Is it possible to find all values of $k$, like a function $k=f(m)?$

Question

Is it possible to find all values of $k$, like a function $k=f(m)$, such that

$$\begin{cases} 81k-41\equiv 0 \mod{2^m} \\ \frac{81k-41}{2^m} \equiv 1 \mod 2 \end{cases}$$

Here, $$\left\{ m,k \right\}\in \mathbb {Z^{+}}$$

I can found a random values: for example

$m=3,k=1$

$m=1,k=3...$

My question is:

Is it possible to find a general form for value of $m$ like a function $k=f(m)?$

If I express my problem with words, for example, $m=10$ and I want to find direct value ,what is value of $k?$

In short, I want to make $k$ dependent on $m$ in general, like a function.

Is it possible?

Answer 1

If $k=1$ and $m=3$ then $40\equiv 0 \mod 2$ but $40 \not\equiv 1 \mod 16$.

So, I don't know to what correspond your "random values", but certainly not to solutions of your system of equations.

In general, if $81k-41 \equiv 1 \mod 2^{m+1}$, then by definition of the modulo, we have $81k-41 \equiv 1 \mod 2^{m}$, and by properties of the modulo relation, $0\equiv 1 \mod 2^m$. Equivalently, this means that $2^m$ divides $1$. It can happen only if $m=0$.

In this case, the first equation is satisfied for all $k$, and the second one may be rewritten as $k\equiv 1 \mod 2$.

To sum up: if $k$ is odd, the equations of your system are not satisfied for any $m$.

If $k$ is even, the only solution of your equation is $m=0$.

Answer 2

This is an answer to the edited question.

Your equations mean that $81k-41=(2n+1)2^m$ for some integer $n\ge0$, or equivalently, $81k\equiv2^m+41 \mod{2^{m+1}}$.

Let $a$ be a modular multiplicative inverse of $81$ mod $2^{m+1}$, i.e., an integer such that $81a\equiv1\mod{2^{m+1}}$. (This can be computed using the extended Euclidean algorithm. Alternatively, Euler's theorem gives $a\equiv81^{2^m-1}\mod{2^{m+1}}$ as an explicit solution, but this is relatively inefficient.) Then we have $k\equiv(2^m+41)a\mod{2^{m+1}}$. This gives all the solutions for $k$.