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I want to find a solution to

$$ \frac{x_1}{x_2 + x_3} +\frac{x_2}{x_1 + x_3} + \frac{x_3}{x_1+x_2} = 4 $$

for $x_1,x_2,x_3>0$, and $x_1+x_2+x_3=1$.

We have $$\frac{x_1}{1-x_1}+\frac{x_2}{1-x_2}+\frac{1-x_1-x_2}{x_1+x_2}=4.$$

Let $\dfrac{x_1}{1-x_1}=s,\dfrac{x_2}{1-x_2}=t,$then $x_1=\dfrac{s}{1+s},x_2=\dfrac{t}{1+t}.$ We then have $$s+t+\dfrac{1-st}{s+t+2st}=4,$$ and $s=x+y,t=x-y,$then$$\frac{4 x^3+3 x^2-4 x y^2+y^2+1}{2 \left(x^2+x-y^2\right)}=4$$ $$y^2=\frac{4 x^3-5 x^2-8 x+1}{4 x-9}$$ which eventually simplifies to solving $$y^2 = x^3 + 121 x^2 + 1144 x + 2704.$$ How can we solve this? I am unfamiliar with eliptic curves, even after reading some basic material, it seems each equation is tackled differently. Is there a routine method for equations like this?

Cheese Cake
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CCZ23
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    You seem to also assume $x_1+x_2+x_3=1$. – mathcounterexamples.net May 06 '22 at 09:43
  • Assuming $x_1+x_2+x_3=1$, we can factorise $x^3+121x^2+1144x+2704$ - hint: substitute $x=-4$, and divide the equation by (x+4). – Cheese Cake May 06 '22 at 09:46
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    Are the $x_i$ supposed to be rational numbers ? – Peter May 06 '22 at 09:58
  • @Peter Yes rational – CCZ23 May 06 '22 at 10:54
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    @CCZ23 There is mistake in equation when making $s,t$ substitution. And you don't need to make $x,y$-substitution. You can express one of $s$ and $t$ in terms of another from quadratic equation. If I haven't mistake in my calculation problem becomes the following: find positive integer pairs $(p,q)$ such that $(q+2p)(6q-p)(2q^2+7pq-2p^2)$ is square of an integer. Here $s=\frac{p}{q}$. – Ivan Kaznacheyeu May 06 '22 at 11:22
  • @IvanKaznacheyeu I get the same; I found a few pairs with $\gcd(p,q) = 1.$ //// 28224 = 2^6 3^2 7^2 q: 5 p: 2 //// 2304 = 2^8 3^2 q: 2 p: 3 //// 1225 = 5^2 7^2 q: 3 p: 11 //// 33124 = 2^2 7^2 13^2 q: 4 p: 11 //// 43277313024 = 2^10 3^2 11^2 197^2 q: 70 p: 541 //// 409478409025 = 5^2 7^2 47^2 389^2 q: 143 p: 1033 //// – Will Jagy May 06 '22 at 20:20
  • There is a formula for any coefficient. It allows for some solution - to get the next solution. https://mathoverflow.net/questions/264754/solution-to-a-diophantine-equation/275193#275193 – individ May 07 '22 at 05:04
  • See the answer by Michael Stoll at this MO-post for finding these points on the elliptic curve. – Dietrich Burde May 08 '22 at 18:51

2 Answers2

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Let $x_3 = 1-x_1-x_2.$ We get $$(6-7x_1)x_2^2+(-7x_1^2+13x_1-6)x_2+6x_1^2+1-6x_1=0$$

Since $x_2$ must be rational number then discriminant must be square number.

Hence we get

$$V^2 = 49U^4-14U^3-59U^2+16U+12\tag{1}$$ with $$U=x_1$$ An equation $(1)$ is birationally equivalent to the elliptic curve
$$Y^2 + XY + Y = X^3 - 234X + 1352 \tag{2}$$ with $$U = \frac{-X-3}{X-23}, V = \frac{-52Y-26X-26}{(X-23)^2}$$ According to LMFDB , elliptic curve has rank $1$ with generator $(8,-1).$

Let $P(X,Y)=(8,-1)$ then $13P$ gives positive solution using group law as follows.

$x_1 = \frac{360580518798960218928732302909807349160733186141387092674568680456952454542061941533770232294581808506248091531}{5049549479055895370585527032143752370420166865650725292672977625843429614339787186642533620730887839277641883695}$

$x_2 = \frac{306595938811259225832173528708283063181310906692288470156330531143167380161534287576910178181384187701055520003}{2212455878539701410315253838580846315686580343082895422240759011653186431824457611838047222821239666672118415365}$

$x_3 = \frac{3852086946706166284829913202913539977730505088064544668668081343553155180751922744069311820918853233969099047039}{4875970872038496234103522347007495376171052920878501712857733921453635116079760054850598705847152516034789194083}$

Tomita
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First notice that x=0,y=52 is a solution. Take a tangent to the curve at this point and it will necessarily intersect at another rational point. By successively taking intersections of the curve with either tangents of rational points or the lines passing through two different rational points you can find more rational points.

Robin Balean
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