Not too hard II

Since $a,b,c$ are inter exchangeable, we can say that they are roots to a cubic equation, denote it to be $f(W) = W^3 - x W^2 + y W - z$

where $x,y,z$ denote the values $a+b+c, ab+ac+bc,abc$ . Thus $z = -8$

Convert the first given equation to $x^2 - 2y = 21$

Similarly using the identity $a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2 - (ab+ac+bc))$

We have

$55 - 3(-8) = x(21-y) \Rightarrow x(y-21) = 31$

Solve these two equations simultaneously

$x(2y - 42) = 62 \Rightarrow x(x^2-21 - 42) = 62 \Rightarrow x(x^2-63) = 62$ which gives $x=-1$ as one of the solution. Factoring out $(x+1)$ yields the other two roots $\frac {1}{2} (1 \pm \sqrt{249} )$

With this, we can solve for $y$ , thus the solutions $\large (x,y) = (-1, 10), \left ( \frac { 1 \pm \sqrt{249}}{2} , \frac {83 \pm \sqrt{249} }{4} \right )$

We now consider the identity

$\begin{aligned} a^4 + b^4 + c^4 & = & (a^2+b^2+c^2)^2 - 2 [ (ab)^2 + (ac)^2 + (bc)^2 ] \\ & = & 441 - 2 [ (ab+ac+bc)^2 - 2abc(a+b+c) ] \\ & = & 441 - 2 (y^2 +16x ) \\ \end{aligned}$

Substitution of different pairs of $x,y$ gives different values, and its minimum value is $- \frac {1869 + 147 \sqrt{249}}{4} \Rightarrow m = 1869, n = 249 \Rightarrow m+n=\boxed{2118}$