Htydjtk

Consider multiplication group tables modulo $n$ with entries $k_{ij} = (icdot j) % n$ visualized according to these principles:

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  Colors are assigned to numbers $0 leq k leq n$ from

  
  
  
  
  

  • $color{black}{textsf{black}}$ for $k=0$ over

  
  

  • $color{red}{textsf{red}}$ for $k=lfloor n/4rfloor$ and

  
  

  • $color{silver}{textsf{white}}$ for $k=lfloor n/2rfloor$ and

  
  

  • $color{blue}{textsf{blue}}$ for $k=lfloor 3n/4rfloor$ back to

  
  

  • $color{black}{textsf{black}}$ for $k = n$

  
  
  
  


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  Sizes are assigned to numbers $0 leq k leq n$ by

  
  
  
  
  

  • $textsf{1.5}$ if $k=lfloor n/4rfloor$ or $lfloor 3n/4rfloor$

  
  

  • $textsf{1.0}$ otherwise

  
  
  
  



• Positions are shifted by $(n/2,n/2)$ modulo $n$ to bring $(0,0)$ to the center of the table.

Visualized this way, you will occasionally find (for some $n$) highly regular multiplication group tables like these (with $n=12,20,28,44,52,68$):

My question is:

Why do these patterns occur exactly when $n = 4p$ with a prime number $p$?

Find here some examples for $n neq 4p$, e.g. $n=61, 62, 63, 64$:

Here for some other prime numbers: $n = 4cdot 31 = 124$ and $n = 4cdot 37 = 148$:

One may observe that for $n = 4m$ and $x,y = m$ or $x,y = 3m$ the "size 1.5" dots are systematically separated by $0$ (= black) and $n/2$ (= white) dots:

For the sake of completeness: the multiplication group table modulo $8 = 4cdot 2$ (which also qualifies, but not so obviously):

If $n=4p$, then for $xy equiv p$ or $3p$ mod $n$ you need $p$ to divide $x$ or $y$ but $2$ to divide neither: thus the "size $1.5$" dots are all on the lines $x = p$, $x = 3p$, $y = p$ and $y = 3p$.

To elaborate a bit on Robert Israels fine answer, first note that:
$$
begin{align}
xy&equiv n/4\
xy&equiv 3n/4
end{align}
$$
implies that $n$ must be divisible by $4$. Hence we only have such situations when $n=4q$ for some $q$. This is true whether $q$ is prime or not. So let us consider $n=4q$ a bit further. Then we look at:
$$
begin{align}
xy&equiv q\
xy&equiv 3q
end{align}
$$
which can be summarized as:
$$
xy=q(2m+1)
$$
for some $m$. This actually reveals why $n=64=4cdot 16$ also appears to somewhat work. Here $q=16$ so any solutions to $xy=16(2m+1)$ will work. We have the lines $x,y=pm2,pm4,pm8,pm16$ with varying density of "size 1.5" points.

When $q$ is prime, the picture becomes simpler since all solutions to $xy=q(2m+1)$ have to lie on one of the four lines $x=pm q,y=pm q$.

I am still a bit unsure of what happens for numbers that are not of the form $n=4q$. One thing is that $xyequivlfloor n/4rfloor$ implies $(n-x)y,x(n-y)equivlceil 3n/4rceil$ so this breaks the symmetry from the cases $n=4q$ to some extend. This breaks parts of the patterns constituting the horizontal and vertical lines.