Reference to a Don Zagier Result and the Congruent Number Problem I was looking for a reference/explanation as to how Don Zagier managed to find the side lengths of a rational right triangle with area 157. There have been many literature references to the fact that Zagier was the first to find the triangle but no hard reference as to where one can look up the ideas used.
Does anyone here know where I could find this result? Thanks!
 A: Don Zagier's own explanation is here, page 4+5 (in German, my translation).

Consider the elliptic curve given by the equation $y^2=x(x+n)(x-n)$.
  If $P=(x,y)$ is an arbitrary nontrivial solution (meaning $y\neq 0$)
  of this equation, the point $P'=(x',y')$ constructed using the
  Diophantine tangent method has the property, that not only the product
  $x'(x'+n)(x'-n)$ ($=y'^2$), but also all three factors $x'$, $x'+n$,
  $x'-n$ are squares, hence $n$ is congruent. 
For the original solution
  $P$ the numbers $x$ and $x\pm n$ need not be squares, but they are
  strongly restricted up to quadratic factors. If for example $n$ is
  prime and $\equiv 5$ (mod 8), then one can easily show that each of
  these three numbers is of the form $\pm\square$, $\pm 2\cdot\square$,
  $\pm n\cdot\square$ or $\pm 2n\cdot\square$ (with $\square$ a
  rational square). This leads to the consideration of a finite number
  of cases, that have to be examined one by one. 
If for example
  $x=-A^2$, $x+n=B^2$, $x-n=-C^2$ (and in our case of $n$ prime,
  $n\equiv 5$ (mod 8) one can easily show that if there is any solution
  if must be of this form), then we need to solve the set of equations
  $C^2-B^2=2A^2$, $C^2-A^2=n$. The first equation can be immediately
  solved using the Diophantine method: it must hold that
  $A=2RS/M$, $B=(R^2-2S^2)/M$, $C=(R^2+2S^2)/M$ for suitable integers
  $R,S,M$. In this way the problem is reduced to the solvability of $M^2
> n=R^4+4S^4$. For $n=5$ the solution is evidently $M=R=S=1$
  ($\Rightarrow x=-4$, $y=6$, $x'=6\frac{97}{144}$). For other prime
  numbers $n$ one must occasionally repeat the descent one or more
  times, when in the first step one applies the Diophantine method to
  the quadratic equation $n=U^2+4V^2$ and tries to find a solution with
  $UV=\square$ ($\Rightarrow U=R^2/M$, $V=S^2/M$). For $n=157$ this
  method leads to a solution in a few steps. 
This is quite remarkable,
  as Fermat himself might have said, but the page is unfortunately too
  small to record the solution: The three rational squares have in
  numerator and denominator each almost 100 digits.

