In my country, there is a math competition per region. In my region's paper, there is a question which I have reduced to solving the following equation for $x, y \in \mathbb{N}$:

$$log_{2}{(2^{x}-2^{y}-2000)} \in \mathbb{N}$$

Now, I solved it by bruteforcing the equation, but I wonder how I can solve it on paper (and without guessing).

I found 2 solutions, but I can't prove that there are only 2 (maybe there's more?), so that's why I want to solve it algebraically:

  • $x = 11, y = 4$
  • $x = 11, y = 5$

I asked my math teacher about it, and it said that he would first remove the $log_{2}$ by exponentiation, and then try to write $2000$ as power of $2$ and reduce, but I got stuck on writing $2000$ as a power of $2$ without a calculator (even if I used one, I wouldn't know how to proceed).

Any ideas on how I can solve this?

  • $\begingroup$ If $x,y\in\mathbb R$, then there are infinitely many such pairs $(x,y)$. $\endgroup$ – mathlove Jan 11 '17 at 7:28
  • $\begingroup$ @mathlove Yes, you are right. I only now understood your point, thank you :) $\endgroup$ – Rakete1111 Jan 24 '17 at 19:30

I guess $x >y$ is assumed for the logarithm to make sense.

Note $$2^x-2^y-2000 = 2^y \cdot (2^{x-y}-1) - 2^4 \cdot 5^3,$$ and we want this to be a power of $2$.

We need $$y \ge 4,$$ otherwise dividing the above by $2^y$ yields $2^{x-y}-1-2^{4-y} \cdot 5^3$, an odd number.

Case 1: $y=4$.

We need $2^{x-4}-1-5^3=2^{x-4}-2 \cdot 63$ to be a power of $2$, say, $2^k$. That is, $2^{x-4}-2^k = 2 \cdot 63$. From this we see $k=1$ necessarily, so $2^{x-5} -1 = 63$, i.e., $$x=11.$$

Case 2: $y>4$.

Then we need $2^{y-4} (2^{x-y}-1) - 5^3$ to be a power of $2$. But this quantity is odd due to $y>4$, so it must equal $1$ (the only odd power of $2$). Thus, $2^{y-4}(2^{x-y}-1) = 126$ so $y-4=1$ and $2^{x-y}-1=63$, yielding $$y=5\text{ and } x=11.$$


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