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!

  • $\begingroup$ He computed it using Heegner points presumably (or maybe Mock Heegner points in this instance, I don't really know the distinction). $\endgroup$
    – kantelope
    Nov 10, 2015 at 18:08
  • 7
    $\begingroup$ I thought he used repeated 2-descents, which are still good enough for the 157 twist. Getting much further does require CM points, as in my paper in the first ANTS conference. (157 is "mock-Heegner" because it doesn't satisfy the "Heegner hypothesis", but ironically is included in Heegner's original application!) $\endgroup$ Nov 10, 2015 at 19:05
  • $\begingroup$ My bad, I had looked and seen it was outside the standard 2-descent range, and so figured Heegner was the obvious method. Are repeated 2-descents the equivalent of 4-descents (as Bremner and Cassels did), made easier in 2-torsion case? $\endgroup$
    – kantelope
    Nov 11, 2015 at 15:37

1 Answer 1


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.


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