Explanation of a regular pattern only occuring for prime numbers
$begingroup$
Consider multiplication group tables modulo $n$ with entries $k_{ij} = (icdot j) % n$ visualized according to these principles:
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 $k = n$
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, i.e. that there are only and exactly $4$ values along these lines.
For the sake of completeness: the multiplication group table modulo $8 = 4cdot 2$ (which also qualifies, but not so obviously):
group-theory number-theory visualization
$endgroup$
add a comment |
$begingroup$
Consider multiplication group tables modulo $n$ with entries $k_{ij} = (icdot j) % n$ visualized according to these principles:
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 $k = n$
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, i.e. that there are only and exactly $4$ values along these lines.
For the sake of completeness: the multiplication group table modulo $8 = 4cdot 2$ (which also qualifies, but not so obviously):
group-theory number-theory visualization
$endgroup$
7
$begingroup$
What do they look like when $n$ is not $4$ times a prime?
$endgroup$
– Robert Israel
Feb 26 at 21:53
$begingroup$
Could you include an example of a not so regular table?
$endgroup$
– Servaes
Feb 26 at 21:54
$begingroup$
To me the $nne 4p$ examples look just about as regular as the $n=4p$ ones, with the exception of the locations of the "size $1.5$" dots.
$endgroup$
– Robert Israel
Feb 26 at 22:02
$begingroup$
@RobertIsrael: To me, too. My question is just about the "size 1.5" dots.
$endgroup$
– Hans-Peter Stricker
Feb 26 at 22:04
add a comment |
$begingroup$
Consider multiplication group tables modulo $n$ with entries $k_{ij} = (icdot j) % n$ visualized according to these principles:
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 $k = n$
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, i.e. that there are only and exactly $4$ values along these lines.
For the sake of completeness: the multiplication group table modulo $8 = 4cdot 2$ (which also qualifies, but not so obviously):
group-theory number-theory visualization
$endgroup$
Consider multiplication group tables modulo $n$ with entries $k_{ij} = (icdot j) % n$ visualized according to these principles:
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 $k = n$
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, i.e. that there are only and exactly $4$ values along these lines.
For the sake of completeness: the multiplication group table modulo $8 = 4cdot 2$ (which also qualifies, but not so obviously):
group-theory number-theory visualization
group-theory number-theory visualization
edited Feb 27 at 0:12
Hans-Peter Stricker
asked Feb 26 at 21:49
Hans-Peter StrickerHans-Peter Stricker
6,74043996
6,74043996
7
$begingroup$
What do they look like when $n$ is not $4$ times a prime?
$endgroup$
– Robert Israel
Feb 26 at 21:53
$begingroup$
Could you include an example of a not so regular table?
$endgroup$
– Servaes
Feb 26 at 21:54
$begingroup$
To me the $nne 4p$ examples look just about as regular as the $n=4p$ ones, with the exception of the locations of the "size $1.5$" dots.
$endgroup$
– Robert Israel
Feb 26 at 22:02
$begingroup$
@RobertIsrael: To me, too. My question is just about the "size 1.5" dots.
$endgroup$
– Hans-Peter Stricker
Feb 26 at 22:04
add a comment |
7
$begingroup$
What do they look like when $n$ is not $4$ times a prime?
$endgroup$
– Robert Israel
Feb 26 at 21:53
$begingroup$
Could you include an example of a not so regular table?
$endgroup$
– Servaes
Feb 26 at 21:54
$begingroup$
To me the $nne 4p$ examples look just about as regular as the $n=4p$ ones, with the exception of the locations of the "size $1.5$" dots.
$endgroup$
– Robert Israel
Feb 26 at 22:02
$begingroup$
@RobertIsrael: To me, too. My question is just about the "size 1.5" dots.
$endgroup$
– Hans-Peter Stricker
Feb 26 at 22:04
7
7
$begingroup$
What do they look like when $n$ is not $4$ times a prime?
$endgroup$
– Robert Israel
Feb 26 at 21:53
$begingroup$
What do they look like when $n$ is not $4$ times a prime?
$endgroup$
– Robert Israel
Feb 26 at 21:53
$begingroup$
Could you include an example of a not so regular table?
