1   Fermat s Method of Infinite Descent  Definition   Rational  primitive triangle   A triangle is rational  if a   b   c       A triangle is primitive  if a   b   c     and are coprime   Lemma 1 1  Assuming that     is a primitive triangle  Then    is of the form  for some integers u   v   0   Proof  Without loss of generality a odd  b even  work modulo 4   This then forces c odd   Then        b 2   2   c   a 2   c   a 2       and note that all the fractions are integers  Also note that the product on the right hand side is a product of positive coprime integers   Unique prime factorisation in   gives that c   a 2   u 2  c   a 2   v 2 for some u   v       Then a   u 2   v 2  b   2 u v  c   u 2   v 2     Definition   Congruent number   D       0 is a congruent number  if there exists a rational triangle   with area         D   Note   It suffices to consider D       0 square free    Example  D   5   6 are congruent    Lemma 1 2  Assuming that   D       0  Then  D is congruent  if and only if D y 2   x 3   x for some x   y      y   0   Proof  Lemma 1 1 shows      D congruent   D w 2   u v   u 2   v 2       for some u   v   w      with w   0  Put x   u v and y   w v 2     Fermat showed that 1 is not  a congruent number   Theorem 1 3  There is no solution to      w 2   u v   u   v     u   v           for u   v   w     and w   0   Proof  Without loss of generality u   v are coprime  u   0  w   0  If v   0 then replace   u   v   w   by     v   u   w    If u   v both odd then replace   u   v   w   by   u   v 2   u   v 2   w 2     Then u   v   u   v   u   v are pairwise coprime positive integers with product a square   Unique factorisation in   gives      u   a 2   v   b 2   u   v   c 2   u   v   d 2     for some a   b   c   d       0   Since u     v   m o d 2    both c and d are odd   Then consider         c   d 2   2     c   d 2   2   c 2   d 2 2   u   a 2       This is a primitive triangle  The area is c 2   d 2 8   v 4     b 2   2   Let w 1   b 2  Lemma 1 1  gives w 1 2   u 1 v 1   u 1   v 1     u 1   v 1   for u 1   v 1      Therefore we have a new solution to       But 4 w 1 2   b 2   v   w 2 so w 1   1 2 w   So by Fermat s method of infinite descent  there is no solution to          1 1   A variant for polynomials  In Section 1  K is a field with char K   2   Write K   for the algebraic closure of K   Lemma 1 4  Assuming that   u   v   K   t   coprime    u     v is a square for 4 distinct               1  Then  u   v   K   Proof  Without loss of generality K   K     Changing coordinates on   1  we may assume the ratios           are   1   0       0   1       1     1       1         for some     K     0   1    M bius map       u   a 2 v   b 2 u   v     a   b     a   b   u     v     a     b     a     b        where   is a square root of     Unique factorisation in K   t   gives that a   b   a   b  a     b  a     b are squares  But      max   deg   a     deg   b       1 2 max   deg   u     deg   v           So Fermat s method of infinite descent  we get a contradiction  unless the degrees of u and v are zero  So u   v   K     Definition 1 5   Elliptic curve  temporary definition      i  An elliptive curve E   K is the projective closure of the plane affine curve      y 2   f   x       where f   K   x   is a monic cubic polynomial with distinct roots in K    We call this equation  a Weierstrass equation     ii  For L   K any field extension      E   L         x   y     L 2   y 2   f   x         0         0 is the  point at infinity  that we get because we take the projective closure   Fact  E   L   is naturally an abelian group   In this course  we study E   K   for K being a finite field  local field    K     p        or number field    K              Lemma 1 2 and Theorem 1 3 tells us that if E is y 2   x 3   x  then     E           0     0   0         1   0           Corollary 1 6  Let E   K be an elliptic curve  Then E   K   t       E   K     Proof  Without loss of generality K   K    By a change of coordinates  we may assume      y 2   x   x   1     x           for some     K     0   1    Suppose   x   y     E   K   t      Put x   u v  where u   v   K   t   are coprime   Then w 2   u v   u   v     u     v   for some w   K   t     Unique factorisation in K   t   gives that u   v   u   v   u     v are squares  Hence by Section 1 1   u   v   K  so x   K  so y   K