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Let the multiplication graph $n/m$ be the graph with $m$ points distributed evenly on a circle and a line between two points $a$, $b$ when $an \equiv b\operatorname{mod} m$.

These graphs often look somehow random but by carefully choosing $n$ and $m$ one finds intricate patterns. Let $n = 20, 40, 60, 80, 100, 120$ and

$m = n + 19$

enter image description here

$m = 2n + 19$

enter image description here

$m = 3n + 19$

enter image description here

Note that it seems essential that $19 = 20 -1$ which guarantees that $m,n$ are coprime.

Let $A, B, C$ be arbitrary positive numbers.

Why does the graph $n/m$ have $(C-1)A+1$ petals when $n = AB$ and $m = Cn + B - 1 = B(AC +1) - 1$?

(Note that the number of petals doesn't depend on $B$!)

For example with $A=2$, $B=19$, $C=6$, the graph $38/246$ with $38 = 2\cdot 19$ and $246 = 6\cdot 38 + 18 = 2\cdot 3\cdot 41$ has $11 = 5\cdot 2 + 1$ petals:

...

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    $\begingroup$ This video does not contain an answer but it does contain some hint as to where to look for answers, and also some hints that the author knows more about the subject than he is telling us (so maybe contacting him would be a good idea): youtube.com/watch?v=qhbuKbxJsk8 $\endgroup$
    – Vincent
    Commented Nov 16, 2018 at 16:04
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    $\begingroup$ Thanks, I know this video well (in fact: it did inspire my question), but Burkard Polster seems to be a busy man. $\endgroup$ Commented Nov 16, 2018 at 16:07
  • $\begingroup$ A similar question was asked here. $\endgroup$ Commented Nov 16, 2018 at 18:55
  • $\begingroup$ Have you tried to investigate about the $ K $ -th roots of unity in $ \mathbb{Z}/m\mathbb{Z} $ with $K $ the number of cusps ? $\endgroup$ Commented Nov 18, 2018 at 21:19

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The following does not answer the question about the number of cusps, but might be useful.

Consider a continuous version of your construction: draw a line through the points $(\cos t, \sin t)$ and $(\cos nt, \sin nt)$ for every $t$ and look at its envelope. If $F(x,y,t)=0$ is an equation of this line, then the envelope is found from resolving the system of equations $$ F(x,y,t)=0, \quad \frac{\partial F}{\partial t}(x,y,t)=0 $$ with respect to $t$. The result is $$ \gamma(t) = \frac{1}{n+1} \begin{pmatrix} \cos nt + n\cos t\\ \sin nt + n\sin t \end{pmatrix} $$ which parametrizes the trajectory of a point on a circle of radius $\frac{1}{n+1}$ rolling on a circle of radius $\frac{n-1}{n+1}$.

(That this trajectory has the $t$-$nt$ lines as tangents can also be proved geometrically by looking at the instantaneous motion of the point: it rotates about the point of contact between the two circles.)

The curve is called epicycloid, and the construction is attributed to Cremona, see for example this webpage.

It has $n-1$ cusp, which is different from your result. Probably the reason is that you consider a discrete set of lines. By the way, your pictures look similar to those for epicycloids for rational non-integer ratio of the circles radii, see the Wikipedia page.

Also I found a page discussing Mathematica codes for Cremona construction.

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