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,

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


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

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

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

However, it has come to my attention that this is a specific case which many triples of Pythagorean triangles would not fall under.

Gerry Myerson from this very site, for example, posted a quadruplet of Pythagorean triangles (essentially, 4 overlapping triplets) which for the most part does not take this form:

$(A;B;C) = (111;6160;6161),\; (M;N) = (56;55)$

$(A;B;C) = (231,2960,2969),\; (M;N) = (40;37)$

$(A;B;C) = (1320,518,1418),\; (M;N) = (37;7)$

$(A;B;C) = (280,2442,2458),\; (M;N) = (37,33)$

Q: Is there a more general formula within which all of these are specific solutions?

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    $\begingroup$ Great Question!! $\endgroup$ – MaximusFastidiousIrreverence Sep 19 '17 at 13:23
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    $\begingroup$ Thanks! I've been working on the 3x3 Magic Square of Squares problem, and I learned that a 3x3 SOS would have to be constructed from 3 Pythagorean triangles which share an area. I worked on the formula Zeleny contributed to WA, found that a 3x3 SOS could not be constructed from Pythagorean triangles of that form, and thought that ALL Pythagorean triangles sharing an area would have to take that form (in which case I would've proven that a 3x3 SOS was impossible). Then it turned out that I had just eliminated one category of SOS candidates instead of solving the problem in its entirety. $\endgroup$ – Simpson17866 Sep 19 '17 at 16:00

COMMENT.-Basically your problem is to find many positive integer solutions of the equation $$xy(x^2-y^2)=k$$ where $k$ is a natural integer (the area of a right triangle).

The given parametrization (from Enrique Zeleny) gives an infinity of sets of three triangles sharing the same area. In the case of Gerry Myerson's example corresponding to the equation $$xy(x^2-y^2)=341880$$ the value of $k$ is something large and is not given a similar parametrization for sets of four triangles. I try here to explain this last fact.

The curve $xy(x^2-y^2)=k$ has no singular points so it have genus equal to $\dfrac{(4-1)(4-2)}{2}=3$. We are facing a quartic plane curve of genus $3$ so by the celebrated Faltings's theorem proving the Mordell Conjecture this curve has only a finite number of rational points (nothing in this context about integer points which is a more difficult problem). I want to say that the posed problem is quite difficult. Just for sets of four triangles, certainly that $k=341880$ must be a kind of minimum possible therefore for sets of five,six,etc triangles the problem becomes even more difficult.And extremely difficult or impossible a parametrization like for three triangles.

The difficulty in the determination of integer points in a curve of genus $1$ can be indirectly visualized by reading the article "Integer points on Curves of Genus 1" written by Joseph H. Silverman: Journal of the London Mathematical Society, Volume s2-28, Issue 1, August 1983, p.1-7. So much more then for genus greater than $1$.

  • $\begingroup$ Wow, that certainly looks daunting, thanks. Are there at least other known special cases that give infinities of triples, or is Zeleny's the only one? $\endgroup$ – Simpson17866 Sep 20 '17 at 5:05
  • $\begingroup$ My opinion is that Zeleny's is the only one but I could be wrong. $\endgroup$ – Piquito Sep 20 '17 at 22:40

I found a formula with help which will find Pythagorean triples, if they exist, for a given area. It is the solution (for n) of the cubic equation resulting from multiplying $A\times B/2$ using Euclid's formula where

$$A=m^2-n^2\qquad B=2mn\qquad C=M^2+n^2$$

$$\frac{AB}{2}=\frac{2mn(m^2-n^2)}{2}=m^3n-mn^3=D\quad\implies\quad mn^3-m^3n+D=0$$

$\qquad \text{when solved for n, becomes}$

$$n_0=2\sqrt{\frac{m^2}{3}}\cos\biggl({\biggl(\frac{1}{3}\biggr)\arccos{\biggl(-\frac{3\sqrt{3}D}{2m^4}\biggr)}\biggr)}$$ $$n_1=2\sqrt{\frac{m^2}{3}}\cos\biggl({\biggl(\frac{1}{3}\biggr)\arccos{\biggl(\frac{3\sqrt{3}D}{2m^4}\biggr)}\biggr)}$$ $$n_2=n_1-n_0$$

where $$\lfloor\sqrt[4]{D}\rfloor\le m\le \lceil\sqrt[3]{D}\rceil$$

Any combination of $D\&m$ that yields an integer $n$ provides the $m,n$ needed to generate the triple(s).

For example given area $341880$, $\lfloor\sqrt[4]{341880}\rfloor = 25 \le m\le \space\lceil\sqrt[3]{341880}\rceil=70$ and we find

$$F(341880,37)_0=33\implies F(37,33)=(280,2442,2458)$$ $$F(341880,37)_2=7\implies F(37,7)=(1320,518,1418)$$ $$F(341880,40)_0=36\implies F(40,37)=(231,2960,2969)$$ $$F(341880,56)_0=55\implies F(56,55)=(111,6160,6161)$$

For large values of $D$ the range of $m$-values may be very large but, since $m|D$, we need only test factors of $D$ in the specified range. For example, there are only $13$ factors of $341880$ between $25$ and $70$ and, in one $15$-digit $area$ I tried, there were only $26$ factors in the range for that area.

Caveate: Euclid's formula generates primitives, doubles, and square multiples of triples but will not generate, for example $(9,12,15)$ which is $3\times(3,4,5)$ so this formula will not find an integer $n$ for $D=54$. However, if we factor $54$, we get $2\times3^3$, and the $3^2$ portion indicates that the primitive $(3,4,5)$ is to be multiplied by $3$.

BTW: $$F(840,7)_0=5\implies F(7,5)=(24,70,74)$$ $$F(840,7)_2=3\implies F(7,3)=((40,42,58)$$ $$F(840,8)_0=7\implies F(8,7)=(15,112,113)$$


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