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 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.Item Analytical Proof of Fermat's Last Theorem for n = 3(University of Kelaniya, 2007) Piyadasa, R.A.D.1. Introduction It is well known that the proof of fermat's Last Theorem, in generaL is extremely difficult. It is surprising that the proof of theorem for n = 3, the smallest corresponding number, given by Leonard Euler, which is supposed to be the simplest, is also difficult and erroneous. Paulo Rebenboin claims that he has patched up [1] the Euler's proof, which is very difficult to understand. however. In this article we present a simple and short proof of the Fermat's last theorem. Fermat's Last Theorem The equation z" = y" + x". (x, y) = 1 has no nontrivial integral solutions (x,y,z) for any prime n ;::: 3 . 2. Proof of the Fermat's last theorem for n = 3 In the following, the parametric solution to the problem based on very simple three lemmas is given. 2.1 Lemma If a 3 - b3 is divisible by Y' (p * 0) and (a, 3) = 1 = (b, 3), then (a-b) is divisible by 3~'- 1 and p;::: 2. This lemma can be easily proved substituting a- b = k in a3 - b3 and we assume it without proof. 2.2 Lemma If the equation (1) has a non trivial integral solution (x,y,z), then one of x,y,zis divisible by 3. Proof of this lemma is also simple and we assumed it without proof. Now. (1) takes the form (2) 2.3 Lemma There are two integers a and f3 such that z-x=331Ha3 , (3,a)=l. Proof of this lemma is exactly the same as in the case of analytic solution of Pythagoras' theorem [2] and let us assume it without proof. Now (2) takes the form .:3 = 33fJ a 3ry 3 + x3 , (3, z) = 1 (3) From the equation 3 y s (y + s) = x 3 - s 3 , it follows that s should be of the form 6 3 , where (6, 3) = 1 since s divides x and (s, y) = 1. Also note also that 6 3 = z- y. Then the above equation becomes 3 y63 (y +53)= x3 - 69 Now it is clear that x -15 is divisible by three. Let us consider the expression x + y- z. x + y- z = x- (z- y) = x -153 . Now consider the original equation z 3 = y 3 + x 3 , (x, y) = 1 in the form that z 3 = (x + y) (Cx + y) 2 - 3xy ). It is clear that x + y and the term, (x + y) 2 - 3xy are co-prime and therefore x + y = () 3 , where z = er; and ((), r;) = 1. Now again X+ y- z = () 3 - er; = ()(() 2 - r;) and therefore X -15 3 is divisible (). x + y- z = y- (z- x) = 3fJ ary- 33!3 a 3 = 3fJ a(ry- 32/3-l a 2 ) and therefore x -15 3 is divisible by 3 fJ a. x is divisible by 15, which follows from ( 4) and therefore x -15 3 is divisible by 3fJ a()l5. Now consider (4) m the form 3f3+ 1aryl5 3()r;=(x-15 3 Xx 2 +xl53 +15 9 ). From which one understands that x -15 3 = 3fJ a()l5 and since z- x = 3 3 ~-l a 3 . x = 3fJ aBI5 + 153 (a) y = YaBI5 + 33.LJ-l a 3 (b) z = 33/3-l a 3 + 3fJ aBI5 + 15 3 (c) In addition to this, we have x + y = () 3 and ('7- 32/3-l a 2 ) = ()15, and therefore substituting for 17 in y , we get ()3 -15 3 - 2.3.LJ aBI5- 33.LJ-t a 3 = 0 (d) Therefore by lemma (1), B-15 should be divisible 3tH. Expressing 33fJ- 1a 3 as 8.3 3!3-3 a 3 + 33!3-3 a 3 and () = 3.LJ-t g + 15 ,we obtain from (d) that (g-2a)(152 +3P-1gi5+3 2P-3 (g 2 +2ag+4a2 ))=3 2P- 3 a 3 (5) If X = (g- 2a ), Y = (15 2 + 3/3-l gl5 + 32/3-J (g 2 + 2ag )+ 4a 2 ), then it 1s clear that (3,Y) = 1, Now we prove the Fermat's last theorem for n = 3, showing that (d) is never satisfied. If 3,Li-l 2 a=1, g=2+32.LJ-J and Y=1=(15+--g-)2 +32.LJ-3(_[__+2g+4) which is never 2 4 satisfied. Similarly , the proof of the theorem follows when a = -1 smce f3 ~ 2 . If a 1:- 1, g = 2a. + 32.LJ-J q3 and () = 2a3P-I + (3 3.LJ-4 p 3 + 15). The equation (d) is of the form () 3 -3(2.3P-1a)I5B-8.33,u-3 a 3 -(33P-3 a 3 +15 3)=0 (6) and is also of the form. x 3 - 3.u.vx- u3 - v3 = 0 and therefore we can make use of the well known method of Tatagliya and Cardon (see[3] ).Then u 3 must be a solution of the quadratic x 2 + Gx - H 3 = 0 and the roots of (d) for ()are u + v, u UJ + v UJ 2 , u UJ 2 + v UJ , where UJ is the cube root of unity. It can be easily shown that this occurs only if u = 0. which gives a = 0 and