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?