Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. 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

Let $x,y,z \in \mathbb{N}$ where x is even; x,y are relatively prime and $x^{2}+y^{2}=z^{2}$. It will be tried to show that there exist $u,v \in \mathbb{N}$ relatively prime and $u> v$ and

$x=2uv, y=u^{2}-v^{2}, z=u^{2}+v^{2}$

We look at : $x^{2} = z^{2}-y^{2} = (z-y)(z+y)$ so we get the conditions:

$x= \sqrt{z-y)(z+y)} ; y= \sqrt{z^{2}-x^{2}} ; z=\sqrt{(x+y)(x+y)} $

then plug one of them into the conditions for u,v :

$x = 2uv ; \sqrt{(z-x)}\sqrt{(z+x)} = u^{2}-v^{2}; z=u^{2}+v^{2} $

Original idea was to plug x into

Now there are 4 variables and 3 equations! So this seems to be wrong.

Does anybody see the right way.

Tell me. Please.

share|cite|improve this question
Why is $x$ odd and divisible by 2? – Daniel Freedman Nov 2 '11 at 1:59
This is true IF $\gcd(x,y,z) = 1,$ also it must by $y$ that is odd since you wrote $x=2uv.$ – Will Jagy Nov 2 '11 at 1:59
Additional premise is that x,y are relatively prime – VVV Nov 2 '11 at 2:00
This is in many, many books, often of an expository nature, or beginning number theory. There is no need for any square roots. – Will Jagy Nov 2 '11 at 2:17
@VVV: "Tell me. Please" doesn't sound right to me. May I suggest that you write something else instead? – Arturo Magidin Nov 2 '11 at 3:42
up vote 2 down vote accepted

The easiest way to see this is to know that unique factorization exists in $\mathbb Z[i]$. Then, if $x$ an $y$ are relatively prime, and $x$ is even, you can show that $x+yi$ and $x-yi$ must be relatively prime in $\mathbb Z[i]$. But $(x+yi)(x-yi)=z^2$ is a perfect square, so, since the factors are relatively prime, $x+yi=e(u+vi)^2$ where $e$ is a unit (so $e\in \{ 1,-1,i,-i \}$.)

But $(u+vi)^2 = (u^2-v^2) + 2uv i$, so, since $y$ is necessarily odd, $e$ must be $i$ or $-i$, and you get that $x+yi = \pm 2uv \mp (u^2-v^2)i$

In particular, if $x,y>0$, we can see that $x=2|u||v|$ and $y=|u|^2-|v|^2$ or $|v|^2-|u|^2$.

share|cite|improve this answer

It is trivial to show that, with those expressions for u and v, that $u^2,v^2 \in \mathbb{N}$ as you are simply choosing $u^2$ to be the value halfway between $b^2$ and $c^2$. Also note that $a/2 = uv \in \mathbb{N}$, because a is even by assumption. $u^2$ and $v^2$ are coprime because otherwise (a,b,c) is not a primitive triple. Now, if $u=d\sqrt{e}$ and $v=f\sqrt{g}$ then the only way for $a=2uv$ to be an integer is if e = g = 1, since otherwise $u^2$ and $v^2$ would have a common factor. Therefore, we see that $u,v\in\mathbb{N}$.

Funny that I just spent time figuring this out last week or so in an effort to understand Fermat's descent proof for $n=4$. I'm interested if someone can simplify it.

share|cite|improve this answer

We know that $x,y,z$ can not be all even if $x$ is even and $x,y$ are relatively prime. If x is even, then z can not be even. now with $\frac{1}{2}(z-y)(z+y)=(\frac{x}{2})^{2}$. Assume the factors are not relative primes, and there is a number p dividing them. Then p divides y and z. Then the factors also have to be squares due to prime factorisation, of the form : $\frac{1}{2}(z+y)=u^{2}$ and $(\frac{1}{2}(z-y)=v^{2}$ for $u,v \in \mathbb{N}$. These both factors are relatively prime so u and v must also be relatively prime. The equations can be solved into the conditions: $x=2uv , y=u^{2}-v^{2}, z=u^{2}+v^{2}$

For $y \in \mathbb{N}$, it also holds that with $\frac{1}{2}(z+y)=u^{2}, \frac{1}{2}(z-y)=v^{2}$ $u>v$.

Is this proof correct?

share|cite|improve this answer
This is incorrect - $\frac{1}{2}(z-y)(z+y)=(\frac{x}{2})^{2}$ is not true. – Thomas Andrews Nov 2 '11 at 13:45
But $\frac{1}{2}(z-y)\frac{1}{2}(z+y)=(\frac{x}{2})^{2}$ – VVV Nov 2 '11 at 17:01
Sure, but that's not what you wrote – Thomas Andrews Nov 2 '11 at 17:02
Is it correct if I change this or is it still wrong? – VVV Nov 2 '11 at 17:03
You should really be more clear about what you are saying. But the argument is essentially correct - if $AB=C^2$, $A,B$ positive integers and relatively prime, then $A$ is a square and $B$ is a square. So you've shown that $\frac{z-y}{2}$ and $\frac{z+y}{2}$ are relatively prime, and hence must be squares, and the rest follows. – Thomas Andrews Nov 2 '11 at 17:08

It's not true. In the first place, if $x$ is odd, then $x\ne2uv$. But even if you fix that, the assertion the $u$ and $v$ are relatively prime implies that $x$, $y$, and $z$ are (nearly) relatively prime, which doesn't have to be the case.

share|cite|improve this answer
x,y are relatively prime – VVV Nov 2 '11 at 2:06
and x is even. I mistook even for odd. SoRRy – VVV Nov 2 '11 at 2:33

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.