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

Solution 1:

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.

Solution 2:

There is a counterexample.

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

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

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.$\quad\square$