Prove $s=2^{m-1} \bmod 2^m=2^{n-1}\bmod 2^n$, then $n=m$

Question

How could I prove:

If $s=2^{m-1} \bmod 2^m = 2^{n-1}\bmod 2^n$, then $n=m$

where $k,m,n \in \mathbb{N}$

Answer 1

What is written there means

$$x=2^{m-1}+k2^m=2^{n-1}+j2^n\;,\;\;k,j\in\Bbb Z$$

Suppose $\;n\le m\;$ , then we get

$$2^{n-1}\left(2^{m-n}+k2^{m-n+1}\right)=2^{n-1}\left(1+2j\right)\stackrel{\text{cancellation}}\implies$$

$$2^{m-n}+k2^{m-n+1}=1+2j\implies2^{m-n}+k2^{m-n+1}-2j=1$$

and this is possible iff $\;n=m\;$ and then $\;k=j\;$ (why? Observe that one side is odd, the other is even...)

Complete the argument now.

Answer 2

There exists $(k,l)\in\mathbb{Z}^2$ such that $s=2^{m-1}+k2^m=2^{n-1}+l2^n$. Thus $2^{n-1}-2^{m-1}=k2^m-l2^n$. We can without loss of generality suppose that $n\geqslant m$, thus $$ 2^{n-m}-1=2k-l2^{n-m+1} $$ If $n>m$ then $1=2^{n-m}-2k+l2^{n-m+1}$ is even which is not so $n=m$.