Some time ago I asked this question about MSE here. After placing a bounty, it got some attention, but unfortunately it still has to be resolved. After receiving some advice from MO Meta, I decided to post the question here (note that this is the same as the area just multiplied by two and $ a_n, b_n> 0 $).

Tut,

$$ frac {a_1b_1} {c_1 ^ 2} + frac {a_2b_2} {c_2 ^ 2} = frac {a_3b_3} {c ^ 2_3} $$

for three different primitive Pythagorean triples $ (a_n, b_n, c_n) $?

My personal belief is that this doesn't happen and I'm actively trying to refute it. I would also welcome a counterexample.

**Some results so far:**

User @mathlove on MSE found the following necessary condition

The following is a necessary condition for $ c_i. $

This is necessary for every prime number $ p $. $$ nu_p (c_1)

leq nu_p (c_2) + nu_p (c_3) $$ $$ nu_p (c_2) le nu_p (c_3) + nu_p (c_1) $$

$$ nu_p (c_3) le nu_p (c_1) + nu_p (c_2) $$ Where $ nu_p (c_i) $ is the

Exponent of $ p $ in the prime factorization of $ c_i $,

(You can find proof here)

To search for these values, I have created a comprehensive algorithm to search for these triples with the help of here. I found that

To the $ c ^ 2 <10 ^ {14} $

$$ frac {a_1b_1} {c_1 ^ 2} + frac {a_2b_2} {c_2 ^ 2} neq frac {a_3b_3} {c ^ 2_3} $$

and,

$$ frac {1} {c_1 ^ 2} + frac {1} {c_2 ^ 2} neq frac {1} {c ^ 2_3} $$

Note that the difficulty of finding these triples appears to be due to the division by the square of the hypotenuse, as there are many solutions $ a_1b_1 + a_2b_2 = a_3b_3 $, At first it seemed as if it was extremely unlikely that these solutions would occur (ratios match perfectly), which is why nothing was found, but it looks like there is a little more to it. Due to a bug in my original code, I accidentally looked for solutions to it.

$$ frac {a_1b_1} {c_1} + frac {a_2b_2} {c_2} = frac {a_3b_3} {c_3} $$

What resulted in these very interesting values for $ c <10 ^ 7 $.

$$ frac {3 * 4} {5} + frac {20 * 21} {29} = frac {17 * 144} {145} $$

$$ frac {20 * 21} {29} + frac {119 * 120} {169} = frac {99 * 4900} {4901} $$

$$ frac {119 * 120} {169} + frac {696 * 697} {985} = frac {577 * 166464} {166465} $$

$$ frac {696 * 697} {985} + frac {4059 * 4060} {5741} = frac {3363 * 5654884} {5654885} $$

This pattern has a clearly defined structure. Note the recursive nature, where one of the terms always comes from the sum of the previous ones. In addition, the LHS meters are both individual and on the RHS $ b_3 $ and $ c_3 $ are also a distance from each other.

I found this recently and didn't have much time to learn, but I found that the values all have a corresponding OEIS sequence. How could this help to refute the original statement?

**Background and motivation**

A solution in one way or another to the original question could help solve a few (probably not so important) but annoying open problems in number theory. I am preparing a website that I will link to sometime to get the full background. However, it is too long for this post and I will omit it according to META MO's recommendations to keep it as short as possible. Also, I'm not a research-level mathematician. Please forgive unintentional ignorance when replying to comments.