Prove $\frac{\text{Area}_1}{c_1^2}+\frac{\text{Area}_2}{c_2^2}\neq \frac{\text{Area}_3}{c_3^2}$ for all primitive Pythagorean triples

Question

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.

Answer 1

So you are looking for solutions $$ \bigl( [a_1,b_1,c_1],[a_2,b_2,c_2],[a_3,b_3,c_3]\bigr) \in (\mathbb P^2)^3(\mathbb Q) $$ to the equations $$ a_1^2+b_1^2=c_1^2,\quad a_2^2+b_2^2=c_2^2,\quad a_3^2+b_3^2=c_3^2,\quad a_1b_1c_2^2c_3^2+a_2b_2c_1^2c_3^2+a_3b_3c_1^2c_2^2=0. $$ These equations define a surface $S$ in $(\mathbb P^2)^3$. In general, we don't have very many tools for determining the rational points on algebraic surfaces, but a first question, which you should be able to answer, is to determine the geometry of $S$. For example, it has a singularity at $([0,0,1],[0,0,1],[0,0,1])$. You should find the others, resolve the singularities, and determine the Kodaira dimension of $S$. That will at least suggest what the set of solutions might look like. (A program such as Magma might be able to do this computation for you.)

Another approach is to use some of the obvious symmetries to map to a simpler surface. For example, let $\pi:S\to S$ be the map that swaps $[a_1,b_1,c_1]$ and $[a_2,b_2,c_2]$, i.e., $\pi$ is the restriction to $S$ of the automorphism of $(\mathbb P^2)^3$ that swaps the first two factors. Compute the quotient variety $S/\pi$, see if you can determine its rational points, and then study which ones (if any) lift to $S$. Similarly for the automorphism that cyclically permutes the three factors of $(\mathbb P^2)^3$, and for the full $\mathcal S_3$ permutation group.

Answer 2

Not really an answer, but a suggestion. Did you try to solve the "wrong" problem, where at the denominators you have $c_i$, not $c_i^2$? It seems quite interesting.

If we restrict ourselves to solutions of the type you found, then a Pythagorean triple with $b=a+1$ satisfies $a^2+(a+1)^2=c^2$, which writes as $(2a+1)^2-2c^2=-1$. The fundamental unit of ${\mathbb Z}[\sqrt 2]$ is $1+\sqrt 2$, of norm $-1$. Hence if $(x_n,y_n)$ are the solutions of the Pell equation $x^2-2y^2=\pm 1$, with $x_n+y_n\sqrt 2=(1+\sqrt 2)^n$, then $x_n^2-2y_n^2=(-1)^n$. So we have Pythagorean triplets $(a,a+1,c)$ if we take $2a+1=x_n$ and $c=y_n$, with $n$ odd. Such a triplet looks like $(\frac{x_n-1}2,\frac{x_n+1}2,y_n)$.

If we want a triplet with $c=b+1$ then we have the equation $a^2+b^2=(b+1)^2$ and we get $(a,b,c)=(a,\frac{a^2-1}2,\frac{a^2+1}2)$, with $a$ odd.

Then one can verify that $(a_1,b_1,c_1)=(\frac{x_{2k-1}-1}2,\frac{x_{2k-1}+1}2,y_{2k-1})$, $(a_2,b_2,c_2)=(\frac{x_{2k+1}-1}2,\frac{x_{2k+1}+1}2,y_{2k+1})$ and $(a_3,b_3,c_3)=(x_{2k},\frac{x_{2k}^2-1}2,\frac{x_{2k}^2+1}2)$, with $k\geq 2$, satisfy the equation $\frac{a_1b_1}{c_1}+\frac{a_2b_2}{c_2}=\frac{a_3b_3}{c_3}$. This checks straightforwardly if we write $x_n=\frac{\alpha^n+\beta^n}2$ and $y_n=\frac{\alpha^n-\beta^n}{2\sqrt 2}$, where $\alpha =1+\sqrt 2$ and $\beta =1-\sqrt 2=-\alpha^{-1}$.

A legitimate and nice problem you may think of is whether these are the only solutions or not. It also has some geometric meaning since $\frac{ab}c$ is the height perpendicular on $c$ in the triangle.