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

Important update Yam Mir has found a more general form and Mathlove has found a necessary condition but as of now the problem is still open.

Earlier I posted this pretty gross equality that I was trying to prove, $a,b,c,d,e,f \in \mathbb{N}-0, \gcd(a,b)=1 \ \wedge \ \gcd(c,d) = 1 \ \wedge \ \gcd(e,f) = 1, (a,b) \neq (c,d) \neq (e,f)$

$$\Rightarrow \frac{4a^3b-4ab^3}{a^4+2a^2b^2+b^4} + \frac{4c^3d-4cd^3}{c^4+2c^2d^2+d^4} \neq \frac{4e^3f-4ef^3}{e^4+2e^2f^2+f^4}$$

Which I completely missed some beautiful underlying math for,

I've found that the terms can be rewritten as such,

$$\frac{4m^3n-4mn^3}{m^4+2m^2n^2+n^4}=\frac{4mn(m-n)(m+n)}{(m^2+n^2)(m^2+n^2)} = \frac{4mn(m^2-n^2)}{(m^2+n^2)^2}$$

Since it is parameterized as $\gcd(a,b) = 1 \ \wedge \ a>b>0$. It can be parametrized as a primitive Pythagorean triple!

So now let,

$$a=2mn, b=m^2-n^2,c=m^2+n^2$$

we get,

$$\frac{2a_1b_1}{c_1^2}+\frac{2a_2b_2}{c_2^2} \neq \frac{2a_3b_3}{c^2_3}$$ Where $a_n,b_n,c_n$ form a primitive Pythagorean triple dividing by four yields,

$$\frac{ab}{2c^2} = \text{Area}\cdot\frac{1}{c^2}$$

For terminology sake let's call this the characteristic ratio of a primitive Pythagorean triple. My conjecture is that for all primitive Pythagoreon triples,

$$\frac{a_1b_1}{2c_1^2}+\frac{a_2b_2}{2c_2^2}\neq \frac{a_3b_3}{2c_3^2}$$

Interestingly I've found,

$$\frac{1}{c_n^2} \approx \frac{1}{4n^2\pi^2}$$

plotting ratios from the original equation gives this curve indicating some kind of cyclical phenomenon,

Another thing I've observed,

$$\max{\frac{2a_nb_n}{c_n^2}} = 1$$

Additionally the numerator of the original inequality appears to be all congruent numbers apart of this sequence! So to sum things up I'm trying to show that,

$$\frac{\text{Area}_1}{c_1^2} + \frac{\text{Area}_2}{c_2^2} \neq \frac{\text{Area}_3}{c_3^2}$$

For all primitive Pythagorean triples or find a counter example. I'd also like to know why this may be true and if there is any regularity to the cyclical phenomenon showed? Must these ratios be unique given that primitive triples are rooted in prime factorization? What geometric meaning can be drawn from $\frac{\text{Area}}{c^2}$, why the hypotenuse squared? (note these ratio's might also flirt with the Dirichlet L-function and or elliptic curves.)

Edit @mathlove found a counter example but I unfortunately wrote the wrong parameterization failing to list $a>b>0$ so I am still looking for a different counter example. The problem is still open

Edit for bounty: To be very specific about what I'm asking for, I'd like to 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 or find a counter example. The other questions would be nice but is in no way a requirement to receive the bounty. This bounty will cost me almost $1/3$ of my reputation so even just commenting and sharing thoughts/ideas would go a long way.

Answer 1

There is a counterexample.

For $(a,b,c,d,e,f)=(1,1,1,2,1,3)$, we have

$$\frac{4a^3b-4ab^3}{a^4+2a^2b^2+b^4} + \frac{4c^3d-4cd^3}{c^4+2c^2d^2+d^4}=-\frac{24}{25}=\frac{4e^3f-4ef^3}{e^4+2e^2f^2+f^4}$$

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Added : The following is a necessary condition for $c_i.$

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

Proof : $$\frac{a_1b_1}{c_1^2}+\frac{a_2b_2}{c_2^2}=\frac{a_3b_3}{c_3^2}\implies c_3^2(a_1b_1c_2^2+a_2b_2c_1^2)=a_3b_3c_1^2c_2^2$$ Since $\gcd(c_3,a_3b_3)=1$, we have to have $$\frac{c_1^2c_2^2}{c_3^2}\in\mathbb Z$$ Similarly, we have to have $$\frac{c_2^2c_3^2}{c_1^2}\in\mathbb Z\qquad\text{and}\qquad \frac{c_3^2c_1^2}{c_2^2}\in\mathbb Z$$ The claim follows from these.$\quad\square$

Answer 2

Your question is a special instance of a slightly more general Diophantine problem over $\mathbb Q^3$, because if $$\frac{a_1b_1}{2c_1^2}+\frac{a_2b_2}{2c_2^2}= \frac{a_3b_3}{2c_3^2}$$ is written as $$\frac{a_1b_1}{a_1^2+b_1^2}+\frac{a_2b_2}{a_2^2+b_2^2}= \frac{a_3b_3}{a_3^2+b_3^2}$$ this can be transformed into $$ {\dfrac {1}{\dfrac{a_1^2+b_1^2}{a_1b_1}}+\dfrac {1}{\dfrac{a_2^2+b_2^2}{a_2b_2}}}=\dfrac{1}{\dfrac{a_3^2+b_3^2}{a_3b_3}}$$ and this into $$\dfrac{1}{\dfrac {a_1}{b_1}+\dfrac{b_1}{a_1}}+\dfrac{1}{\dfrac {a_2}{b_2}+\dfrac{b_2}{a_2}}=\dfrac{1}{\dfrac {a_3}{b_3}+\dfrac{b_3}{a_3}}$$

You can see that this is an instance of a more general problem by substitution $r_1=\dfrac{a_1}{b_1}$ and $r_2=\dfrac{a_2}{b_2}$ and $r_3=\dfrac{a_3}{b_3}$ and by pretending that $r_1$ and $r_2$ and $r_3$ are not constrained by the fact that they are ratios of sides of Pythagorean triangles with integer sides.

So the equation becomes $$\dfrac {r_1}{r_1^2+1}+\dfrac{r_2}{r_2^2+1}=\dfrac {r_3}{r_3^2+1}$$ and in a slightly more general interpretation than yours we could view it as it´s over $\mathbb Q^3$

Although the equation is of the simple form and of a small degree it has three variables and, to add to the difficulty in this more general setting, they can all take all rational values.

I am not able at this moment to solve something like this in this generality.

Answer 3

Some helpful parameterizations for your search:

$a^2 + b^2 = c^2$ is completely parameterized by $(a,b,c) = k*(2uv,(u-v)(u+v),u^2 + v^2)$

$a^2 + b^2 = c^2 + d^2$ is completely parameterized by $(a,b,c,d) = k*(uv + rs,ur - vs, ur + vs, uv - rs)$

$ab = cd$ is completely parameterized by $(a,b,c,d) = (pq,rs,pr,qs)$ [This can be easily expanded to more factors, I trust your ability to perceive how, but if you are not sure, do ask]

$a^2 = bc$ is completely parameterized by $(a,b,c) = (uvw, uv^2, uw^2)$