Square Triangles and Fermat's Last Theorem (Wed morning)

The numbers $1$ , $2$ , $3$ , and $4$ are not congruent numbers. In particular, $1$ is not. I.e., there is not rational right triangle whose area is a perfect square, i.e., there are no ``square triangles''.

1. (60 minutes) Watch the Fermat's Last Theorem movie.
2. (10 minutes) Move to computer lab / break.
3. (10 minutes) Recall that we ``know'' $1$ is not a congruent number from Monday, since $L(E_1, 1) = 0.65551\ldots \neq 0$ . But that uses a deep theorem (the proof by Kolyvagin of one direction of the Birch and Swinnerton-Dyer conjecture).
4. (70 minutes) Prove Fermat's Last Theorem for $n=4$ , and use this to deduce that $1$ is not a congruent number.

• COMPUTATION:
  • Via a direct computer search, show there are no right triangles with area $1$ and rational side lengths $(a,b,c)$ (with $c$ the hypotenuse) with the number of digits of the numerators and denominators of $a$ , $b$ , $c$ all $<100$ in absolute value.
  • Look for solutions to $y^2=x^3-x$ with both $x,y$ rational and $y\neq 0$ . Fail to find any.
• THEORY: You definitely want to work together on these problems, even possibly dividing up into groups and assigning parts to different groups, and also having people search online for help (which is allowed). Make solving all these nicely a community effort. [Credit: This sequence of problems was originally written by the Baur Bektemirov (the first ever winner of the International Math Olympiad gold medal from Kazakhstan), while he was a freshman at Harvard working with me.]
  1. Suppose $a$ , $b$ , $c$ are relatively prime integers with $a^2 + b^2 = c^2$ . Then there exist integers $x$ and $y$ with $x>y$ such that $c=x^2+y^2$ and either $a=x^2-y^2$ , $b=2xy$ or $a=2xy$ , $b=x^2-y^2$ . Hint: Recall what we did on day 1 where we parametrized Pythagorean triples $a,b,c$ :
     

  2. Fermat's Last Theorem for exponent $4$ asserts that any solution to the equation $x^4+y^4=z^4$ with $x,y,z\in\mathbb{Z}$ satisfies $xyz=0$ . Prove Fermat's Last Theorem for exponent $4$ , as follows.
     1. Show that if the equation $x^2+y^4=z^4$ has no integer solutions with $xyz\neq 0$ , then Fermat's Last Theorem for exponent $4$ is true. (Note - the power of $x$ is $2$ .)
     2. Prove that $x^2+y^4=z^4$ has no integer solutions with $xyz\neq 0$ as follows. Suppose $n^2+k^4=m^4$ is a solution with $m>0$ minimal amongst all solutions. Show that there exists a solution with $m$ smaller using the previous exercise above (consider two cases).

  3. Prove that 1 is not a congruent number by showing that the cubic curve $y^2=x^3-x$ has no rational solutions except $(0,\pm 1)$ and $(0,0)$ :
     1. Write $y=\frac{p}{q}$ and $x=\frac{r}{s}$ , where $p,q,r,s$ are all positive integers and $\gcd(p,q)=\gcd(r,s)=1$ . Prove that $s\mid q$ , so $q=sk$ for some $k\in\mathbb{Z}$ .
     2. Prove that $s=k^2$ , and substitute to see that $p^2=r^3-rk^4$ .
     3. Prove that $r$ is a perfect square by supposing that there is a prime $\ell$ such that ${\mathrm{ord}}_{\ell}(r)$ is odd and analyzing ${\mathrm{ord}}_{\ell}$ of both sides of $p^2=r^3-rk^4$ .
     4. Write $r=m^2$ , and substitute to see that $p^2=m^6-m^2k^4$ . Prove that $m\mid p$ .
     5. Divide through by $m^2$ and deduce a contradiction to the previous exercise.