Hint:
It is better to expres $n$ in terms of $m$ and then use divisibility properties $$n = {m^2+8\over m+11} \implies m+11\mid m^2+8$$
Notice $m+11\mid m^2-121$ so $m+11\mid 129=3\cdot 43$ so $m+11\in\{1,3,43,129\}$ so $m=32$ or $m=118$...
Call the peak color $p$. Because the peak is adjacent to every base vertex, no base vertex can have color $p$. On the base, note that no color can be used more than twice without being forced to be a neighbor of itself. So there are 3 possible cases:
- 5 colors used: $c_1, c_2, c_3, c_4, c_5$, each used once.
- 4 colors used: $c_2$ is used twice, $c_1, c_3, c_4$ each used once.
- 3 colors used: $c_1$ used once, $c_2, c_3$ each used twice.
In the 5 base color case, We fix a vertex as being $c_1$ and proceed to the right to label the other colors. All orderings are distinct, so we have $6!$ ways to assign actual colors to $p$ and $c_1$ through $c_5$. Each ordering has 5 rotations, so the number of distinct colorings is $\frac {6!}5$.
In the 4 base color case, we set $c_1$ to be the color of the vertex which is between the two vertices that are the same, so the colors (preceeding to the right) are $c_1 - c_2 - c_3 - c_4 - c_2$. This assignment doesn't allow for rotation. So there are $5!$ ways of coloring the vertices with 1 peak + 4 base colors. So there are $\frac {6!}{1!}$ ways of coloring the vertices with 1 peak + 4 base colors.
In the 3 base color case, we set $c_1$ to be the color that is used only once. Again, this uniquely identifies the $c_1$ color, so rotations are not possible. There are $4!$ ways to assign the colors. There are $\frac{6!}{2!}$ ways to assign the colors.
So I get a total of $$\frac {6!}5 + \frac {6!}{1!} + \frac{6!}{2!} = 144 + 720 + 360 = 1224$$
If you make the same mistake I had made and forget that you are still picking from 6 colors, even when you are only picking 5 colors or 4 colors total, then you get $\frac {6!}5 + 5! + 4! = 288$. So apparently that was the source of the error in the "solution" you found.
Best Answer
Hint. Consider the perpendicular to $BC$ at $B$ and let $G$ be its intersection with line $AC$. Since triangle $ABC$ is isosceles, we have that its height relative to side $BC$ intersects $BC$ in its midpoint. Thus, $BG = 2DF$ and triangle $BGE$ is isosceles.