Ah! That is Dancing

We require that $\dfrac{3n - 5}{n + 1} = 3 - \dfrac{8}{n + 1}$ be a perfect square. Now for this expression to be a perfect square we will first require that $\dfrac{8}{n + 1}$ be an integer. This will be the case when $n + 1$ is a divisor of $8$ , i.e., when

$n + 1 = \pm 1, \pm 2, \pm 4, \pm 8 \Longrightarrow \dfrac{8}{n + 1} = \pm 8, \pm 4, \pm 2, \pm 1$ .

The range of $3 - \dfrac{8}{n + 1}$ will then be $3 - 8 = -5$ to $3 + 8 = 11$ . The only perfect squares in this range are $1, 4$ and $9$ .

Now $3 - \dfrac{8}{n + 1} = 1$ when $n + 1 = 2 \Longrightarrow n = 3$ and

$3 - \dfrac{8}{n + 1} = 4$ when $n + 1 = -8 \Longrightarrow n = -9$ .

Since $-6$ is not one of the possible values for $\dfrac{8}{n + 1}$ we can't have $3 - \dfrac{8}{n + 1} = 9$ , so the only possible values for $n$ are $3$ and $-9$ , which sum to $\boxed{-6}$ .

Such that $\sqrt{\frac{3n-5}{n+1}}$ is a integer $\frac{3n-5}{n+1}=k^2$ for someone $k$ . Then if $v=n+1 \Longrightarrow 3n-5=3v-8$ and substituting in the fraction $\frac{3v-8}{v}=k^2 \Longrightarrow 3v-8=vk^2 \Longrightarrow v(3-k^2)=8$ . We have two options the 1st is: If $v$ is positive, then $3-k^2>0 \Longrightarrow k^2<3 \therefore k^2 = 1$ or $k^2=2$ but $\sqrt2$ is not a integer $\Longrightarrow$ only take $k^2=1 \Longrightarrow v=4 \therefore n=3$

The 2nd option: if If $v$ is negative, then $3-k^2<0 \Longrightarrow k^2>3 \therefore k^2 = 4$ or $k^2=9$ but $v$ should be minor to 8 given that $3-k^2$ shouldn´t be a fraction. Then only take $k^2=4 \Longrightarrow v(3-4)=8$ , $v=-8 \therefore n=-9$

$\Longrightarrow$ exist 2 values for $n$ , $n=3$ and $n=-9$ $\therefore 3-9=\boxed{-6}$