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

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?

• Something I tried (and didn't finish) this morning: your opinion on the following argument? Find all $(a,b)$ such that \begin{align} &\ \ \ \ \ \ \ xy(x-y)(x+y) \\ &=(x+a)(y+b)(x+a+b-y)(x+a+b+y) \\ &\implies \\ &0=\left[b\right] \cdot x^3+\left[3a+b\right]\cdot x^2y+\left[-b\right] \cdot xy^2+\left[-a\right] \cdot y^3 \\ &+ \left[b(3a+2b)\right] \cdot x^2+\left[(a+b)(3a+b)\right]\cdot xy + \left[-ab\right] \cdot y^2 \\ &+ \left[b(a+b)(3a+b)\right]\cdot x+\left[a(a+b)^2\right]\cdot y \\ &+\left[ab(a+b)^2\right]\end{align} and solving the cubic? Maybe the $D$ would be too nightmareish – AmateurMathPirate Sep 28 '17 at 15:12
• Well I'm away from that paper so I can't check. All I did was set $x \to x+a$ and likewise for y and b. Perhaps I did err on that. – AmateurMathPirate Sep 28 '17 at 18:26
• Yes I think I messed that one up – AmateurMathPirate Sep 28 '17 at 19:02
• more data: pastebin.com/rRBDJSGA – user326210 Apr 11 '18 at 23:41
• @user326210: Some of that data belongs to an infinite family. Kindly see answer below. – Tito Piezas III Apr 12 '18 at 16:07

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

• That is so beautiful :) I'd been trying to find a formula for the second family for so long, and thank you so much for showing me! Should we call it "The Piezas Family," or did you get it from somewhere else? – Simpson17866 Apr 12 '18 at 17:09
• Just to clarify: this infinite family accounts for all 11 examples. – Will Orrick Apr 12 '18 at 19:12
• Does the formula account for all possible equi-area pythagorean triples with the form $M_1 + N_1 = M_2 + N_2 = M_3-N_3$? Is there a proof of where the formula comes from? – user326210 Apr 12 '18 at 19:30
• @Simpson17866: It is by yours truly. :) – Tito Piezas III Apr 13 '18 at 1:07
• @WillOrrick: The case $u=1$ explains the red numbers. But I had a feeling more general $u$ could account for the other examples. I'll revise the post after I investigate it further. – Tito Piezas III Apr 13 '18 at 1:20

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.