Can someone check if this is correct?
Apply Proposition 1 to find two new pairs of Pythagorean triples with identical legs.
Proposition 1: (original image proposition 1) $$ (a,b,c)= \begin{cases} (4(2k+1),4k(k+1)-3,4k(k+1)+5)\\ (4(2k+1),16k(k+1)+3,16k(k+1)+5) \end{cases} $$
$4(2k+1)$, assume $k = 2 $
$2nm=20$, so $a=20$
$4k(k+1)–3$ = so $b=21$
$4k(k+1)+5$ so $c=29$
$n^2-m^2=21$
$n^2+m^2=29$
Solving the equations $n=5, m=2$ where $n>m>0$ $\gcd(n,m)=1$
Your proposition does find those two triples and perhaps it works for all values of $k$ but sometimes, there are more than $2$ triples with a given side length.
To find triples with matching sides A or B, see my answer to another question here. You may notice that there are more than $2$ triples with the same even side in the examples below taken from my answer link.
$$f(6,5)=(11,60,61)\qquad f(10,3)=(91,60,109)\qquad f(15,2)=(221,60,229)\qquad f(30,1)=(899,60,901)\qquad $$
$$f(7,6)=(13,84,85)\qquad f(14,3)=(187,84,205)\qquad f(21,2)=(437,84,445)\qquad f(42,1)=(1763,84,1765)\qquad $$
The subset of triples where $GCD(A,B,C)=(2x-1)^2, x\in\mathbb{N}$ includes all primitives where $GCD(A,B,C)=1$ and here is a sample where $n$ is a set number and $k$ is a count or "depth" within a set. How does your formula fit in?
$$\begin{array}{c|c|c|c|c|c|c|} n & k=1 & k=2 & k=3 & k=4 & k=5 & k=6 \\ \hline Set_1 & 3,4,5 & 5,12,13& 7,24,25& 9,40,41& 11,60,61 & 13,84,85 \\ \hline Set_2 & 15,8,17 & 21,20,29 &27,36,45 &33,56,65 & 39,80,89 & 45,108,117 \\ \hline Set_3 & 35,12,37 & 45,28,53 &55,48,73 &65,72,97 & 75,100,125 & 85,132,157 \\ \hline Set_{4} &63,16,65 &77,36,85 &91,60,109 &105,88,137 &119,120,169 & 133,156,205 \\ \hline Set_{5} &99,20,101 &117,44,125 &135,72,153 &153,104,185 &171,140,221 & 189,180,261 \\ \hline Set_{6} &143,24,145 &165,52,173 &187,84,205 &209,120,241 &231,160,281 & 253,204,325 \\ \hline \end{array}$$ In every set, all triples have $C-B=(2n-1)^2$ and the increment between values of $A$ is always $2(2n-1)$ where $n$ is also the set number. My formula for generating them is: $$A=(2n-1)^2+2(2n-1)k\quad B=2(2n-1)k+2k^2\quad C=(2n-1)^2+2(2n-1)k+2k^2$$
I gather you are trying to generate all primitives with a single variable formula. Is that true?