TITLE: Is there a formula for this specific pattern of Pythagorean Triangles sharing an area?
QUESTION [6 upvotes]: As mentioned in my earlier question,

The basic formula for generating a Pythagorean triangle $A^2 + B^2 = C^2$ is,
$A = M^2 - N^2;\quad  B = 2MN ;\quad  C = M^2 + N^2$
And Wolfram Alpha gave me a solution (credited to an Enrique Zeleny) for three triangles which share a common area (calculated as $\frac{AB}{2}$), hence,
$$M_1 N_1 (M_1^2-N_1^2)=M_2 N_2 (M_2^2-N_2^2)=M_3 N_3 (M_3^2-N_3^2)$$

The parametric solution discussed in that previous question was based on the special case where
$M_1 = M_2 = N_3 = r^2 + rs + s^2$
but it was determined that there was no way to expand the parametric equation in a way that would cover all possible triplets.
Since then, I have identified $11$ same-area triplets which take the form
$M_1+N_1 = M_2+N_2 = M_3-N_3$
but I have been unable to identify a specific formula which would generate this relation.
The primitive data points that I have identified so far are

$(10,4),(12,2),(15,1)$
$(20,6),(21,5),(28,2)$
$(24,14),(35,3),(40,2)$
$(42,20),(55,7),(66,4)$
$(44,30),(70,4),(77,3)$
$(56,30),(78,8),(91,5)$
$(65,33),(88,10),(104,6)$
$(70,52),(117,5),(126,4)$
$(99,35),(112,22),(144,10)$
$(90,56),(136,10),(153,7)$
$(130,28),(119,39),(170,12)$

But I can't figure out how to turn these into a parametric function of $(r,s)$. When I first tried looking at the first 4, I guessed that

$(10,4),(12,2),(15,1)$ went with $(r,s) = (2,1)$
$(20,6),(21,5),(28,2)$ went with $(r,s) = (3,1)$
$(24,14),(35,3),(40,2)$ went with $(r,s) = (3,2)$
$(42,20),(55,7),(66,4)$ went with $(r,s) = (4,1)$

but quickly found that this didn't work because the first partial solution for $M_1$ that works for $(r,s) = (2,1), (3,1),$ and $(4,1)$
$M_1 = 6r^2 - 20rs + 26s^2$
didn't work for $(r,s) = (3,2)$
Clearly, I did not guess which $(r,s)$ went with which triplets of $(M,N)$ correctly, but guessing at every single possible combination – and then testing each possibile combination individually – doesn't seem feasible.

Q: Is there a way to figure out which $(r,s)$ goes with which triplets of $(M,N)$ so that I can find the formula that generates each?

REPLY [0 votes]: Updated 13.04.18
The area of the Pythagorean triangles is
$$S = M_iN_i(M_i^2-N_i^2).\tag1$$

At first, let us consider the case
$$\mathbf{M_1=M_2=N_3}.$$

Solution for $\mathbf{M_1=M_2}.$
Let 
$$M_1 = M_2 = C,\quad N_2\not=N_1,\quad C, N_1, N_2\in\mathbb N,\quad C > N_1, \quad C > N_2,\tag2$$
then, taking in account $(1),$ 
$$CN_1(C^2-N_1^2) = CN_2(C^2-N_2^2),$$
$$(N_1-N_2)C^2 = N_1^3-N_2^3,$$
$$N_2^2+N_1N_2+N_1^2-C^2 = 0,\tag3$$
The discriminant of the quadratic equation $(3)$ must be a square,
$$N_1^2 + 4(C^2-N_1^2) = d^2,$$
$$3N_1^2+d^2 = (2C)^2.\tag4$$
If $N_1$ is odd, then the solution of the Diophantine equation $(4)$ is
$$d=\dfrac{3N_1^2-1}{2},\quad C = \dfrac{3N_1^2+1}{4}.$$
$$N_1=2k+1,\quad d=6k^2+6k+1,\quad C=3k^2+3k+1,\quad N_2 = \dfrac{d-N_1}{2}=3k^2+2k,\tag5$$
If $N_1$ is even, then identity
$$48k^2+(6k^2-2)^2=(6k^2+2)^2$$
leads to the solution
$$N_1=4k,\quad d=6k^2-2,\quad C=3k^2+1,\quad N_2 = \dfrac{d-N_1}{2}=3k^2-2k-1,\tag{5a}$$

