Generating Pythagorean Triples from Others via Dissections Roger Alperin's paper Modular Tree of Pythagoras shows it is possible to generate Pythagorean triples from others.  If $a,b,c$ are the sides of a right triangle $a^2 + b^2 = c^2$ then we can derive another one using the formula:

$$
\begin{array}{ccrcrcc} 
a  &amp;\mapsto &amp; a &amp;+&amp; 2b  &amp;+&amp; 2c \\
b  &amp;\mapsto &amp; 2a &amp;+&amp; b &amp;+&amp; 2c \\
c  &amp;\mapsto &amp; 2a &amp;+&amp; 2b &amp; +&amp; 3c
\end{array}
$$

The other two matrices can be found in Wikipedia.  Is there any way to illustrate how to generate one integer right-triangle from another using a dissection?  





Another possibility uses parametrization $(a,b,c) = (m^2 - n^2, 2mn, m^2 + n^2)$.  In this case the substitution above reduces to:

$$
\begin{array}{ccrcrcc} 
m  &amp;\mapsto &amp; m &amp;+&amp; n \\
n  &amp;\mapsto &amp;  &amp;+&amp; n 
\end{array}
$$

which is an element of $\mathrm{SL}(2,\mathbb{Z})$.  How can we visualize this shift as an action of the triangles on the plane?
You can generate all triples where $|b-a|=1$ such as $(3,4,5)\quad (20,21,29)  ...$ if you start with a seed of $(a_0,b_0,c_0)=(0,0,1)$ and apply the following formula $$a_{n+1}=3a_n+2c_n+1\quad b_{n+1}=3a_n+2c_n+2\quad c_{n+1}=4a_n+3c_n+2$$

You can also generate all triples of the subset where $GCD(a,b,c)$ is an odd square (which contains all primatives and with no trivials) for all natural numbers $(m,n)\in\mathbb{N}$ if you replace $m$ in Euclid's formula with $2m-1+n$.
Solutions Pythagorean triples: $X^2+Y^2=Z^2$

You can also submit through another Pythagorean triples: $a^2+b^2=c^2$

Here is some formula, although they can write an infinite amount. All a matter of taste.

$a=p^2-4ps+3s^2$

$b=2s(2s-p)$

$c=p^2-4ps+5s^2$

Or these:

$a=2s(s-p)$

$b=p(p-2s)$

$c=p^2-2ps+2s^2$

Then using these triples can come to others.

$X=ak^2-at^2$

$Y=bk^2-2ckt+bt^2$

$Z=ck^2-2bkt+ct^2$

$p,s,k,t$ - what some integers.