this corresponds to the trivial solution x.y.z = 0 .Hence the proof of the theorem.Item Physical Interpretation of Anomalous Absorption of Partial Waves by Nuclear Optical Potentials(University of Kelaniya, 2007) Piyadasa, R.A.D.; Karunatileke, N.G.A.; Munasinghe, J.M.A formula for semi-classical elastic S-matrix element has been derived by Brink and Takigawa for a potential having three turning points with a potential barrier (see [1] ) . If S, 1 denotes the S-matrix element corresponding to angular momentum l and total angular momentum j, S1 J is given, in the usual notation, by . s: {1 + N(iE;)exp(2iS32 )} Su =exp(2zu1) N(z'c) exp( 2z.8 ) , 32 (l) where N(z) is defined by N(z) = f rxp(zln(~)) and E = -i S21 . r 1 +z n If k = ~2:~£ is the wave number corresponding to a zero of semi-classical S-matrix element, it can be shown that 1 + 1 + exp(2nc) exp (2 1· s 31 ) -_ 0 N(ic) and one obtains s31 = (2n +I) H + __!__ ln(--N_(_ic_)_J 2 2i I + exp(2nc) (2) which is a necessary and sufficient condition for the semi-classical S-matrix element to be zero. Now, S U = 0 means the absence of an outgoing wave. Since the asymptotic wave boundary condition for the corresponding partial wave U U (k,r) is given by U lJ (k,r) ~ U1H (k,r)- SuUt) (k,r), (3) where U 1(- l and U j + l stand for the incoming and outgoing Coulomb wave functions respectively. A new phenomenon was discovered by M. Kawai and Y. Iresi (See[2]) in case of elastic scattering of nucleons on composite nuclei. They found that elastic S-matrix element becomes very small for special combinations of energy (E), orbital angular momentum (!), total angular momentum (j) and target nucleus. It has been found that this phenomenon is universal for light ion elastic scattering (see[3 ]). To the zero S-matrix element corresponding to this phenomenon, we have found that 2_ ln __!!ii~) __ ~ 0 both in case of deuterons scattering on nuclei and 2i 1 + exp(27r£) 4 He scattering on 40 Ni ,which means Su = (2n + 1) 7l'. It can be shown [1] that th~ S- 2 2i.~ 2iS1 matrix element can be put into the form Su :::::: _e_ + -=---z = 178 + 171 assuming that N N I e ZiS32 I :-:; I N 12 , where 17 B and 171 stand for the amplitude of the reflected wave at the external turning point and the amplitude of the reflected wave at the innermost turning point, respectively. Then it is clear that Su = 0 is due to the fact that the destructive interference of these waves in the asymptotic region.Item Mean Value Theorem and Fermat's Last Theorem for n=3(University of Kelaniya, 2007) Piyadasa, R.A.D.Fermat's last theorem can be stated as that the equation zn =yn +xn,(x,y)=1 (A) has no non-trivial integral solutions for (x, y, z) except for n=2, and therefore we have carefully examined all primitive Pythagorean triples [1],[2] and we solved Pythagoras' equation analytically resulting in a new generators for Primitive Pythagorean triples and a simple conjecture which is explained and proved for n = 3 in the following. 1. Conjecture When y is divisible by 2, all primitive Pythagorean triples (x,y,z) are related by z 2 -x2 = y 2=2.(z-x)(x+Bh) (1) where h = z - x = 2 21H a 2 and e = .I_ . This resembles the Mean value theorem 2 f(z)- f(x) = (z- x).f'(r;), where f(x) = x 2 , / (r;) = 2.r; and r; = z + x ,which is a 2 perfect square. If (A) has a non trivial integral solutions for a prime n-:;; 2, then there may be such solutions that one of x, y, z is divisible by n which is well known .Then the equation (A) can be put into the form (2) with z- x = nf3n-Jan assuming that y is divisible by n ,where all letters except r; stand for integers .. This also resembles the Mean vale theorem f(z)- f(x) = (z- x)./ (r;) withf(x) = xn. It is conjectured [1]that if (A) is true, then rn = r;n-l does not hold witli integral y, r; except for n = 2 . 2.Proof of the Conjecture In this contribution, the conjecture in the previous section is proved for n = 3. Proof is based on the following lemmas. 2.1 Lemma If has an integral solution for x,y,z ,then one ofx, y, z is divisible by 3. The proof of this lemma is simple and it is assumed without proof. (3) Since the above equation holds for - x,-y,-z, without loss of generality, one may assume that y is divisible by 3. 