Square Triangles Revisited

Since there was so much trouble with the proof of Fermat for exponent $4$ , let's revisit it. First, Fermat's proof:

``if the area of a rational right triangle were a square, then there would also be a smaller one with the same property, and so on, which is impossible.''

OK, that wasn't so helpful. Anyway, we did everything on Thursday except 2b, i.e., suppose that $n^2+k^4=m^4$ is a solution in positive integers to the equation $x^2+y^4=z^4$ with $m>0$ minimal among all possible solutions. Show that there is a a smaller solution, hence deduce a contradiction.

1. Case 1: If $n$ is even then there exist integers $x,y\geq 0$ (with $x>y$ ) such that
   $$n = 2xy\qquad k^2 = x^2 - y^2 \qquad m^2 = x^2 + y^2.$$
   Then
   $$(mk)^2 = x^4 - y^4,$$
   from which we obtain a smaller solution.

2. Case 2: Next assume $n$ is odd. The trick is to proceed as above, but to apply exercise 1 again to the Pythagorean triple $m^2 = x^2 + y^2$ and note that various numbers are perfect squares. We give full details below.

   There are integers $x,y$ with $x>y$ such that

   $$n = x^2 - y^2\qquad k^2 = 2xy \qquad m^2 = x^2 + y^2.$$

   Since $2xy=k^2$ is a perfect square we can write either $x=u^2$ , $y=2v^2$ or $x=2u^2$ , $y=v^2$ for integers $u,v$ .

   1. Case: We have $x=u^2, y=2v^2$ , i.e., $y$ is even and $x$ is odd (note that $x$ must be odd if $y$ is even since $n=x^2-y^2$ is odd). Since $m^2 = x^2 + y^2$ is itself a Pythagorean triple with $x$ odd, we can again apply exercise 1 to find coprime integers $a,b$ such that
      $$x = a^2 - b^2\qquad y = 2ab\qquad z = a^2 + b^2.$$
      Then $2v^2 = y = 2ab$ , so since $a,b$ are coprime we have $a = d^2$ , $b=e^2$ for integers $d,e$ . Since $x=u^2$ and $x=a^2-b^2$ , we have
      $$u^2 = d^4 - b^4,$$
      which yields a smaller solution to $x^2+y^4=z^4$ .
   2. Case: We have $x=2u^2, y=v^2$ , i.e., $y$ is odd and $x$ is even (note that $y$ must be odd if $x$ is even since $n=x^2-y^2$ is odd). Since $m^2 = x^2 + y^2$ is itself a Pythagorean triple with $y$ odd, we can again apply exercise 1 to find coprime integers $a,b$ such that
      $$y = a^2 - b^2\qquad x = 2ab\qquad z = a^2 + b^2.$$
      Then $2u^2 = x = 2ab$ , so since $a,b$ are coprime and have square product we have $a = d^2$ , $b=e^2$ for integers $d,e$ . Since $y=v^2$ and $y=a^2-b^2$ , we have
      $$v^2 = d^4 - b^4,$$
      which yields a smaller solution to $x^2+y^4=z^4$ .