Don't try to complicate it!

Since $a$ is one of the roots of the equation $x^2-x-4=0$ , then, $a^2 -a = 4$ .

We can easily see that $a^3 +1 = a^3 -a^2 + a^2+1$ .

We will do some tricky things. The expression $a^3 -a^2 + a^2 +1$ can be written as $a(a^2-a) + a^2 +1 = 4a+a^2 + 1$ .

Again, another tricky thing. $4a + a^2 + 1 = (a^2 - a) + 5a + 1 = 4 + 5a + 1$ .

Now, we have $a^3 + 1 = 5a + 5 = 5(a+1)$ .

For the denominator, we simplify, $a^5 - a^4 -a^3 + a^2 = a^3(a^2-a) - a(a^2 -a)$ .

Another tricky thing, $4a^3 - 4a = 4a(a^2 - 1) = 4a(a^2 -a+a-1)$

Simplify, we have $4a(4+a-1) = 4a(a+3) = 4a^2 + 12a = 4a^2 - 4a + 16a = 16 + 16a$ .

So, we have $a^5 - a^4 - a^3 + a^2 = 16(a+1)$ .

Hence, we have $\dfrac{a^3+1}{a^5-a^4-a^3+a^2} = \dfrac{5}{16} = \dfrac{m}{n}$ .

$m+n = \boxed{21}$

Since $a$ is a root of $x^2 - x - 4 = 0$ , we have $a^2 = a+4$ .

By repeatedly multiplying this equation by $a$ , we may work out the powers of $a$ , like so: $a^3 = a\cdot a^2 = a\cdot(a+4) = a^2 + 4a = (a+4) + 4a = 5a+4$ $a^4 = a\cdot a^3 = a\cdot(5a+4) = 5a^2 + 4a = 5(a+4) + 4a = 9a+20$ $a^5 = a\cdot a^4 = a\cdot(9a+20) = 9a^2 + 20a = 9(a+4)+20a = 29a + 36$

Using this, we may simplify the numerator and denominator: $a^3 + 1 = (5a+4) + 1 = 5a+5$ $a^5-a^4-a^3+a^2 = (29a+36)-(9a+20)-(5a+4)+(a+4) = 16a + 16,$ so that $\frac{a^3+1}{a^5-a^4-a^3+a^2} = \frac{5a+5}{16a+16} = \frac{5}{16}$ where we note, as a sanity check, that $a\neq -1$ , so the last step above is actually valid.