August 2022

Problem:

Find positive whole values for $a,b,c$ such that $\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b}=4. \quad(*)$

Solution:

Given

$y^2=x^3 +(4n^2+12n-3)x^2+32(n+3)x\quad\quad\quad(1)$

where $n$ is an integer; $x,y$ are rationals, the map

$\begin{cases} a=\frac{8(n+3)-x+y}{2(4-x)(n+3)}, \\ b=\frac{8(n+3)-x-y}{2(4-x)(n+3)}, \\ c=\frac{-4(n+3)-(n+2)x}{(4-x)(n+3)} \end{cases}\;(2)\implies \;\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} - n=0.$

i.e.,

for a fixed $n$ and $(x,y)$ on $(1), (a,b,c)$ generated by $(2) \implies \frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} = n\quad(3)$

Notice also

$f(\square,\blacksquare,\circ)\overset{\Delta}{=}\frac{\square}{\blacksquare+\circ} + \frac{\blacksquare}{\square+\circ} + \frac{\circ}{\square+\blacksquare} \implies f(t\cdot a, \; t\cdot b, \; t\cdot c) = f(a, b, c).\quad\quad\quad(4)$

Moreover, starting with $(x_1, y_1)$ on

$y^2 = x^3+px^2+qx+r,\quad\quad\quad(5)$

the following map generates successive rational points $(x_i, y_i)$ on (5):

$\begin{cases} i=2:x_2 = (\frac{3x_1^2+2px_1+q}{2y_1})^2-p-2x_1, \; y_2 = -y_1 + \frac{(3{x_1}^2+2px_1+q)}{2y_1}(x_1-x_2) \\ i>2: x_i = \left(\frac{y_{i-1}-y_1}{x_{i-1}-x_1}\right)^2-p-x_1-x_{i-1}, \; y_i= -y_1+\frac{y_{i-1}-y_1}{x_{i-1}-x_1}(x_1-x_i)\; \end{cases}(6)$

Let $n=4$ , we have

$y^2=x^3+109x+224x\quad\quad(7)$

$\begin{cases} a=\frac{56-x-y}{14(4-x)} \\ b=\frac{8(56-x-y)}{14(4-x)} \\ c=\frac{-28-6x}{7(4-x)} \end{cases}\quad\quad\quad(8)$

and

$\begin{cases} i=2:x_2 = (\frac{3x_1^2+218x_1+224}{2y_1})^2-109-2x_1, \; y_2 = -y_1 + \frac{(3{x_1}^2+218x_1+224)}{2y_1}(x_1-x_2) \\ i>2: x_i = \left(\frac{y_{i-1}-y_1}{x_{i-1}-x_1}\right)^2-109-x_1-x_{i-1}, \; y_i= -y_1+\frac{y_{i-1}-y_1}{x_{i-1}-x_1}(x_1-x_i)\; \end{cases}(9)$