Showing that there are infinitely many integer solutions to $x^2+y^2=z^2$ Let $x,y,z \in \mathbb{N}$ with $x,y$ relatively prime, $x$ even and $x^{2}+y^{2}=z^{2}$. Show that there are infinitely many $(x,y,z)$ triplets which satisfy these conditions.
 A: It is easy to notice that $2s+1=(s+1)^2-s^2$, i.e., every odd number is a difference of two squares. Now if I take any odd square, I can get a Pythagorean triangle from this. There are infinitely many odd squares.
If I want to write this explicitly, I can put $(2k+1)^2=4k^2+4k+1=2[2k(k+1)]+1$, so I just plug $s=2k(k+1)$ into above formula and I get
$$(2k+1)^2=(2k^2+2k+1)^2-(2k^2+2k)^2$$
i.e.,
$$(2k+1)^2+(2k^2+2k)^2=(2k^2+2k+1)^2.$$
Of course I do not obtain all Pythagorean triples in this way. This are only triples where the difference of the hypotenuse and one of the legs is one. (Note that all such triples are primitive since $\gcd(s,s+1)=1$.)
A: Euclid's formula for generating Pythagorean triples states that:
$x=2mn , y=m^2-n^2 , z=m^2+n^2$
Let's show that if $\gcd(x,y)=1 \Rightarrow \gcd(m,n)=1$
suppose that $\gcd(m,n)>1$ , so we may write $m=k\cdot n$ , where $k>1 , n>1$ ,therefore:
$x=2kn^2 , y=n^2(k^2-1) \Rightarrow n^2 \mid x$ and $ n^2 \mid y \Rightarrow \gcd(x,y)>1$ which is false since we know 
that $\gcd(x,y)=1$ . This means that our assumption $\gcd(m,n)>1$ is incorrect so  $\gcd(m,n)=1$
Now let's show that $m-n$ is an odd number:
Since: $\gcd(x,y)=1$ and $x$ is even $\Rightarrow$ $m^2-n^2=(m-n)(m+n)$ must be an odd number,
therefore $m-n$ is odd number.
So we have this two conditions satisfied :
$\gcd (m,n)=1$ and $m-n$ is odd number
So we may conclude that $(x,y,z)$ is primitive 
Pythagorean triple and if you check a list of elementary properties of primitive Pythagorean triples
you will find statement that there exist infinitely many primitive Pythagorean triples whose hypotenuses are squares of natural numbers so there are infinitely many $(x,y,z)$ triples.
A: There are an infinite amount of u,v relative prime pairs with u>v , thus there are also infinite amount of numbers satisfying $\frac{1}{2}(z+y)\frac{1}{2}(z-y) = (\frac{x}{2})^{2}$ and that means there are infinite solutions of the form (x,y,z) for x even , x,y relatively prime.  ? ? ? 
A: I we use Euclid's formula
$ \quad A=m^2-k^2,\quad B=2mk,\quad C=m^2+k^2\quad$
we can see that
$$
A^2+B^2=\big((m^2-k^2)+(2mk)^2\\
=(m^4-2m^2k^2+k^4)+4m^2k^2\\
=(m^4+2m^2k^2+k^4)\\
=(m^2+k^2)^2=C^2
$$
One problem with Euclid's formula is that it generates trivial triples
like $f(1,0)=(1,0,1)\quad f(3,0)=(9,0,9)\quad \cdots\qquad$
I developed a formula that generate non-trivial Pythagorean triples for every pair of natural number $(n,k)$ and produces only the subset of triples where $GCD(A,B,C)$ is an odd square -- meaning it includes all primitives where
$GCD(A,B,C)=1^2$.
\begin{equation}
  A=(2n-1)^2+2(2n-1)k\quad B=2(2n-1)k+2k^2\quad C=(2n-1)^2+2(2n-1)k+2k^2
\end{equation}
$$A^2=(2n-1)^4+4(2n-1)^3 k+4(2n-1)^2 k^2$$
$$B^2=4(2n-1)^2 k^2+8(2n-1) k^3+4k^4$$
$$C^2=(2n-1)^4+4(2n-1)^3 k+8(2n-1)^2 k^2+8(2n-1) k^3+4k^4$$
$$A^2+B^2=(2n-1)^4+4(2n-1)^3 k+8(2n-1)^2 k^2+8(2n-1) k^3+4k^4=C^2$$
