# 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

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.

• It would perhaps motivate people to think about your question if you told us a bit about where it came from. – TonyK Nov 1 '19 at 21:46
• I did not find partial results, just more general formulation. – user720692 Nov 7 '19 at 2:49
• And you can give bounty to mathlove if some better answers do not appear. – user720692 Nov 7 '19 at 2:51
• Here's a suggestion. Take some value of $c$ that is a hypotenuse of many primitive triangles, e.g., $c=5\times13\times17\times29\times37\times41$ is the hypotenuse of $32$ primitive triangles. Calculate all the $(a,b)$ values for that $c$, and see whether any two of them add up to a third one. If so, you have a counterexample. – Gerry Myerson Dec 3 '19 at 11:25
• Well, sure. Since $2a_nb_nc_n^{-2}$ can't be more than $1$, at least one of the terms on the left must be less than $1/2$, which forces the smaller of $a_n,b_n$ to be small compared to $c_n$, and the other to be close to $c_n$. That's long & thin. – Gerry Myerson Dec 4 '19 at 1:00

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.

• Thank you I very much appreciate your answer here and will look into this further (I'm not entirely familiar with Diophantine equations). When you say, "pretending that $r_n$ are not constrained by the ratios..." what exactly do you mean by this? Are you saying that what you've shown removes the need to look at the constraint and the problems are equivalent? The reason I ask is because I want the equation to be constrained as such. – PMaynard Nov 7 '19 at 2:18
• Also do you disagree with the proof above by Mathlove? – PMaynard Nov 7 '19 at 2:19
• No, the problems are not equivalent, because more general equation has at least some solutions and yours could have none, but all the solutions of your equation are also solutions of a more general equation. I think that Mathlove gave necessary condition for the hypotenuses, not the solution of your problem. – user720692 Nov 7 '19 at 2:24
• :D You can ask questions here, some guys know some stuff, but, generally, we´re all wasting time here, since there are so many so much more important world-problems than math-problems. But if you really like doing math and you enjoy it, then you can do it. I understand you much. I am not sure exactly now, I am just looking at my more general formulation: there are only three variables, the highest degree of a variable is 2, there are no much terms,but all three variables are allowed to attain all rational values so I am not sure at this moment. What do you don´t understand about that equation? – user720692 Nov 7 '19 at 3:22
• If you set $r_1=x$ and $r_2=y$ and $r_3=z$ and pretend that it´s an equation over $\mathbb R^3$ then it´s an equation of a differentiable (so also continuous) surface. My formulation is about determining points on that surface that have all three rational coordinates. – user720692 Nov 7 '19 at 3:38

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.$$\quad\square$$

• I'm sorry I messed up the question a little bit. Thanks for finding this! but I should have added a>b and the quotient > 0 I'll edit it – PMaynard Nov 2 '19 at 5:35
• I've edited the question to reflect some new findings also – PMaynard Nov 2 '19 at 19:10
• @PMaynard: I've just added a necessary condition for $c_i$. The claim includes the fact that if a prime number $p$ divides one of $c_i$, then $p$ divides at least one of the other two. – mathlove Nov 6 '19 at 6:36