$\frac {2 \cdot 3^{m + 1}}{k} - \frac {2 \cdot 3^{m}}{k + 1} \in \Bbb N_+$, find all possible values of $k, m$. If $$\frac {2 \cdot 3^{m + 1}}{k} - \frac {2 \cdot 3^{m}}{k + 1} \in \Bbb N_+$$ and $$\frac {2 \cdot 3^{m}}{k} - \frac {2 \cdot 3^{m}}{k + 1} \le 1$$ where $k, m \in \Bbb N_+$ and $k \ge 2$, find all possible values of $k, m$.
This comes from a problem in graph theory, which may be simplified to this (though I don't know if this method works). According to the solution to the original question, $(k, m) = (2, 1)$.
Can I say that $\frac {2 \cdot 3^{m + 1}}{k} \in \Bbb N_+$ and $\frac {2 \cdot 3^{m}}{k + 1} \in \Bbb N_+$? I found that if that's true then $(k, m)$ has only one solution, i.e., $(2, 1)$. But I'm unsure whether I can get that result. If I can't, how can I solve the problem above?
This is the original question:
Consider a graph $G(V, E)$ of order $n$ such that $$\forall x, y \in V ((x \not = y) \implies (|E(G - \{ x, y \})| = 3^m)),$$ where $m \in \Bbb N_+$, find all possible values of $n$.
I observed that (by double counting) $$|E(G)| = \frac {\binom {n}{n - 2}}{\binom {n - 2}{n - 4}} 3^m \le \binom {n}{2},$$ which is equivalent to this question (with $k = n - 3$).
 A: \begin{align*}
\frac {2 \cdot 3^{m + 1}}{k} - \frac {2 \cdot 3^{m}}{k + 1} \in \Bbb N_+ \\
\implies 2\cdot 3^m \Bigg(\frac {3}{k} - \frac {1}{k + 1}\Bigg) \in \Bbb N_+ \\
\implies 2\cdot 3^m\frac {2k+3}{k(k+1)} \in \Bbb N_+ \\
\implies 2\cdot 3^m\frac {2t+1}{t(t-1)} \in \Bbb N_+ \tag{$t\to k+1$} \\
\end{align*}
Clearly, $t\not = 1$.
Now, $t \nmid 2t+1$ since $\gcd(2t+1,t)=\gcd(2t+1-2t,t)=\gcd(1,t)=1$. Therefore $t$ must divide $2\cdot3^m$ So $t$ has to be of the form $2^\alpha 3^\beta$ where $\alpha \in \{0,1\}$ and $\beta \in \{0,1,\dots m\}$
If $t$ is odd ($\alpha=0$), $t-1$ is even and does not divide $3^m(2k+1)$ which is always odd. But $t-1$ divides $2\cdot3^m(2t+1)$ for the expression to be $\mathbb{N}_+$ Hence, $t-1|2\implies t-1=2\implies t=3$ and $(\alpha, \beta) = (0,1)$. $k=t-1=2$. Then, $m\in\{1,2,\dots \infty\}$. ($m\not =0$ as $t\not =1$)
But then $$\frac{1}{k}-\frac{1}{k+1}=\frac{1}{12}\implies 2\cdot 3^m\Bigg(\frac{1}{k}-\frac{1}{k+1}\Bigg) = \frac{3^m}{6}\le 1 $$ and your inequality holds only when $m\in\{0,1\}$. Taking the intersection of the two solutions for $m$, we find $m=1$. Hence, $(k,m)=(2,1)$ 
If $t$ is even ($\alpha=1$), $t-1$ is odd and for $t-1$ to divide $3^m(2t+1)$, $$\gcd(3^m(2t+1),t-1)=t-1=\gcd(3^m(2t+1)-2\cdot3^m(t-1),t-1)= \gcd(3^{m+1},t-1)$$
Either $t-1=1\implies t=2$ or $t-1=3^{m+1} \implies t=1+3^{m+1} \implies k=3^m$. But $1+k$ should divide $2\cdot 3^m$ as the other fraction $\Big(\frac {2 \cdot 3^{m + 1}}{k}\Big)$ is an integer. Hence, $1+k=2 \implies k=1$ and $m=0$. But you need $k\ge2$ and $m\in \mathbb{N}_+$
Thus the only acceptable solution is $(k,m)=(2,1)$.
