MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
"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
up vote 5 down vote accepted

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.

share|cite|improve this answer
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 $m=2$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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