13

Fermat solved the Diophantine equation $(x^2)^2 + (y^2)^2 = z^2$ using descent, the key step was using the Pythagorean triples:

  • $x^2 = u^2 - v^2$
  • $y^2 = 2 u v$
  • $z = u^2 + v^2$

but then it is seen that $x,v,u$ is another Pythagorean triple,

  • $x = m^2 - n^2$
  • $v = 2 m n$
  • $u = m^2 + n^2$

then the observation that $y^2 = 2^2 u m n$ (along with some coprimality conditions) implies that $u$,$m$ and $n$ are squares - which gives us the descent step needed.

I wanted to use this technique to solve $(x^2)^2 + (y^2)^2 = 2 z^2$ but it does not seem to work out. Parameterizing the curve through $(1,1,1)$ gives:

  • $x^2 = -m^2 - 2 m n + n^2$
  • $y^2 = m^2 - 2 m n - n^2$
  • $z = m^2 + n^2$

Nothing here seems to match up so well with the original problem (as it did in Fermat's case) and parameterizing through different points doesn't help either.

Is there any way to make descent work here? Was Fermat just lucky or is there a reason why descent can't work out here?

update: I found the parametrization

  • $x^2 = 2m^2 - n^2$
  • $y^2 = 2m^2 - 4mn + n^2$
  • $z = 2m^2 - 2mn + n^2$

which looks like a better candidate for descent but I haven't been able to actually perform it. Since I cannot see how to show that $n$ and $x$ are both squares.

Bart Michels
  • 26,985
  • 6
  • 59
  • 123
quanta
  • 12,733
  • 1
    @quanta: I remembered that it seems to appear in other places such as the book by A.Weil; however, I am not very sure, and maybe using some identities like the variations of the multiplicative formulae for pell's equations? – awllower Mar 12 '11 at 15:24
  • 1
    And +1 for the attempt to use the thoughts of group theory to other interesting Diophantine equations :). – awllower Mar 12 '11 at 15:26
  • Are you looking for a solution that uses descent or any solution (possibly using a somewhat similar idea)? – milcak Mar 13 '11 at 05:38
  • @milcak, in this thread I am asking how to do descent directly on this curve. In this one I am just asking for any approaches to solving it. – quanta Mar 13 '11 at 12:54
  • @quanta, I'm really not sure what you want. You've got a method for relating it to another equation and then solving that other equation by descent; can't you just "pull back" the descent on the other equation to a descent on this one? – Gerry Myerson May 07 '11 at 11:14
  • @Gerry, I don't know. It's not clear to me why descent should be so natural on the circle but not any other conics. – quanta May 07 '11 at 11:18
  • Fair enough. If you don't get an answer here by the time the bounty runs out, you might consider posting it to MathOverflow. – Gerry Myerson May 07 '11 at 12:41

1 Answers1

4

You can't show that $x$ is a square because $x$ needn't be a square. $x=y=17$, $z=289$ is a solution.

But Mordell, Diophantine Equations, page 18 proves that the only solution of $x^4+y^4=2z^2$ with $\gcd(x,y)=1$ is $x=\pm1$, $y=\pm1$. Sketch: $x$ and $y$ are both odd; $$z^4-x^4y^4=\left({x^4-y^4\over2}\right)^2;$$ now it follows from his Theorem 2, namely, $x^4-y^4=z^2$, $\gcd(x,y)=1$, has only $x=\pm1$, $y=\pm1$, or $x=\pm1$, $y=0$. He does Theorem 2 by descent.

Gerry Myerson
  • 185,413