Only Integers

Since $t$ is an integer, it follows that $\frac{t^3+8}{t^2-4} = t + \frac{4(t+2)}{(t-2)(t+2)}$ is an integer if and only if $\frac{4}{t-2}$ is an integer and $t\neq -2.$ Therefore, $t-2$ is one of the integral factors of $4:$ $t-2 = -4, -2, -1, 1, 2, 4 \quad \iff \quad t = -2, 0, 1, 3, 4, 6$ but since $t\neq -2,$ the full list of values is $t = 0, 1, 3, 4, 6$ giving an answer of $0 + 1 + 3 + 4 + 6 = \boxed{14}$

Let $k$ be an integer with $k = \frac{t^3+8}{t^2-4} = \frac{(t+2)(t^2-2t+4)}{(t+2)(t-2)}$ Thus, $\begin{aligned} \frac{t^2-2t+4}{t-2} &= k \\ t^2 -2t + 4 &= kt - 2k \\ t^2 -(k+2)t + (4+2k) &= 0 \\ t &= \frac{k+2 \pm \sqrt{k^2 + 4k + 4 - 4(4+2k)}}{2} \\ t &= \frac{k+2 \pm \sqrt{k^2 - 4k - 12}}{2} \\ t &= \frac{k+2 \pm \sqrt{(k-6)(k+2)}}{2} \end{aligned}$

Note here that $\sqrt{(k-6)(k+2)}$ , if it's an integer, will be even precisely when $k$ is even and odd precisely when $k$ is odd. So the numerator of $t$ will always be even (again, when it's an integer). Thus, if we get an integer in the numerator, $t$ will be an integer. So we need only to find all integers $k$ for which $(k-6)(k+2)$ is a perfect square. There are two cases:

• $(k-6)(k+2) = 0$ . Thus $k=-2$ and $k=6$ are potential solutions.
• $(k-6)(k+2) = m^2$ for some non-zero integer $m$ . Rearrange this equation to read $k^2 - 4k - (m^2 + 12) = 0$ . This is a quadratic in $k$ with discriminant $16 + 4(m^2 + 12) = 4(m^2 + 16)$ . Thus, $k$ will be an integer precisely when $m^2 + 16 = m^2 + 4^2$ is a perfect square. Since $m$ is a non-zero integer, this is a Pythagorean Triple! The only Pythagorean Triple with a leg of $4$ is $3,4,5$ . Thus $m=\pm 3$ and we have $k^2 - 4k - 21 = 0$ , so $k=7$ and $k=-3$ are potential solutions.

We now have $-3$ , $-2$ , $6$ and $7$ as possible values for $k$ . Plugging them back into our solution for $t$ yields $-2$ and $1$ ; $0$ ; $4$ ; and $3$ and $6$ ; respectively. Note that $t=-2$ is not a solution, since it would make the denominator of our original fraction $0$ . But all others are solutions. Thus the sum of the possible integer values of $t$ is $1 + 0 + 4 + 3 + 6 = \boxed{14}$

Introduce new variable $T=t+2$ .

$\frac{(T-2)^3+8}{(T-2)^2-4}=\frac{T^3-6T^2+12T}{T^2-4T}=\frac{T^2-6T+12}{T-4}$

then take the GCD of the terms in the nominator and the denominator.

$GCD(T^2-6T+12,T-4)=(T-4,4)$

we need the GCD to be equal to $T-4$ . Therefore, these are the possibilities for $T-4$

$T-4=1,2,4,-1,-2,-4$

$T=5,6,8,3,2,0$

So,

$t=3,4,6,1,0,-2$

we plug the possibilities into the main expression to they actually work. The final result is $0,1,3,4,6$