Browsing by Author "Piyadasa, R.A.D."

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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.
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].
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.
A Different Look at the Primitive Integral Triads of z n = y n + xn (n = 2) and a Conjecture on z n - xn for any n(¹ 2)
(University of Kelaniya, 2005) Piyadasa, R.A.D.; Mallawa Arachchi, D.K.; Munasinghe, J.
The primitive Pythagorean triples (x, y, z) are now well understood [1]. However, we believe that a closer look at the solution is needed along new directions to understand the terrible difficulty in giving a simple proof for the Fermat’s last theorem. Keeping this fact in mind we look at the solutions of z 2 = y 2 + x2 , (x, y) = 1 in the following manner. (x, y, z) is a primitive Pythagorean triple if and only if x2 + y 2 = z 2 , (x, y) = 1 (1)It is obvious that one of (x, y, z) is even and it can be shown that z is never even by using (1) and substituting z = y + p, p ³ 1, in it. Now either x or y is even. If we suppose that y is even, z 2 - x2 = y 2 and then it follows that z - x = 22b -1 or z - x = 22b -1a 2 where a ,b ³ 1 and are integers. The following are examples for the justification of our point. 2 2 2 3 2 2 2 13 12 5 , 2 , 2 17 15 8 , 1 , 2 = + = - = = + = - = z x z x b b 1132 = 1122 +152 ,b = 1,a = 7 z - x = 2´ 72 Now we apply the mean value theorem of the form a2 - b2 = 2(a - b)x where a n for any prime n ³ 3 . Then, z n - xn = n (z - x)x n-1 by the mean value theorem and we conjecture that x is irrational when z - x =a nnbn-1 .
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.
The Equality of Schrödinger’s Theory and Heisenberg’s S-matrix Theory
(Research Symposium 2010 - Faculty of Graduate Studies, University of Kelaniya, 2010) Silva, H. I. R. U.; Piyadasa, R.A.D.
It is well known that the Schrödinger’s equation can be solved in few cases of physical importance [1] . Nevertheless, S-matrix theory can be used in general to describe physically important variables such as differential cross section, total cross section, etc….[2]. Since there’s no any justification of theoretical work to the best of our knowledge to verify that the Schrödinger’s theory and Heisenberg’s S-matrix theory are equivalent in case of important interacting potentials for which the Schrödinger’s equation can be solved analytically, we have used Heisenberg’s S-matrix theory and Schrödinger’s wave mechanics to justify that the two theories give exactly the same eigenvalues in cases which we have examined. To obtain them, we were able to find the discrete energy eigenvalues in closed form in Heisenberg’s theory without graphical methods.
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.
Exact Formula for the Sum of the Squares of the Bessel Function and the Neumann Function of the Same Order of Half-Odd Integer
(University of Kelaniya, 2008) Piyadasa, R.A.D.; Mallawa Arachchi, D.K.
Sum of the squares of the Bessel function and the Neumann function of the same order of half-odd integer has been found to be very useful in addressing a puzzle in nuclear physics. One approximate formula available in the literature is valid for the complex argument whose real part is greater than zero, and the absolute value of error term is undefined for half-odd integers. Another approximate formula which is valid for all complex arguments has been obtained using sophisticated mathematical method called Barnes' method. However, the error in the formula is very difficult to calculate. We have obtained exact formula for the sum of the squares of Bessel and Neumann functions of the same order of half-odd integers which is valid for all complex arguments, and its proof is also given.
Ground water regime in Tsunami affected Southern coastal area of Sri Lanka
(Sabaragamuwa University of Sri Lanka, 2006) Piyadasa, R.A.D.; Weerasinghe, K.D.N.; Liyanage, J.A.
The identity of Fermat equation and simple proof of Fermat’s last theorem for n=5
(Sri Lanka Association for the Advancement of Science, 2013) Pallewatta, P.G.M.O.; Piyadasa, R.A.D.
Integer roots of a polynomial equation and Fermat’s last theorem for n=3.5
(Research Symposium 2009 - Faculty of Graduate Studies, University of Kelaniya, 2009) Piyadasa, R.A.D.
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 m1   , 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 m1  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 .
The inverse square potential and relativistic bound states
(12th Annual Research Symposium, University of Kelaniya, 2011) Karunarathne, Sanjeewa; Piyadasa, R.A.D.
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.
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.
New Interpretation Of Primitive Pythagorean Triples And A Conjecture Related To Fermat’s Last Theorem
(University of Kelaniya, 2007) Piyadasa, R.A.D.; Munasinghe, J.; Mallawa Arachchi, D.K.; Kumara, K.H.
In this study primitive Pythagorean triples have been carefully examined and found that all of them satisfy a simple rule related to mean value theorem. It is pointed out that integral triples satisfying the equation on Fermat’s Last Theorem should satisfy a special rule related to mean value theorem. A conjecture is proposed which may lead to find a simple proof of Fermat’s Last Theorem.
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.
New Theorem on Primitive Pythagorean Triples
(University of Kelaniya, 2005) Piyadasa, R.A.D.; Karunathilake, N.G.A.
As a result of our survey on primitive Pythagorean triples, we were able to prove the following theorem: All primitive Pythagorean triples can be generated by almost one parametera , satisfyinga > 2 +1. Furthermore, a is either an integer or of the form h a = g where g and h (> 1) are relatively prime numbers. The proof of the theorem can be briefly outlined as follows: Taking z = y + p for some p ³ 1, z 2 = y 2 + x2 can be put into the form 2 2 1 1 + = + y x y p If p x a = , then the above equation can be put into the form ( )2 2 2 1+b = 1+a b ........................................................................ (1), where 2 1 = a 2 -1 b . Then the above equation can be reduced into 2 2 2 2 2 2 1 2 1 1 a a a + - = - + . In order to generate primitive triples, the above equation has to be multiplied by 4 if a is even and h 4 if h a = g . Now we are able to generate all the primitive Pythagorean triples if a satisfies the conditions of our theorem and 2 a 2 -1 is reduced to cancel 2 in the denominator whenever necessary. The condition a > 2 +1 and a is either integer or of the form = (h > 1) h a g with g and h are relatively prime odd be imposed after a careful study of the equation . In conclusion, an algorithm can be developed to determine p and y so that (( y + p), y, x) is a primitive Pythagorean triple in the order x < y < y + p for given x. A new theorem on primitive Pythagorean triples is found and it may be useful in understanding the Fermat’s Last Theorem.
On the integer roots of a special class of prime degree polynomial equations
(Research Symposium 2009 - Faculty of Graduate Studies, University of Kelaniya, 2009) Piyadasa, R.A.D.
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 1m , 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 m1   , 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.