How to prove $\frac{2^a+3}{2^a-9}$ is not a natural number

Question

How can I prove that

$$\frac{2^a+3}{2^a-9}$$

for $a \in \mathbb N$ is never a natural number?

Answer 1

Hint: $$\frac{2^a+3}{2^a-9}=1+\frac{12}{2^a-9}$$

So $\frac{12}{2^a-9}$ must be a natural number as well.

Answer 2

The expression is equivalent to $$1+\frac{12}{2^n-9}$$ so there aren't many cases to consider.

Answer 3

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 $\mathbb{N}=\{0,1,2,3,\ldots\}$.

Answer 4

$$1+\frac{12}{2^a-9}$$

which means that $2^a$ should equal either $10,11,12,13,15$ or $21$, but neither of them is a power of $2$, so it's never a natural number

Answer 5

Alternatively...

If $\displaystyle \frac{2^a+3}{2^a-9} = k$ for some integer $k$, then

$$(k-1)2^a = -12$$ or $$2^a = 12/(1-k).$$ As $2^a = 2, 4, 8, ...$, the only possible values for $k$ are $k = -5, -2$. But these are not natural numbers as they are negative integers.