A while ago I asked this question on MSE here. After placing a bounty it got quite a bit of attention but unfortunately it has yet to be resolved. After getting some advice from MO Meta I have decided to post the question here (note this is the same as area just multiplied by two and $a_n,b_n > 0$).

Does,

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

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

My personal belief, for what it's worth, is that this does not occur and I am actively trying to disprove it. However, I would appreciate a counter example just as much.

**Some results so far:**

User @mathlove on MSE has found the following necessary condition,

The following is a necessary condition for $c_i.$

It is necessary that for every prime $p$, $$ \nu_p(c_1) \leq \nu_p(c_2)+\nu_p(c_3)$$ $$\nu_p(c_2)\le \nu_p(c_3)+\nu_p(c_1)$$ $$\nu_p(c_3)\le \nu_p(c_1)+\nu_p(c_2)$$ where $\nu_p(c_i)$ is the exponent of $p$ in the prime factorization of $c_i$.

(Proof can be found here)

In order to search for these values I created an exhaustive algorithm to search for these triples with help from here. This is what I've found,

For $c^2 < 10^{14}$

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

and,

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

Note that the trouble finding these triples appears to come from dividing by the square of the hypotenuse as there are many solutions to $a_1b_1 + a_2b_2 = a_3b_3$. It seemed at first that these solutions may just be extremely unlikely to occur (ratios lining up perfectly) hence why nothing was found but it looks like there is a little more to this. A bug in my initial code made me accidentally search for solutions to this,

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

Which yielded 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 well defined structure to it. Notice the recursive nature where one of the terms always comes from the sum of the previous. Additionally the LHS numerators are both one apart and on the RHS the $b_3$ and $c_3$ are also a distance one from each other.

**Background and motivation**
A resolution one way or another to the original question could help to resolve a couple of (presumably not so important) but pesky open problems in number theory. I am preparing a website that I will link to eventually that will give the full background. However, it is too lengthy for this post and in accordance with advice from META MO I will omit it to keep this as brief as possible. Additionally I am not a research level mathematician so please forgive any unintended ignorance when responding to comments.

**Edit for bounty**
Joe Silverman and Constantin-Nicolae Beli have already given some good insight into the problem, I am putting a bounty on this with the hope that it will get more attention. I don't have much reputation so doing anything, even just commenting would go a long way for me. Looking at the problem as it stands I see one main problem and a subproblem that may help answer the main problem.

Main problem

Prove the original statement or find a counter example.

Subproblem

Is the solution set mentioned by Constantin-Nicolae Beli the only solutions to $\frac{a_1b_1}{c_1}+\frac{a_2b_2}{c_2} = \frac{a_3b_3}{c_3}$ and would that also be the same solution set for when the hypotenuse is squared?

**Important update**

Going back and looking at the background of where this comes from. I found that what I am asking for is equivalent to this,

$$\frac{\left(c_{1}-x_{1}\right)\left(c_{1}+x_{1}\right)}{c_{1}^2}+\frac{\left(c_{2}-x_{2}\right)\left(c_{2}+x_{2}\right)}{c_{2}^2}=\frac{\left(c_{3}-x_{3}\right)\left(c_{3}+x_{3}\right)}{c_{3}^2}$$

Where $c_n$ is the associated hypotenuse of the primitive triple and $x_n$ is an integer solution to the circle,

$$x^2+y^2=2c^2$$

I wanted to mention this connection as it is related to the solution set mentioned by Constantin-Nicolae Beli.