Zi Song's Gargantuan Root

Given: $x=\frac{1}{29}(3-\sqrt[2013]{29x-1})$

Let $29x-1=t^{2013}$

$\implies t^{2013}+t-2=0$

Let $f(t)=t^{2013}+t-2$ . By Descartes' Rule of Signs , we get,

$\bullet \text{At max 1 positive real root since number of sign changes in} \: f(t)=t^{2013}+t-2 \: \text{is 1}$

$\bullet \text{At max 0 negative real roots since number of sign changes in} \: f(-t)=-t^{2013}-t-2 \: \text{is 0}$

Also, $t=1$ is a root of $f(t)$ and it must be the only real root (by Descartes' Rule of Signs).

$\implies x=\frac{2}{29}$

$\implies a+b+c=\boxed{32}$

$Let~f(a,b,c)=3 - \Big (29*(a*sqrt(b)/c) - 1 \Big)^(1/2013) )-29* ( a*sqrt(b)/c ).\\ I~varied~values~of~a,~b,~c. ~~a=1, ~~b=1, ~~~f(1,1,10)={\color{#D61F06}{-}}0.031...., ~~f(1,1,20)={\color{#3D99F6}{+}}.089....\\ So~close~ up,~~f(1,1,15)={\color{#3D99F6}{+}}.0023....~~~~~~~~f(1,1,14)={\color{#D61F06}{-}}.246422...\\ Zero~between~c=14~and~c=15.~~No~zero~at~an ~integer.~~So~b\neq 1.\\ b=2~to~5~~also~gave~no~solution.\\ Next~tried~a=2,~b=1, ~~c=1,10,20,29~!!!!~got~0, ~the~ solution.\\ So~a=2,~~b=1,~~c=29.\\ Still~required~to~prove~that ~this~is~ the~only~solution.\\ Note~(a*sqrt(b)/c)=x.$