TITLE: Product of prime powers equal to $1$ in the ring of integers modulo $2^{127}$
QUESTION [8 upvotes]: I am working with the multiplicative ring of integers modulo $2^{127}$.
Consider the set  $E=\{(k,l) \mid 5^k \cdot 3^l \equiv 1\mod 2^{127}, k > 0, l> 0\}$.
I wonder if anybody knows or has an idea where to look for a result related to a lower bound for $M=\min\{k+l \mid (k,l)\in E \}$.
We have that $0<M\leq \mathrm{ord}_{\mathbb{Z}_{2^{127}}}(5)+\mathrm{ord}_{\mathbb{Z}_{2^{127}}}(3)$ where $\mathrm{ord}_{\mathbb{Z}_{2^{127}}}(5)=2^{125}$ and $\mathrm{ord}_{\mathbb{Z}_{2^{127}}}(3)=2^{125}$ (orders of these primes in the multiplicative ring $\mathbb{Z}_{2^{127}}$).
I also would like to generalize the above for primes other than 5 and 3.
Is there a result about a tighter lower bound for $M$?

REPLY [3 votes]: Claim:

Let $c=40647290924413185736448652556727923386$, then the set of solutions is
  given by $$A = \{(k,l) \in (\Bbb Z/2^{126} \Bbb Z)^2 \mid 5^k3^l \equiv 1\bmod 2^{127}\}=
 \{(cn, 2 n) \mid n \in \Bbb Z\}$$

This gives an explicit formula for all solutions of $0\leq l+k < 2^{126}$, namely $$l+k = (2n \bmod 2^{126})+(cn \bmod 2^{126})$$ with $0<n<2^{125}$. I find it hard to calculate the minimum of that expression.
Reason:
The set $A = \{(k,l) \mid 5^k3^l \equiv 1\bmod 2^{127}\}$ can be interpreted as a subspace of $(\Bbb Z/2^{126} \Bbb Z)^2$. We only need to find a generator of that subspace to give an explicit characterization of $A$.
This is hard in general, but in the case of the modulus $2^{n}$ we can do the following: Given $k,l \in \Bbb Z/ 2^{n-1} \Bbb Z$ such that  $5^k3^l \equiv 1 \bmod 2^{n}$ we also know that $5^k3^l \equiv 1 \bmod 2^{n-1}$. Conversely, given a solution to the second equation, we can try to lift $(k,l)$ from $ \Bbb Z/ 2^{n-2} \Bbb Z$ to $ \Bbb Z/ 2^{n-1} \Bbb Z$.
Now by consideration $\bmod \,2^3$ it is clear that both $k$ and $l$ are even. We can therefore lift a solution of $5^k3^2 \bmod 2^3$ (my little python code did that pretty instantly) to find that $$5^{c}3^2\equiv 1 \bmod 2^{127}$$ which is of order $2^{125}$ in $\Bbb Z/2^{126} \Bbb Z$ and therefore a generator of $A$. 

EDIT: One can use continued fractions of $c/2^{m}$ to obtain small
 values of $(cn \bmod 2^{m})$ and thus also of $l+k$. The lowest I found so far is $$\begin{eqnarray}
 l&=&11726533429350798020\\
 k&=&\;\,\;\;391079140617450804\\l+k&=&12117612569968248824<\sqrt{2^{127}}<2^{64}.
 \end{eqnarray}$$