How to prove $\frac{2^a+3}{2^a-9}$ is not a natural number How can I prove that
$$\frac{2^a+3}{2^a-9}$$
for $a \in \mathbb N$ is never a natural number?

Define the natural numbers to be the set $\mathbb{N}=\{1,2,3,\ldots\}$. Assume $\frac{2^a+3}{2^a-9}$ is a natural number, then so is $$1+\frac{12}{2^a-9},$$ and 
$$\frac{2^a+3}{2^a-9}-1=\frac{12}{2^a-9}\in\mathbb{N}\cup\{0\}.$$
But $$\frac{12}{2^a-9}\not\in\mathbb{N}\cup\{0\}.$$ Contradiction. Similar argument holds if you assume $\mathbb{N}=\{0,1,2,3,\ldots\}$.