2.2 Lemma If y is divisible by 3, then z- x = 3'11 1 a~ ,where a is an integer including ± I. The proof of this lemma is exactly the same as as in the case of analytic solution of the Pythagoras" equation for primitive Pythagorean triples and is also assumed without proof. ~ow, (3) takes the form (4) 2.3 Lemma If (cd) = l=(hJ). then it follmvs from the Fermat's little theorem that a' ±h1 IS divisible by 3 and since a' ±h' = (a±h)((a±h)2 ±3ah),the least power of 3 that divides a~ ± h' is 2. Substituting z-x=3'13 1a' m (4),oneobtains :> - 31>{1-1 6 + '11/1-1 + 2 )I - a .) .X X (5) and (5) can readily he put into the form 4 r' =36fl3a6 +(2x+ 3 ,;r-la')2 = 4( If ¢ is an integer it can be expressed as x +!',where ,u IS an integer, follovvs from (6) that (6) and then it 3 1' fH a 6 = ( 4 x + 2 p + 3 ' 11 1 a 1 )( 2 JL - 3 ' 11 -1 a 3 ) (7) Let us assume a is a prime for simplicity .If (2p- 3111 - 1 a') is not equal to I , then it cannot be 3611 1 a 6 since then 3x +x-I+ 2p + 33.8-l a 1 = 0 from which and (5) it follows that(3.x) :t:: I Therefore 2p- 3311 -1 a 3 is equal to a 6 or 31'!1-1 and hence or or 2p-3'11-1 a' = l and (c) gi \"CS 3611 1a' =4x+2.3'11 1a 1 +1 (a) (b) (c) (d) Since x-I or x + 1 is divisible by 32 due to (5) and since a3 + h3 is divisible by 32 we conclude that (a) or (b) or (d) is never satisfied since (3,x) :f:: 1. Hence our assumption that s and y are integers never holds . If a is a composite number it can be expressed as a product of primes and the proof of the conjecture follows in the similar manner as above.Item Anomalous Absorption of Deuteron Partial Waves by Nuclear Optical Potential(University of Kelaniya, 2006) Piyadasa, R.A.D.; Kawai, M.; Munasinghe, J.M.Kawai and Y. Iseri (2), (3) found an interesting phenomenon in nuclear physics, motivated by the work of (1), in case of nucleon-nucleus elastic scattering. In the following this phenomenon is discussed in case of neutron (n)-nucleus (A) elastic scattering. In elastic scattering of neutron on (A), the elastic S – matrix element for a particular combination of j l E A cm , , , becomes very small (almost zero), and they called this phenomenon anomalous absorption of neutron partial waves by nuclear optical potential, where j is the total angular momentum, l is the angular momentum, cm E the centre of mass energy and A the mass of the nucleus. The striking feature of this phenomenon is systematic in various parameter ( f ,l, E , A) cm planes. Among them, systematic in ⎟⎠ ⎞ ⎜⎝ ⎛ 3 1 , A k l plane is actually remarkable, which consists of straight lines. All straight lines correspond to a definite node of wave functions associated with A k E l j cm , , , . It is quite interesting to examine whether this phenomenon occurs in case of composite projectiles such as d , He , etc. Now, it has been shown (4) that this phenomenon is universal. The main purpose of this paper is to report results of the case d − A after being rescrutinized by us. It is striking that the systematic in ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + + + 3 1 2 , ( 1) A k η η l l plane is remarkably clearer than the case of neutron. Here, k η + η 2 + l (l +1) , the closest approach is physically meaningful in case of the presence of the Coulomb potential.Item On the validity of a practical three – body model(University of Kelaniya, 2006) Piyadasa, R.A.D.; Karunathilake, N.G.A.It is well known among Physicists that the classical three – body problem is not solvable, whereas, the Quantum Mechanical three–body problem is solvable due to the famous Faddeev’s work [1]. However, the problem in Faddeev’s method is not a practical method since it is not directly applicable to the simplest three–body problem. In particular, the important Coulomb potential cannot be included in a mathematically rigorous manner. Kushu group [2] developed a practical method based on [3] which has been remarkably well [4] in producing experimental results, and is now used all over the