Is there a formula for this specific pattern of Pythagorean Triangles sharing an area?

Question

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?

Answer 1

A subset of your 11 data points comprise another infinite family that solves,

$$M_1 N_1 (M_1^2-N_1^2)=M_2 N_2 (M_2^2-N_2^2)=M_3 N_3 (M_3^2-N_3^2)$$

distinct from the one given by Enrique Zeleny. You didn't notice that,

• $(10,4),(12,\color{red}2),(15,\color{red}1)$
• $(20,6),(21,5),(28,2)$
• $(24,14),(35,\color{red}3),(40,\color{red}2)$
• $(42,20),(55,7),(66,4)$
• $(44,30),(70,\color{red}4),(77,\color{red}3)$
• $(56,30),(78,8),(91,5)$
• $(65,33),(88,10),(104,6)$
• $(70,52),(117,\color{red}5),(126,\color{red}4)$

The high-lighted numbers then give us a clue there is a second family. Thus, to summarize,

1st family (Zeleny's). Special case: $M_1 = M_2 = N_3$

$M_1 = r^2 + rs + s^2;\quad N_1 = r^2 - s^2$

$M_2 = r^2 + rs + s^2;\quad N_2 = 2rs + s^2$

$M_3 = r^2 + 2rs;\quad\quad\; N_3 = r^2 + rs + s^2$

2nd family. Special case: $M_1+N_1 = M_2+N_2 = M_3-N_3$

$M_1 = (u + v) (2u + 3 v);\quad N_1 = v (u + 3 v)$

$M_2 = (u + 2 v) (u + 3 v);\quad N_2 = u(u + v)$

$M_3 = (u + 2 v) (2u + 3 v);\quad N_3 = uv$

where $M_1+N_1 = M_2+N_2 = M_3-N_3 = 2 (u^2 + 3 u v + 3 v^2)$.

Answer 2

Updated 13.04.18

The area of the Pythagorean triangles is $$S = M_iN_i(M_i^2-N_i^2).\tag1$$

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

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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 $$N_1=4k,\quad d=6k^2-2,\quad C=3k^2+1,\quad N_2 = \dfrac{d-N_1}{2}=3k^2-2k-1,\tag{5a}$$

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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.

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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.