$endgroup$
– Servaes
Feb 26 at 21:54
$begingroup$
Could you include an example of a not so regular table?
$endgroup$
– Servaes
Feb 26 at 21:54
$begingroup$
To me the $nne 4p$ examples look just about as regular as the $n=4p$ ones, with the exception of the locations of the "size $1.5$" dots.
$endgroup$
– Robert Israel
Feb 26 at 22:02
$begingroup$
To me the $nne 4p$ examples look just about as regular as the $n=4p$ ones, with the exception of the locations of the "size $1.5$" dots.
$endgroup$
– Robert Israel
Feb 26 at 22:02
$begingroup$
@RobertIsrael: To me, too. My question is just about the "size 1.5" dots.
$endgroup$
– Hans-Peter Stricker
Feb 26 at 22:04
$begingroup$
@RobertIsrael: To me, too. My question is just about the "size 1.5" dots.
$endgroup$
– Hans-Peter Stricker
Feb 26 at 22:04
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 $x=pm q,y=pm q$.
UPDATE: Some insights on the general case here. Write $n=4q+t$ for $tin{-1,0,1,2}$. Then we have:
$$
begin{align}
lfloor n/4rfloor &= q+lfloor t/4rfloor\
lfloor 3n/4rfloor &= 3q+lfloor 3t/4rfloor
end{align}
$$
where we can make the following table:
$$
begin{array}{c|r|r|r|r}
t & -1 & 0 & 1 & 2\
hline
lfloor t/4rfloor & -1 & 0 & 0 & 0\
lfloor 3t/4rfloor & -1 & 0 & 0 & 1
end{array}
$$
and so we can cover each $t$-case based on that.
CASE $t=0$
Let us first reconsider the case $t=0$ so that $n=4q$, which was already covered above.
Then:
$$
xy=q(2mu+1)
$$
will yield "size 1.5" dots. If $q=ab$ is composite then:
$$
begin{align}
x&=a\
y&=b(2mu+1)
end{align}
$$
will yield a vertical line of points $2b$ apart that are all of "size 1.5". If on the other hand $q$ is prime, then:
$$
begin{align}
x&=q\
y&=2mu+1
end{align}
$$
will yield a very visible vertical line of points only $2$ apart.
CASE $t=1$
Suppose $t=1$. Then $n=4q+1$. For this case, having $xyequivlfloor n/4rfloor=q$ implies:
$$
begin{align}
xy &=q+mu n
end{align}
$$
Now, since $q,n$ are relatively prime, $mu$ must be a multiple of $q$ for two such situations to have any shared factor. So assume $mu=gamma q$. Then:
$$
xy=(gamma n+1)q
$$
So if $q=ab$ we have:
$$
begin{align}
x &= a\
y &= (gamma n+1)b
end{align}
$$
showing that no vertical lines can exist, since the $y$-values must be far more than $n$ units apart. On the other hand, plugging in $mu=0$ in the $y$-expression above shows why points tend to lie on the hyperbola:
$$
xy=q
$$
I think the other cases can be broken down in a similar fashion, so I will stop here.
$endgroup$
$begingroup$
Thanks for this answer! May I help with specific visualizations? Where and why are you "still a bit unsure"?
$endgroup$
– Hans-Peter Stricker
Feb 26 at 23:43
$begingroup$
@HansStricker: Thank you. Well, it is just that the "patternless" cases, ie. no horizontal or vertical lines, behave in a more chaotic-seeming manner regarding the 1.5 dots.
$endgroup$
– String
Feb 26 at 23:47
$begingroup$
@HansStricker: If you are able to draw more clear conclusions for those "chaotic" cases where $nneq 4q$ or to visualize them in order to see different principles at play, please let me know. I think the floor function is what throws in the "chaos" in the first place :o)
$endgroup$
– String
Feb 26 at 23:50
1
$begingroup$
@HansStricker: I would put my money on Robert Israel or the like for deeper insights. I merely took his work and explained some helpful details, I guess :o) I regret my comment about the floor function. The main players are the factorizations themselves. The floor function merely compensates for the unsuccesful division of $n$. Any rounding technique would be equally chaotic, I think.
