Do All Primitive Triples Belong to the Plato, Pythagoras and Fermat Families? After running into this terminology on the "Formulas for generating Pythagorean triples" Wikipedia page, I was curious whether all triples fit into these categories.

The article states:


  Plato: c - b = 1, Pythagoras: c - a = 2, Fermat: |a - b| = 1


Another page lists these as the first primitive triples where c&lt;100:

(3, 4, 5)       (5, 12, 13)     (8, 15, 17)     (7, 24, 25)
(20, 21, 29)    (12, 35, 37)    (9, 40, 41)     (28, 45, 53)
(11, 60, 61)    (16, 63, 65)    (33, 56, 65)    (48, 55, 73)
(13, 84, 85)    (36, 77, 85)    (39, 80, 89)    (65, 72, 97)


However, I noticed that some sets like the (36, 77, 85) triple have a very large difference between each integer. Does this mean that not all pythagorean triples are part of one of those families? And if so, is there a formula for generating all the pythagorean triples?
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: $C_1-B_1=1^2, C_2-B_2=3^2, C_3-B_3=5^2\text{ and }C_{25}-B_{25}=49^2=2401$.

$$\begin{array}{c|c|c|c|c|} 
 \text{$Set_n$}&amp; \text{$Triplet_1$} &amp; \text{$Triplet_2$} &amp; \text{$Triplet_3$} &amp; \text{$Triplet_4$}\\ \hline
\text{$Set_1$} &amp; 3,4,5 &amp; 5,12,13&amp; 7,24,25&amp; 9,40,41\\ \hline
\text{$Set_2$} &amp; 15,8,17 &amp; 21,20,29 &amp;27,36,45 &amp;33,56,65\\ \hline
\text{$Set_3$} &amp; 35,12,37 &amp; 45,28,53 &amp;55,48,73 &amp;65,72,97 \\ \hline
\text{$Set_{25}$} &amp;2499,100,2501 &amp;2597,204,2605  &amp;2695,312,2713 &amp;2793,424,2825\\ \hline
\end{array}$$
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.
Yes,

For $~a~$ is an even number then $~b= \frac{1}{2i}(a^2-i^2)~$ , where $~i~$ is an positive even integer such that $~i&lt;[a/2]~$, $~b~$ belong to $~N~$ , and $~c= b+i~$
e.g., $~a=36~$

For $~i=~$ even positive number $~2n^2; ~~~n~$ belong to $~N~$

$~~~~~~= 2,8&lt;[36/2]$

$i=2,~~~~ (36,323,325)~$

$i=8, ~~~~(36,77,85)~$

From this it is clear that for even number a, we can find more than one primitive triples, it may have Plato, pythagorean , and Fermat family triples....this is the generating formula for even primitive triples.... 

For odd generating primitive Pythagorean Triples, I will send it later on....
Regards,
Shailesh