Given the cycle group $Zn$ such that $n=2^w$, where $w$ is an integer. Consider a $\delta$, $0<\delta<n$, not coprime with $n$. So, this will not generate every element in the group. From the fundamental theorem of arithmetic, we know that $\delta$ can be written as $2^p\times\delta'$. Let $e=2^p$. Can we prove that for an $x$, if $x+k\delta\equiv f(mod\:n)$ holds for some $k$ where $0\leq f<n$, then for any integer $i$ there exists a $k'$ such that $x+ie+k'\delta\equiv f (mod\:n)$.
It seems very natural to me, but I couldn't prove it, and I would be so grateful if anyone can help me out with this.
I assume that when you say
you mean that $\gcd(2,\delta')=1$. If this is indeed the case, $\delta'$ has an inverse in the cyclic group $Z_{2^{w-p}}$. Let $\gamma$ be an integer equivalent to this inverse modulo $2^{w-p}$. Then, choosing $k' = k - i \gamma$ works. This works out if you just do the algebra.
Improved notation would indeed make this fact more 'natural' to see.