# Fermats proof that a rational square is never congruent

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?

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"a rational square is never congruent" what does it mean?! – Fedor Petrov Dec 8 '10 at 22:20
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". – Gerry Myerson Dec 8 '10 at 22:25
I changed tags, as this is not logic or algebraic geometry. – Andrés E. Caicedo Dec 8 '10 at 22:26

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.

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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. – Kevin Buzzard Dec 8 '10 at 22:53
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. – Lucas K. Dec 8 '10 at 22:53
@Lucas, you mean $x=(3/2)z$. But this would be ruled out if we take Kevin's relative primality suggestion. – Gerry Myerson Dec 8 '10 at 23:00
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. – Kevin Buzzard Dec 8 '10 at 23:19
@Keven, I was typing that at the same time as your comment. If you add the coprime restriction, you get indeed another story. – Lucas K. Dec 8 '10 at 23:28

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 $m=2$.

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