$$n$$ is an integer and $$\sqrt{\frac{3n-5}{n+1}}$$ is also an integer. The sum of all such $$n$$ is

Setting $$\frac{3n-5}{n+1}=k^2$$ and rewriting gives $$n=-1+\frac{8}{3-k^2}$$, so $$3-k^2$$ must divide 8. Checking divisors of 8 for values where $$3-k^2$$ gives a perfect square $$k^2$$, the valid cases are $$k=1$$ (giving $$n=3$$) and $$k=2$$ (giving $$n=-9$$). The sum of all such $$n$$ is $$3+(-9)=-6$$.