Unique solution of equation on $Z$

Question

I am a computer science student and in one of our (online, given the circumstances) class, we met an equation that has only 1 solution. I am unable to find it analytically, I was only able to solve it using numerical tools. Does anyone knows how to show that there are no other solutions that $m = 1$, $i = 1$ ?

$$ 2^m = 1 + 2^{1-i} $$ for $i \in \mathbb{N} \setminus \{0\}$ and $m$ an integer s.t. $m > -i$

Answer 1

First note that $2^{1-i}>0$, so $2^{m}>1$ and $m>0$. Thus, $m$ is a positive integer, and $2^{m}$ is a positive even integer. However, this means that $2^{1-i}$ must also be a positive integer, and since $2^{m}$ is even, it must be odd. This only happens when $i=1$, and thus the only solution is $\boxed{i = 1, m=1.}$

Answer 2

Guide:

Note that $0 < 2^{1-i}\le 1$

• If $m>1$, then $2^m-1 \ge 3 > 1 \ge 2^{1-i}$.

• I will leave the exercise to examine what happens when $m<1$ as an exercise.