2
$\begingroup$

I've been going through Fermats proof that a rational square is never congruent. And I've stumbled upon something I can't see why is. Fermat says: ''If a square is made up of a square and the double of another square, its side is also made up of a square and the double of another square'' Im having difficulties understanding why this is. Can anyone help me understand this?

$\endgroup$
3
  • $\begingroup$ "a rational square is never congruent" what does it mean?! $\endgroup$ Dec 8, 2010 at 22:20
  • 1
    $\begingroup$ It means it is never a "congruent number," which is equivalent to saying it is never the area of a right triangle with rational sides. For more, search the web for "congruent number". $\endgroup$ Dec 8, 2010 at 22:25
  • 2
    $\begingroup$ I changed tags, as this is not logic or algebraic geometry. $\endgroup$ Dec 8, 2010 at 22:26

2 Answers 2

5
$\begingroup$

In other words, Fermat is saying that if $x^2=y^2+2z^2$, then $x=c^2+2d^2$ for some $c$, $d$. I take it you know how to show that the solutions of $x^2+y^2=z^2$ are given by $x=2kmn$, $y=(m^2-n^2)k$, $z=(m^2+n^2)k$. Maybe if you subject $x^2=y^2+2z^2$ to the same kind of analysis, you get Fermat's claim.

$\endgroup$
5
  • 2
    $\begingroup$ One will surely need some assumptions on $x$ for this to be the case. Stupid comment: don't want $x<0$. Slightly less stupid comment: if $x^2=y^2+2z^2$ then I can multiply $x$, $y$ and $z$ by some random large number and get another solution (and it's certainly not the case that all multiples of a given number will be of the form $c^2+2d^2$). So perhaps some "$x$, $y$ and $z$ are coprime positive integers" assumption is needed as well. Hmm...and I bet that does it. $\endgroup$ Dec 8, 2010 at 22:53
  • $\begingroup$ I doubt if this statement of Fermat is read correctly. If you take y = z/2, then x = 2/3z. So, that is not a limitation at all. $\endgroup$
    – Lucas K.
    Dec 8, 2010 at 22:53
  • $\begingroup$ @Lucas, you mean $x=(3/2)z$. But this would be ruled out if we take Kevin's relative primality suggestion. $\endgroup$ Dec 8, 2010 at 23:00
  • $\begingroup$ Right! Lucas: you are suggesting $(x,y,z)=(3z/2,z/2,z)$ for any $z$, but if $x,y,z$ also have to be coprime positive integers then you'd better have $z=2$ and so $(x,y,z)=(3,1,2)$ which is OK. $\endgroup$ Dec 8, 2010 at 23:19
  • $\begingroup$ @Keven, I was typing that at the same time as your comment. If you add the coprime restriction, you get indeed another story. $\endgroup$
    – Lucas K.
    Dec 8, 2010 at 23:28
5
$\begingroup$

The result Fermat is using here is the following: if a number $n$ is represented primitively by the quadratic form $x^2 + my^2$, where $m = 1, \pm 2, 3$, then so is any (positive) divisor of $n$ (primitively represented means $\gcd(x,y) = 1$). Fermat had descent proofs for these claims. Lagrange later showed that if a number $n$ is represented primitively by the quadratic form $x^2 + my^2$, then any prime divisor of $n$ is represented by some (reduced) form with the same discriminant $-4m$. In Fermat's examples, the class number is $1$, and the only reduced form with discriminant $-4m$ is then the principal form.

Gerry's idea will also work, and it is instructive to find out where the method of parametrization for $m=5$ differs from the case $m=2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.