Generating primitive Pythagorean triples Given  this equation $a^2 + b^2 = c^2$ , in order to generate all primitive Pythagorean triples all we nedd to do is : write (a,b,c) as :
$a = 2mn $
$b = m^2-n^2$
$c= m^2+n^2$
with conditions : $\gcd(m,n)=1$ and $(m+n)$%$2= 1 $
But what about if we add a parameter $K$ to the  equation such that : $a^2 + b^2 = c^2 - K$ , with $K$ being a positive integer , how can we generate  $(a,b,c)$  in this case satisfying the equation above such that $\gcd(a,b,c) =1$  ?

We need only show how side $C$ of one triple is the same as side $A$ of another triple. For example, if we have $(3,4,5)$ and $(5,12,13)$, then
$$A_1^2+B_1^2=C_2^2-K_2^2\quad\text{ is simply }\quad 3^2+4^2=13^2-12^2$$
In this "easiest" case, $GCD(A_1,B_1,C_2)=1$.
There is a formula to identify these matches, if they exist. Let us begin with $(33,56,65)\text{ and }(63,16,65)\quad $ and find matches where $A_2=C_1=65$.
\begin{equation}
A=m^2-n^2\implies n=\sqrt{m^2-A}\qquad\text{for}\qquad  \sqrt{A+1} \le m \le \frac{A+1}{2}
\end{equation}
$$A=65\implies \lfloor\sqrt{65+1}\rfloor=8\le m \le \frac{65+1}{2} =33\quad\land\quad m\in\{9,33\}\implies n \in\{4,32\} $$
$$F(9,4)=(65,72,97)\qquad  \qquad F(33,32)=(65,2112,2113)$$
What this means is $$(33^2+56^2)=(63^2+16^2)=(97^2-72^2)=(2113^1-2112^2)$$
There are an infinite number of these because side $A$ includes any odd number greater than $1$. Also, I haven't proven it yet but all of the examples I have seen have shown that the $GCD(A,B,C)=1$ requirement is also met.
