0
$\begingroup$

I know some ways to find some Pythagorean triples.

And I understand that if $a^2 + b^2 = c^2$ then $(a-b)^2 + (a+b)^2 = 2c^2$.

I feel like that suggests a way forward, but I cannot find that way.

Is there an algorithm to pick four whole numbers to serve as the legs of two right triangles such that the triangles are not congruent but the hypotenuses are the same whole number?

$\endgroup$
1
$\begingroup$

Let $(a_1,b_1,c_1)$ and $(a_2,b_2,c_2)$ be arbitrary non-proportional pythagotrean triples, and let $d:={\rm gcd}(c_1,c_2)$. Then $${c_2\over d}\left(a_1,b_1,c_1\right),\qquad{c_1\over d}\left(a_2,b_2,c_2\right)$$ are pythagorean triples to noncongruent right triangles with the same hypotenuse $c={c_1c_2\over d}$.

$\endgroup$
  • $\begingroup$ Awesome! It even works! Did you just work this out or did you already have it on hand? $\endgroup$ – Chaim Jan 31 at 17:23
  • $\begingroup$ @Chaim: I'm sorry, but this approach is more or less obvious. It would be more difficult to produce two primitive triples $(a_i,b_i, c)$. The above triples are not primitive. $\endgroup$ – Christian Blatter Jan 31 at 18:57
1
$\begingroup$

It is possible to find examples of two or more non-congruent right triangles with the same hypotenuse. I don't know of a systematic way to search for them (I used brute force). A few thoughts directed me to a choice of a hypotenuse to research: the hypotenuse of a primitive right triangle with integer sides must be odd; and that odd number may not have an odd number of prime factors congruent to $3\mod(4)$ if it can be represented as $m^2+n^2$. I chose to look at $1105=5\cdot 13\cdot 17$, which yielded the following three primitive Pythagorean triples: $(1105,1104,47),\ (1105,1073,264),\ (1105,943,576)$.

In the $m^2-n^2,\ 2mn,\ m^2+n^2$ representation of such triples, the first corresponds to $m=24,\ n=23$; the second to $m=33,\ n=4$; and the third to $m=32,\ n=9$.

Added by edit: Fermat proved that every prime number of the form $4n+1$ can be expressed as the sum of two squares in only one way. Primes of the form $4n+3$ cannot be expressed as the sum of two squares.

The Brahmagupta–Fibonacci identity says that $$(a^2 + b^2) (c^2 + d^2) = (a c − b d)^2 + (a d + b c)^2 = (a c + b d)^2 + (a d − b c)^2$$

This means that two suitable primes, each of which can be represented uniquely as the sum of two squares, have a product that can be represented in two distinct ways as the sum of two squares. Thus any odd number that is the product of two distinct primes, each of which has the form $4n+1$, can be the hypotenuse of two distinct primitive Pythagorean triples (as exemplified in the comment by David K). The inclusion of more suitable prime factors in the number corresponding to the hypotenuse will increase the number of ways that the resulting product can be represented as the sum of two squares (as exemplified in my earlier given example).

$\endgroup$
  • $\begingroup$ I strongly suspect that you have in fact found a method that can be made systematic. You found three triangles because your hypotenuse had three prime factors congruent to $1$ mod $4$. What happens when you try adding $29$ as a factor? Is there a pattern you can guess, maybe prove? $\endgroup$ – Ethan Bolker Feb 1 at 21:54
  • $\begingroup$ @Ethan Bolker Perhaps the method can be made algorhithmic, but the numbers get so large so fast that it out paces my ability to calculate with access to a calculator only. I also bet you don't have to include every possible prime to find examples. $\endgroup$ – Keith Backman Feb 1 at 21:57
  • $\begingroup$ There are also the primitive triangles $(33,56,65)$ and $(16,63,65)$, which I think is the smallest such example. $\endgroup$ – David K Feb 1 at 22:09
0
$\begingroup$

You know Euclide's formula that gives all primitives pythagorean triples starting from two integers $m>n>0$, and this formula say that we can find only one value $c=m^2+n^2$. Non primitive triples can be found multiplying a primitive by an integer $k$ so, for any $k$ we have different triples and we conclude that we can have only one pythagorean triple for any $c$.

$\endgroup$
  • $\begingroup$ I think that you are saying this: regardless of whether we limit ourselves to primitives, we cannot find two different triples with the same hypotenuse. Right? $\endgroup$ – Chaim Jan 31 at 17:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.