Can one find explicitly $a_i,b_i,c_i\in\Bbb N,i=1,2,3,4$ so that $$ a_i<b_i, \qquad \text{ and } \qquad a_i^2+b_i^2=c^2_i \qquad\text{for } i=1,2,3,4$$ and $$a_1b_1=a_2b_2=a_3b_3=a_4b_4, \qquad c_1<c_2<c_3<c_4.$$

Some context:

A Pythagorean triple is a triple $(a,b,c)\in\Bbb N$ so that $a^2+b^2=c^2$. We say that $(a,b,c)$ is primitive if $a,b$ and $c$ are coprime. In the dedicated wikipedia article the following is written:

  • $\big((20, 21, 29), (12, 35, 37)\big)$ is the first pair of primitive Pythagorean triples such that the induced triangles have same area $=210$.

  • $\big( (4485, 5852, 7373), (3059, 8580, 9109), (1380, 19019, 19069)\big)$ is a triple of primitive Pythagorean triples such that the induced triangles have same area $=13123110$.

  • For each natural number $n$, there exist $n$ Pythagorean triples with different hypotenuses and the same area.

An equivalent formulation of the question above is: Is there any explicitly known quadruple of Pythagorean triples such that the induced triangles have same area?

Note: A093536 claims that for such a quadruple, the area in question will be $\geq 10^{17}$.

Note: These problem are equivalent for the following reasons: It follows from the basic fact that the area of a triangle associated to a Pythagorean triple $(a,b,c)$ is given by $ab/2$ and $2n^2=k^2$ has only $(0,0)$ as integer solution. BTW note that $a_i,b_i,c_i$ are coprime if and only if $\gcd(a_i,\gcd(b_i,c_i))=1$ for $i=1,\ldots,4$.

  • $\begingroup$ Write a system of nonlinear Diophantine equations and solve. For 4 it would be difficult to solve. Need to start small. $\endgroup$
    – individ
    May 8, 2015 at 17:59
  • 1
    $\begingroup$ @individ I don't even a clue on how to solve such a system for 2 (I never worked on Diophantine equations, so nonlinear ones...). Do you have some references maybe? $\endgroup$
    – idm
    May 8, 2015 at 18:26
  • 1
    $\begingroup$ Note that A093536 makes that claim only for primitive triples. $\endgroup$ Jun 13, 2017 at 23:14

3 Answers 3


It says here, "One can also find quartets of right triangles with the same area. The quartet having the smallest known area is $$(111, 6160, 6161), (231, 2960, 2969), (518, 1320, 1418), (280, 2442, 2458)$$ with area $341880$ (Beiler 1966, p. 127). Guy (1994) gives additional information."

The references are Beiler, A. H., "The Eternal Triangle," Ch. 14 in Recreations in the Theory of Numbers: The Queen of Mathematics Entertains. New York: Dover, 1966, and Guy, R. K., "Triangles with Integer Sides, Medians, and Area." §D21 in Unsolved Problems in Number Theory, 2nd ed. New York: Springer-Verlag, pp. 188-190, 1994. But there is a 3rd edition of Guy's book.

I note that these triangles are not all primitive, but you didn't ask for that.


Look at A055193 i oeis.org. I added a little in EXAMPLE about the area that gives five triangles with the same area Sture Sjöstedt sture.sjostedt(at)spray.se


As per the comment by Sture Sjöstedt, there are $5$ triples with the same area of $71831760$. You can find all triples that exist for the same area by plugging in values of $D$(area) and a range of values of 'm' using the following formula that I developed (with help) here:

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

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

For the case of $71831760$, where $$\lfloor\sqrt[4]{71831760}\rfloor=92\le m\le \lceil\sqrt[3]{71831760}\space \rceil=416$$

$n_0=f(71831760,169)=161\qquad F(169,161)=(2640,54418,54482)$ $n_2=f(71831760,169)=15\qquad F(169,15)=(28336,5070,28786)$ $n_1=f(71831760,176)=169\qquad F(176,169)=(2415,59488,59537)$

The other two non-primitive triples with the same area can be found by testing the factors of $71831760$. This is tedious and it would seem that there are no more than $3$ solutions to this cubic equation but, for the example provided by Gerry Myerson, we have $D= 341880$.

For the case of $341880$, where $$\lfloor\sqrt[4]{341880}\rfloor=24\le m\le \lceil\sqrt[3]{341880}\space \rceil=70$$ we we find

$n_0=f(341880,37)=33\qquad F(37,33)=(280,2442,2458)$ $n_2=f(341880,37)=7\qquad F(37,7)=(1320,518,1418)$ $n_1=f(341880,40)=37\qquad F(40,37)=(231,2960,2969)$ $n_1=f(341880,56)=55\qquad F(56,155)=(111,6160,6161)$


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