Generating Pythagorean Triples from Others via Dissections

Question

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:

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

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?

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

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

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?

Answer 1

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.

Answer 2

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 $2m-1+n$.