 ,z = r2 + s2, where 0 < s < r and r,s are of opposite parity, that (3) has no non trivial integral solution for d,b, w.To prove the same in an easy manner consider the equation d2 + b2 = 2a2 in the form d2 -a2 = a2 -b2 and writing it as ( d -a)( d + a) = (a -b)( a +b) use the technique used in Ref.3 to obtain the parametric solution for d andb If d -a= a -b., then d + b = 2a , from we deduce db= a2 .This never holds since (d,a)= 1 = ( b,a) by (1) and (2).1f (d -a)!!=( a -b) , where (u, v) =1 , V then V ( d + a)-= (a +b) . From these two relations, one derives the simultaneous u equations vd -ub = a( u -v) ud+vb=a(u+v) (4a) (4b) From ( 4a),( 4b ),it is easy to deduce the relations that we need to prove the theorem as (v2 + u2 )d = [2uv + u2 -v2 ]a, (v2 + u2 )b = [2uv -(u2 -v2 )]a, (v2 +u2)(d+b))=4uva, v2 +u2 =2a, assuming that uand v are odd. Hence d -b=(u2 -v2),(d+b)=2uv. Therefore d2 -b2 =2(u2 -v2)uv=2c2 and hence u, v are perfect squares and we can find two integers g,h such that. g4 -h4 = w21 < w�� .Now, proof of the last part of the theorem follows from the method of infinite descends of Fermat. Even if u and v are of opposite parity proof of the theorem can be done in the same way. To complete the proof of a Fermat's theorem that g4 -h4 = w�� is not satisfied by any non-trivial integers, we write (g2 + h2 )(g2 -h2) = w�� , where g,h are of opposite parity, to obtain g2 + h2 = x2, g2 -h2 = y2 and x4 -y4 = 4g2 h2 = z�� , where x and y are odd and eo-prime. But, in the case of the main theorem, we have shown that this is not satisfied by any non-trivial odd x, y and even z0 numbers . This completes the proof of the Fermat's theorem we mentioned above.Item Exact formula for the sum of the squares of spherical Bessel and Neumann function of the same order(University of Kelaniya, 2008) Jayasinghe, W.J.M.L.P.; Piyadasa, R.A.D.The sum of the squares of the spherical Bessel and Neumann function of the same order (SSSBN)is the square of the modulus of the Hankel function when the argument of all function are real, and is very important in theoretical physics. However, there is no exact formula for SSSBN.Corresponding formula, which has been derived by G.N.Watson[l] is an approximate formula[!], [2] valid for Re(z) > 0 ,and it can be eo (2 k -l)!! r(v + k + !) written as J ,; (z) + N; (z) �� 2 L ( 2 ) and the error term RP satisfies JrZ k=O 2k z 2k k! f V- k + !_ 2 IR I cosvJr p! I (R(v) ,p) I 2 2P • h ?p < sm - t P cosR(vJr) (2p)! p-1 where cosh2vt = �� m! (v,m) 2m sinh2m t+ R cosht �� (2m)! P m=O Upper bound of RP in the important case when v = n + !_ , is undefined since 2 r(n+l+ m cos R(vJr) =cos VJZ' = 0 ,where R stands for the real part and m!(v, m)= ) r (n +l-m ) The same formula has been derived [l]by the method called Barne's method but the error tern is very difficult to calculate. In this contribution, we will show that an exact formula exists for SSSBN when the order of the Bessel and the Neumann function is 1 . . 2 () 2 () 2 L n (2k-l) !r (n+ k+ l) n + - ,and It can be wntten as J 1 z + N 1 z = - k 2k ( ) • 2 n+-- n+- JrZ 2 z k' r n - k + 1 Proof of the formula 2 2 k=O ' In order to show that the above formula is exact, one has to establish the identity, cosh(2n+ l)t = I r (n+l+m) 2 2111sinh2111t (1) . cosht m��of(n+ 1-m) 2m! It is an easy task to show that the equation (1) holds for n= 0 and n= 1. Now, assume that the equation ( 1) is true for n �� p. It can be easily shown that cosh(2 p + 3)t = 4 cosh(2 p + 1)t. sinh 2 t + 2 cosh(2 p + 1)- cosh(2 p -I)t (2) and hence the following formula holds. cosh(2 p + 3)t " r(p +I+ m) 2 2111+2 sinh 2111+2 t " r(p +I +m) 2 2111 sinh 2111 t p-I r(p +m) 2 2111 sinh 2111 t _ ::_:_ = :L + 2 :L - :L ----7- ---7 - - - cosh t lll=o r(p + I -m) 2m! lll=o r(p + I -m) 2m! lll=o r(p -m) 2m! =l+L...J +LJ +LJ m=l r(p-m+2) (2m-2). m=l r(p+l-m) 2m! m=l r(p-m+l) (2m-1) +P 2r(2p + 1) )22P sinh2P t r(2p) 22P sinh2p t 22(p+l) sinh2(p+l) t where p = + (p v +r(2p+1) 2p! r(2) 2 -1} 2p! It can be shown that Q = I22m sinh2m t (p +m+ 1). and P = (2p + 1)22P sinh 2P t + 22(p+I) sinh 2(p+t) t m=l (p-m+ 1) !(2m) ! Hence , cosh(2p + 3)t = f r(p +m+ 2)2m sinh2m t COSht v=O r(p + 2-m) 2(m)! Since (1) is now true for n �� p + 1 , by the mathematical induction , the equation (1) is true for all n .By Nicholson's formula[1], Jv2(z)+Nv2(z)= :2 J K0(2zsinht)cosh 2wd t (3) 0 where K0 (z) = �� Je-zcosht dt is the modified Bessel function of the second kind of the -00 zero order. Substituting for cosh(2v.t) from (1) and using we obtain n (2k-1J!r(v+k+J-) J 12(z)+ N /(z)= 2 �� 2 n+-2 n+-2 71Z L..J k 2k ( 1 ) k=O 2 Z k!r V -k + 2 from which the square of the modulus of the Hankel function follows immediately.Item Singularities of the elastic S-matrix element(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+n��J 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 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]) .