Congruent Numbers and Elliptic Curves: II

1. The Bijection Revisited Conceptually: (75 minutes)
   1. (2 minute) Definition of congruent number.
   2. (2 minute) Definition of elliptic curve.
   3. (40 minutes) Bijection between points and congruent triangles: revisited. In particular, here is a conceptual way to think about it. The set of rational right triangles with area $n$ is the same as the set of simultaneous solutions to the system of equations
      $$a^2 + b^2 = c^2, \qquad \frac{ab}{2} = n.$$
      This is the intersection of two quadratic surfaces in $3$ -space, hence a curve. That curve turns out, after an appropriate change of variables, to be the elliptic curve $y^2 = x^3 - n^2 x$ . Exercise: ...6 steps... (Solution presentation: 30 minute).

2. Break: (10 minute).

3. Perimeter: (15 minute) Discuss the problem of which integers are the perimeter of a right triangle with rational side lengths. Again, the set of such triangles for a given $m$ is the set of simultaneous solutions to 2 equations:
   $$a^2 + b^2 = c^2, \qquad a+b+c = m.$$
   This is the intersection of a quadratic surface and a plane, hence a quadratic curve. (Exercise - 15 minutes trying this further; presentation of solution.)

4. Generating New Points on Elliptic Curves from Other Points (30 minutes): Explain how it works. Show how to use SAGE to do it. (Exercise - 30 minutes doing this (and getting new triangles); presentations.)

• COMPUTATION:
  • Find a triangle with perimeter $6$ . Hint: rescale the $3,4,5$ triangle. How does rescaling change the perimeter?

  • Find four distinct right triangles with area $6$ .

  • Let $E$ be the elliptic curve defined by $y^2 = x^3 - 36x$ , and let $P=$ Compute $P, 2P, 3P, 4P$ by enter $E$ and $P$ into SAGE and compute the given sums. Trying computing lots of other points too.

    sage: E = EllipticCurve([-36,0])
    sage: P = E([...])
    sage: 2*P
    

• THEORY:
  • Fix an integer $n$ . Show that the rational right triangles with area $n$ are in bijection with the solutions to $y^2 = x^3 - n^2 x$ with $y\neq 0$ as follows:
    1. Think purely conceptually (don't write anything down!):
       1. Imagine the surface
          $$a^2 + b^2 = c^2$$
          in three-dimensional space.
       2. Next imagine the surface
          $$\frac{ab}{2} = n$$
          in three-dimensional space.
       3. Finally, imagine the intersection of those two surfaces. You should be visualizing a curve in three dimensional space.

    2. Solve for $a$ in the equation $ab/2 = n$ and substitute it into $a^2 + b^2 = c^2$ to obtain an equation of the curve you visualized in three space in the previous problem. You should be able to put this curve in the form
       $$4n^2 + X^4 = Y^2.$$
       (You'll have to let $X$ and $Y$ equal something involving $a$ and $b$ .)
    3. Replace $X$ by $y_1$ and $Y$ by $x_1+y_1^2$ in the equation for your curve. Simplify and get another curve (but in the variables $x_1$ and $y_1$ ):
       $$4n^2 = x_1^2 + 2 y_1^2 x_1$$
       You should be able to do all this by hand. But if you want to use a computer, here's an example of how in SAGE:

       sage: X = gp('X'); Y = gp('Y'); x_1 = gp('x_1')
       sage: y_1 = gp('y_1'); n = gp('n')
       sage: print (Y^2).subst(Y,x_1+y_1^2).subst(X, y_1)
       sage: print (4*n^2 + X^4).subst(Y,x_1+y_1^2).subst(X, y_1)
       x_1^2 + 2*y_1^2*x_1 + y_1^4
       y_1^4 + 4*n^2
       

       (In SAGE the command gp('stuff') makes a formal variable, and subst allows you to do formal substitutions.)
    4. Multiply both sides of the equation you obtain by $x_1$ , then replace $x_1$ by $x_2$ and $y_1$ by $y_2/x_2$ , to obtain:
       $$4n^2 x_2 = x_2^3 + 2y_2^2.$$

    5. Do a few additional manipulations to finally obtain the equation
       $$y^2 = x^3 - n^2x.$$

    6. If you combine everything you've done so far you get a bijection from yesterday (you do not have to show this). Moreover, you can now go back and forth between solutions to $y^2 = x^3 - n^2 x$ and rational right triangles with area $n$ . For example, try this with $n=6$ .
       1. Show that the point on the cubic $y^2 = x^3 - 36x$ that corresponds to the $3,4,5$ triangle is $(-3,9)$ .
       2. Use SAGE to find another point $Q$ on $y^2 = x^3 - 36x$ .
       3. Find the triangle corresponding to $Q$ . Verify that it really does have area $6$ .

• RESEARCH:
  • Say something about perimeters of rational right triangles. Convert this question to a question about solutions to some equation (as bove). Solve this other equation. Give a systematic way to enumerate all rational right triangles with given perimeter.