Repository logo
Communities & Collections
All of DSpace
  • English
  • العربية
  • বাংলা
  • Català
  • Čeština
  • Deutsch
  • Ελληνικά
  • Español
  • Suomi
  • Français
  • Gàidhlig
  • हिंदी
  • Magyar
  • Italiano
  • Қазақ
  • Latviešu
  • Nederlands
  • Polski
  • Português
  • Português do Brasil
  • Srpski (lat)
  • Српски
  • Svenska
  • Türkçe
  • Yкраї́нська
  • Tiếng Việt
Log In
New user? Click here to register.Have you forgotten your password?
  1. Home
  2. Browse by Author

Browsing by Author "Perera, B.B.U.P."

Filter results by typing the first few letters
Now showing 1 - 3 of 3
  • Results Per Page
  • Sort Options
  • Thumbnail Image
    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 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 .
  • Thumbnail Image
    Item
    Uniqueness of roots of a cubic and proof of Fermat‟s last theorem for n=3
    (12th Annual Research Symposium, University of Kelaniya, 2011) Shadini, A.M.D.M.; Piyadasa, R.A.D.; Perera, B.B.U.P.
  • Thumbnail Image
    Item
    Useful identities in finding a simple proof for Fermat’s last theorem
    (Research Symposium 2010 - Faculty of Graduate Studies, University of Kelaniya, 2010) Piyadasa, R.A.D.; Shadini, A.M.D.M.; Perera, B.B.U.P.
    Fermat’s last theorem, very famous and difficult theorem in mathematics, has been proved by Andrew Wiles and Taylor in 1995 after 358 years later the theorem was stated However, their proof is extremely difficult and lengthy. Possibility of finding s simple proof, first indicated by Fermat himself in a margin of his notes , has been still baffled and main objective of this paper is, however, to point out important identities which will certainly be useful to find a simple proof for the theorem.

DSpace software copyright © 2002-2025 LYRASIS

  • Privacy policy
  • End User Agreement
  • Send Feedback
Repository logo COAR Notify