The slope of $nx\ \%\ m$

Question

(There is a follow-up question at MO.)

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Let $x\ \%\ m$ be the residue of $x$ modulo $m$, i.e.

$$x \equiv x\ \%\ m\pmod{m}$$

Let $\mu^n_m(x)$ denote multiplication by $n$ modulo $m$, i.e.

$$\mu^n_m(x) = nx\ \%\ m$$

Plotting $\mu^n_m(x)$ for $0 < n < m$ yields patterns with characteristic "slopes", here for $m=64$:

_{[click image to enlarge]}

While one could define the slope of $\mu^n_m(x)$ to be just $n$, this is not, what the patterns reveal. So I was looking for a more appropriate definition of the slope of $\mu^n_m(x)$ and came up with the following definition:

Let $n_0$ be the number greater than $1$ which minimizes $(n_0 - 1)^2 + (\mu^n_m(n_0) - n)^2$. Define then the slope $s^n_m$ of $\mu^n_m(x)$ to be the number

$$s^n_m = (\mu^n_m(n_0) - n)/(n_0 - 1)\pmod{m},$$

the latter being a shortcut for

$$((n_0 - 1)s^n_m - \mu^n_m(n_0) + n)\ \%\ m = 0$$

I've chosen this definition for two reasons:

1. Part 1 (the definition of $n_0$) considers the fact, that the dominant lines in the plot of $\mu^n_m(x)$ (which "show" the slope of the pattern) are those with a minimal number of parallels, i.e. a maximal distance between parallels, and thus the highest number of points on them, i.e. a minimal distance of points on them.

2. You can apply the same definition for the non-modular case: the number $n_0>1$ that minimizes $(n_0 - 1)^2 + (n\cdot n_0 - n)^2 = (n_0 - 1)^2(1 + n^2)$ is always $2$ – independently of $n$ –, and the slope is accordingly $(n\cdot n_0 - n)/(n_0 - 1) = n$.

Find here the values for $n_0$ and $s^n_m$ for $m = 64$ and $0 < n < m$ together with the "visual" slopes of the dominant lines:

My question is:

Are there closed expressions for the values of $n_0$ and $s^n_m$ as defined above – as number theoretic functions of $n$ and $m$?

Furthermore:

What can be generally said about the slope $s^n_m$? When is it $0$ or $1$? When does it "jump" when going from $n$ to $n+1$?

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To make clear that no regular pattern and no simple closed expression for $s^n_m$ is to be expected (except for $m$ prime), I plotted $s^n_{256}$ as shades of gray, from white for $s^n_{256} = 0$ to black for $s^{255}_{256} = 255$:

Answer 1

Observations for $m=64$:

$s^n_m = n$ with these exceptions

• $s^n_m = 0$ for $n = k\cdot m/8$ with $k = 1,\dots,7$

• $s^n_m = 1$ for $n = k\cdot m/4 + 1$ with $k = 0,\dots,3$

• $s^n_m = n - m/2$ for $n = 34, 35, 47, 53,54$

Not very surprisingly one finds for prime numbers $p$ that $s^n_p = n$ for all $n$, e.g. for $p=61$:

Stated otherwise: $s^n_m = n$ for all $n < m$ iff $\mathbb{Z}/m\mathbb{Z}$ is a field.

In this respect, prime numbers behave like integers in non-modular arithemtic:

$$s^{n_1}_p = s^{n_2}_p \equiv n_1 = n_2$$

What's missing is a proof!