The views of these people only apply to a small number of triples and sometimes depend on which of $A or B$ is odd or even. Below are original functions gleaned from 8 million spreadsheet formulas.
These generate all triples where GCD(A,B,C) is the square of an odd number. $\mathbf {\text{This includes all primitives}}$ and excludes all non-odd-square multiples of primitives. In the following sample of sets of triplets ($Set_1, Set_2, Set_3, \text{ and }Set_{25}$), we can see a great variation in the differences between $A,B,C$. We can also see that $\mathbf {(C-B) \text{ is the }n^{th} \text{odd square}}$. In the example: $C_1-B_1=1^2, C_2-B_2=3^2, C_3-B_3=5^2\text{ and }C_{25}-B_{25}=49^2=2401$.
$$\begin{array}{c|c|c|c|c|}
\text{$Set_n$}& \text{$Triplet_1$} & \text{$Triplet_2$} & \text{$Triplet_3$} & \text{$Triplet_4$}\\ \hline
\text{$Set_1$} & 3,4,5 & 5,12,13& 7,24,25& 9,40,41\\ \hline
\text{$Set_2$} & 15,8,17 & 21,20,29 &27,36,45 &33,56,65\\ \hline
\text{$Set_3$} & 35,12,37 & 45,28,53 &55,48,73 &65,72,97 \\ \hline
\text{$Set_{25}$} &2499,100,2501 &2597,204,2605 &2695,312,2713 &2793,424,2825\\ \hline
\end{array}$$
Note: These triples can be generated by a variation of Euclid's formula where $A=(2m-1+n)^2-n^2\quad B=2(2m-1+n)n\quad C=(2m-1+n)^2+n^2\quad$ but we will use a formula developed from observations of these sets because it is easier to see that $n$ is a set number and $k$ is a member number.
Theorem:
$$\forall n,k \in \mathbb{N}, \exists A,B,C\in \mathbb{N}:A^2+B^2=C^2 \iff A=(2n-1)^2+2(2n-1)k$$
Proof: Let $$A=(2n-1)^2+2(2n-1)k$$
Solving $A^2+B^2=C^2$ for $B$ and $C$, respectively, and substituting $A$, we find that
$$B=2(2n-1)k+2 k^2$$and$$C=(2n-1)^2+2(2n-1)k+2k^2$$
We can then show that
$$A^2=(2n-1)^4+4(2n-1)^3 k+4(2n-1)^2 k^2$$
$$B^2=4(2n-1)^2 k^2+8(2n-1) k^3+4k^4$$
$$C^2=(2n-1)^4+4(2n-1)^3 k+8(2n-1)^2 k^2+8(2n-1) k^3+4k^4$$
$$A^2+B^2=(2n-1)^4+4(2n-1)^3 k+8(2n-1)^2 k^2+8(2n-1) k^3+4k^4=C^2$$
$\therefore \forall n,k \in \mathbb{N},\exists A,B,C\in \mathbb{N}:A^2+B^2=C^2 \iff A=(2n-1)^2+2(2n-1)k\text{ } \blacksquare$
The Plato family $C-B=1$ applies only to members of $Set_1$ as we can see in the sample above. The Pythagoras family $C-A=2$ applies only to the first members of all sets. I haven't heard of the Fermat family $| A-B |=1$ and it seems to apply at random as in these examples of $f(n,k)=A,B,C$.
$$f(1,1)=3,4,5$$
$$f(2,2)=21,20,29$$
$$f(4,5)=119,120,169$$
$$f(9,12)=697,696,985$$
$$f(21,29)=4059,4060,5741$$
$$f(50,70)=23661,23660,33461$$
I stopped checking at $Set_{50}$ and I don't see a pattern, do you? Do note that the first two families are nothing more than the members of $Set_1$: f(1,k) or the first members of all sets: f(n,1). I hope I have shown you there is much more to explore than the families mentioned.
If you wanted to explore other configurations, I have another set of equations for side $A$ even:$$A=2n^2+2{N}n+4kn-4n\text{, }B=2n(2k-1)+(2k-1)^2\text{, }C=2n^2+2n(2k-1)+(2k-1)^2$$ and it generates the same triplets rotated $90$ degrees in the example with $A$ and $B$ reversed.
Update: I have recently inferred, but not proven, that triples where $GCD(A,B,C)\ne 2$ and $GCD(A,B,C)\ne x^2,x\in\mathbb{N}$ cannot be generated by Euclid's formula $(A=m^2-n^2\quad B=2mn\quad C=M^2+n^2)$ alone without a multiplier, as shown here.
The only thing I don't [understand] is the last part where it's given as "One may thus equate numerators with numerators and denominators with denominators, giving Euclid's formula"
It uses unique fractionization $\Rightarrow$ uniqueness of reduced fractions (with denominators $> 0),\,$ i.e.
$\qquad\qquad \begin{align}\gcd(\color{#c00}{c,b})=1\\ \gcd(j,k)= 1\end{align}$, $\ \ \dfrac{c}b = \dfrac{j}k\ \Rightarrow\ \begin{align} c&\,=\,j\\ b &\,=\, k\end{align},\ \ \ {\rm for}\ \ b,c,j,k\in \Bbb Z,\ b,k > 0$
Follow the link for a simple proof using Euclid's Lemma (hint: $\,j = ck/b\,\Rightarrow\,\color{#c00}{b\mid c}\,k\,\Rightarrow\,b\mid k)$
Remark $ $ A more conceptual way to derive this parametrization of Pythagorean triples is to employ arithmetic of Gaussian integers $\,\Bbb Z[i] = \{ a + b\,i\,: a,b\in\Bbb Z\}$. Like integers they enjoy (Euclidean) division (with smaller remainder) and this implies they too satisfy the analog of the Fundamental Theorem of Arithmetic = existence and uniqueness of factorization into primes (= irreducibles). This implies that coprime factors of a square must themselves be squares (up to unit factors $\,\pm1,\pm i)$
Thus if $\ z^2 = x^2 + y^2 = (x-y\,i) (x+ y\,i) $ and $\,x,y\,$ are coprime then one easily checks that $\,x-y\,i,\,x+y\,i\,$ are coprime, so being coprime factors of the square $\,z^2$ they must themselves be squares (up to a unit factor). Thus e.g. $\ x + y\ i\, =\, (m + n\ i)^2 =\ m^2 - n^2 + 2mn\, i,\,$ hence $\,x = m^2-n^2,\ y = 2mn\,$ (using the unit factor $1$; using the other unit factors $\, -1,\pm i\,$ merely changes signs or swaps $\,x,y\,$ values). Notice how very simple the solution is from this perspective.
This is a simple prototypical (arithmetical) example of the simplification that results from the transformation of nonlinear problems into linear problems by working in larger algebraic extension rings. See here for some further discussion of such.
Best Answer
Starting with $(165,52,173)$, we attempt to find the $p,q$ pair which generates this triple, i.e., $p^2-q^2=165,2pq=52,p^2+q^2=173$. Clearly, we have $2q^2=173-165=8\implies q=2$ and thus $p=13$.
The ancestor of this triple arises from $(|p-2q|,q)\text{ or }(q,|p-2q|)$, whichever places $p',q'$ in largest-to-smallest order. In this case, we have $p-2q=9$ and thus the ancestor pair is $(p',q')=(9,2)$ and therefore the ancestor triple is $(p'^2-q'^2,2p'q',p'^2+q'^2)=(77,36,85)$.