Solution for $\mathbf{M_1=N_3}.$
Let 
$$M_1 = N_3 = C,\quad C, N_1, M_3\in\mathbb N,\quad C > N_1, \quad M_3 > C,\tag6$$
then, taking in account $(1),$ 
$$CN_1(C^2-N_1^2) = M_3C(M_3^2-C^2),$$
$$(M_3+N_1)C^2 = M_3^3+N_1^3,$$
$$M_3^2 - N_1M_3 + N_1^2-C^2 = 0,\tag7$$
The discriminant of the quadratic equation $(7)$ must be a square,
$$N_1^2 + 4(C^2-N_1^2) = d^2,$$
$$3N_1^2+d^2 = (2C)^2.\tag8$$
The solution of the Diophantine equation $(8)$ is
$$d=\dfrac{3N_1^2-1}{2},\quad C = \dfrac{3N_1^2+1}{4}.$$
If $N_1$ is odd, then
$$N_1=2k+1,\quad d=6k^2+6k+1,\quad C=3k^2+3k+1,\quad M_3 = \dfrac{d+N_1}{2}=3k^2+4k+1.\tag9$$
If $N_1$ is even, then identity
$$48k^2+(6k^2-2)^2=(6k^2+2)^2$$
leads to the solution
$$N_1=4k,\quad d=6k^2-2,\quad C=3k^2+1,\quad M_3 = \dfrac{d+N_1}{2}=3k^2+2k-1,\tag{9a}$$
The formulas $(5)$ and $(9)$ can be combined, and this gives
$$\begin{cases}
M_1=M_2 = N_3 = 3k^2+3k+1\\
N_1=2k+1\\
N_2 = 3k^2+2k\\
M_3 = 3k^2+4k+1\\
k\in\mathbb N.
\end{cases}\tag{10}$$
Easy to show that the triangles have the same area.
Also the formulas $(5a)$ and $(9a)$ can be combined, and this gives
$$\begin{cases}
M_1=M_2 = N_3 = 3k^2+1\\
N_1=4k\\
N_2 = 3k^2-2k-1\\
M_3 = 3k^2+2k-1\\
k\in\mathbb N.
\end{cases}\tag{10a}$$
Easy to show that the triangles have the same area.

The family $\mathbf{M_1+N_1=M_2+N_2=M_3+N_3}.$
Let 
$$A_i = M_i+N_i,\quad B_i = M_i - N_i\tag{11}$$ then
$$4S = A_iB_i(A_i^2-B_i^2)].\tag{12}$$
Comparison of $(12)$ with $(1)$ shows that this family can be obtained using formulas $(10)-(11).$
Using formulas $(9),(9a),(10),(10a),$ one can obtain the table
$$\begin{matrix}
(M_1,N_1)&(M_2,N_2)&(M_3,N_3)&\rightarrow&(A_1,B_1)&(A_2,B_2)&(A_3,B_3)\\
(7,3)&(7,5)&(8,7)&\rightarrow&(10,4)&(12,2)&(15,1)\\
(19,5)&(19,16)&(21,19)&\rightarrow&(24,14)&(35,3)&(40,2)\\
(37,7)&(37,33)&(40,37)&\rightarrow&(44,30)&(70,4)&(77,3)\\
(13,8)&(13,7)&(15,13)&\rightarrow&(21,5)&(20,6)&(28,2)\\
(61,9)&(61,56)&(65,61)&\rightarrow&(70,52)&(117,5)&(126,4)\\
(91,11)&(91,85)&(96,91)&\rightarrow&(102,80)&(176,6)&(187,5)\\
(127,13)&(127,120)&(133,127)&\rightarrow&(140,114)&(247,7)&(260,6)\\
(169,15)&(169,161)&(176,169)&\rightarrow&(184,154)&(330,8)&(345,7)\\
(49,16)&(49,39)&(55,49)&\rightarrow&(65,33)&(88,10)&(104,6)\\
(109,24)&(109,95)&(119,109)&\rightarrow&(113,85)&(204,14)&(228,10)\\
\end{matrix}$$
Easy to see that this table contains a lot of another solutions for these two families.
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