world, in case of elastic scattering of lights ions such as d, Li, etc. Which are easily breakable in scattering on composite nuclei. This method (CDCC) is simpler and it solves quantum mechanical Shrödinger equation corresponding to the three–body problem concerned. Very concise recap of this model is given in case of the three–body model n – p – A (d – A) in the following. The total wave function Ψ of the 3–body n–p–A system associated with the model, Hamiltonian H is expressed by =Σ − J M J M J M H a R 1ψ in the usual notation. Now J M ψ is expanded in the complete set of eigen functions of the deuteron sub Hamiltonians H K V (r) n p r n p = + in the usual notations. Here the ground–state wave function of (r) d φ and continuum set of wave functions { (k, r)} l φ play a vital role. Now ( ) [ ]J M l l L L l L J l l L J J M J M d J M (P , R) (r) Y (Rˆ) Y (rˆ) i (k, r) P(k),R dr Y (Rˆ) i Y (rˆ) i 0 0 0 00 = × +ΣΣ∫ × ∞ = ∞ ψ χ φ φ χ in the usual notation. t E , the total centre of mass energy, is given by d d N t m E P P k k 2 2 2 2 0 2 0 2 2 ( ) 2 h h h = + = + μ ε μ in the usual notation. Some assumptions, further, are needed. One of which in the cut off of the continuum and consider the Riemann sum over [0, k1 ], [k, k2 ],...... [ki , ki+1 ],...... [kN−1 − kN = km ]. Proceedings of the Annual Research Symposium 2006 - Faculty of Graduate Studies, University of Kelaniya 80 Still further, the following assumption is needed to do computer calculations. ( , ) ( ( ), ) ( ˆ ( ), ) ˆ ( ) 1 k r P k R dk P k R r i il J il L k k J l l L i i ∫φ χ = Δ χ φ + where ∫ + = Δ 1 ˆ ( ) ( , ) i i k k il l φ r φ k r dk This averaging procedure was drastically criticised by the experts [4] of Faddeev theory. The criticism was so drastic and that one had to answer at least on Physical grounds and which was done in [5]. The above criticism was fully answered, mainly on physical grounds, by the authors of [6] doing a the then gigantic numerical calculation. It has been now shown [7] also that CDCC method is the first order approximation to the Faddeev method. Then the question is why the first order method work so well. Answer to this question is mainly [8] and [9]. The main purpose of this paper is to justify, to a certain extent, CDCC, in a mathematical rigorous manner, by producing the correct form, which has been scrutinized by the authors, of the potential tails of CDCC and numerical support as in the following. Continuum – Continuum coupling potential ( ) , V R k k , in the usual notation, can be written as ∫ ∞ ′ = ′ 0 , 0 0 0 V (R) U (k, r) V (R, r)U (k , r) dr k k (1) in the usual notation for the simplest case of CDCC, where ( , ) ( , ) 0 0 U k r = r φ k r (2) Here ( , ) 0 φ k r defines the deuteron S – state breakup wave function of linear momentumk . Now ( , ) 2 sin( ( )) 0 U k r kr δ k π = + (3) Neglectingδ (k) , the phase shift, for the sake of simplicity, one writes V R [ (k k )r (k k )r ]V R r dr k k cos cos ( , ) 2 ( ) 1 , = ∫ − ′ − + ′ λ ′ (4) V (R, r) λ here has the usual meaning. In case of square well potential ( ) dr R r k R r k R r a r TR V V R a a k k ∫ − ′ + + ′ + − ( ) = sin 2 ( ) sin 2 ( ) ( ) 2 2 0,0 , (5) This can be readily simplified to ⎥⎦ ⎤ ⎢⎣ ⎡ ⎟⎠ ⎞ ⎜⎝ − ⎛ ⎟⎠ ⎞ ⎜⎝ = + ⎛ ′ c kR ka c kR ka R V a Vk k R cos 4 sin 4 64 cos 4 cos 4 3 16 1 3 3 2 ( ) 2 3 2 3 0,0 , π Proceedings of the Annual Research Symposium 2006 - Faculty of Graduate Studies, University of Kelaniya 81 (6) when k = k′ , under the assumption R >> a . Here c given by cka = 1. If ka >> 1, 2 3 0,0 , 3 2 ( ) R V a Vk k R π ′ = (A) When k = k′ ⎪⎭ ⎪⎬ ⎫ ⎪⎩ ⎪⎨ ⎧ ⎥⎦ ⎤ ⎢⎣ ⎡ − ′ ′ ′ − ⎥⎦ ⎤ ⎢⎣ ⎡ − ′ ′ ′ ′ = 2 2 2 3 3 3 0,0 , (2 ) cos 2 sin 2 (2 ) cos 2 sin 2 (2 ) cos 2 cos 2 (2 ) 2 cos 2 cos 2 ( ) ka kR ka k a k R k a ka kR ka k a k R k a R V a Vk k R π (B) where K′ = k′ − k and K = k′ + k (A) and (B) agrees with numerical calculations very nicely, which is depicted by the figures attached, in case of realistic potentials. In the figure 1, the diagonal potential (1 – 1) , (6 – 6) agree exactly the form, mathematically established, and figure 2 in case of non-diagonal potentials.