Eq. (\[lineq\]) and they are all positive constants, the solutions are not asymptotically flat, but asymptotically AdS spacetimes.

 Black hole solution with $M=0$
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The spacetime (\[m-2metric\]) describes a black hole provided $f(r_+)=0$ and $f(r)>0$ with $r>r_+$. Here $r=r_+$ is called black hole horizon. We can see from (\[Fequ\]) that even if the “mass" $M$ of the solution vanishes, black hole horizon can still exist. To show this, let us discuss the solution $F=F_0$ of Eq. (\[Fequ\]) with $M=0$: $$\sum_{i=0}^{s}\widehat{\alpha}_{i}F^i_0=\frac{Q}{r^{m}}$$ with horizons. Assume $f_0(\bar{r})=0$ at some positive $\bar{r}$, and we have $$\bar{r}^2F_0(\bar{r})= \delta \quad \mathrm{or}\quad
F_0(\bar{r})=\frac{\delta}{\bar{r}^2}\, .$$ That is, $\bar{r}$ must satisfy the following equation $$\label{2eq27}
\bar{r}^{m}\sum_{i=0}^{s}\widehat{\alpha}_{i}\left(\frac{\delta}{\bar{r}^2}\right)^i=Q\,
.$$ In order for the equation to hold, the constant $Q$ must satisfy some constraints. We will discuss these constraints in the cases of $\delta=0$ and $\delta=\pm1$, respectively.

(i). $\delta=0$. This case is simple. In this case, only one term in (\[2eq27\]) remains. We have $$\bar{r}^{m}\widehat{\alpha}_{0}=Q\, ,$$ Since $\widehat{\alpha}_{0}>0$, this indicates a positive $\bar{r}$ exists provided $Q>0$, and $\bar{r}$ is just the horizon radius $r_+$.

(ii). $\delta=\pm 1$. If $m$ is odd, one then has $s=(m-1)/2$, and $$\label{2eq29}
\bar{r}^{m}\widehat{\alpha}_{0}\pm\bar{r}^{m-2}\widehat{\alpha}_{1}+\cdots
+(\pm1)^s\bar{r}\widehat{\alpha}_{s}=Q\, .$$ Obviously, because all coefficients $\{\widehat{\alpha}_{0},
\widehat{\alpha}_{1}\cdots \widehat{\alpha}_{s}\}$ are positive, and especially $\widehat{\alpha}_{0}>0$, 