$endgroup$
– String
Feb 27 at 0:00
1
$begingroup$
@HansStricker: I updated my answer. It has turned into more of a full distinct answer now. I hope it makes sense and sheds some light!
$endgroup$
– String
Feb 27 at 10:44
|
show 1 more comment
$begingroup$
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 $y = 3p$.
$endgroup$
1
$begingroup$
Where and how does the primeness of $p$ come into play?
$endgroup$
– Hans-Peter Stricker
Feb 26 at 22:14
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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oldest
votes
$begingroup$
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 $x=pm q,y=pm q$.
UPDATE: Some insights on the general case here. Write $n=4q+t$ for $tin{-1,0,1,2}$. Then we have:
$$
begin{align}
lfloor n/4rfloor &= q+lfloor t/4rfloor\
lfloor 3n/4rfloor &= 3q+lfloor 3t/4rfloor
end{align}
$$
where we can make the following table:
$$
begin{array}{c|r|r|r|r}
t & -1 & 0 & 1 & 2\
hline
lfloor t/4rfloor & -1 & 0 & 0 & 0\
lfloor 3t/4rfloor & -1 & 0 & 0 & 1
end{array}
$$
and so we can cover each $t$-case based on that.
CASE $t=0$
Let us first reconsider the case $t=0$ so that $n=4q$, which was already covered above.
Then:
$$
xy=q(2mu+1)
$$
will yield "size 1.5" dots. If $q=ab$ is composite then:
$$
begin{align}
x&=a\
y&=b(2mu+1)
end{align}
$$
will yield a vertical line of points $2b$ apart that are all of "size 1.5". If on the other hand $q$ is prime, then:
$$
begin{align}
x&=q\
y&=2mu+1
end{align}
$$
will yield a very visible vertical line of points only $2$ apart.
CASE $t=1$
Suppose $t=1$. Then $n=4q+1$. For this case, having $xyequivlfloor n/4rfloor=q$ implies:
$$
begin{align}
xy &=q+mu n
end{align}
$$
Now, since $q,n$ are relatively prime, $mu$ must be a multiple of $q$ for two such situations to have any shared factor. So assume $mu=gamma q$. Then:
$$
xy=(gamma n+1)q
$$
So if $q=ab$ we have:
$$
begin{align}
x &= a\
y &= (gamma n+1)b
end{align}
$$
showing that no vertical lines can exist, since the $y$-values must be far more than $n$ units apart. On the other hand, plugging in $mu=0$ in the $y$-expression above shows why points tend to lie on the hyperbola:
$$
xy=q
$$
I think the other cases can be broken down in a similar fashion, so I will stop here.
$endgroup$
$begingroup$
Thanks for this answer! May I help with specific visualizations? Where and why are you "still a bit unsure"?
$endgroup$
– Hans-Peter Stricker
Feb 26 at 23:43
$begingroup$
@HansStricker: Thank you. Well, it is just that the "patternless" cases, ie. no horizontal or vertical lines, behave in a more chaotic-seeming manner regarding the 1.5 dots.
$endgroup$
– String
Feb 26 at 23:47
$begingroup$
@HansStricker: If you are able to draw more clear conclusions for those "chaotic" cases where $nneq 4q$ or to visualize them in order to see different principles at play, please let me know. I think the floor function is what throws in the "chaos" in the first place :o)
$endgroup$
– String
Feb 26 at 23:50
1
$begingroup$
@HansStricker: I would put my money on Robert Israel or the like for deeper insights. I merely took his work and explained some helpful details, I guess :o) I regret my comment about the floor function. The main players are the factorizations themselves. The floor function merely compensates for the unsuccesful division of $n$. Any rounding technique would be equally chaotic, I think.
$endgroup$
– String
Feb 27 at 0:00
1
$begingroup$
@HansStricker: I updated my answer. It has turned into more of a full distinct answer now. I hope it makes sense and sheds some light!
$endgroup$
– String
Feb 27 at 10:44
|
show 1 more comment
$begingroup$
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 $x=pm q,y=pm q$.
UPDATE: Some insights on the general case here. Write $n=4q+t$ for $tin{-1,0,1,2}$. Then we have:
$$
begin{align}
lfloor n/4rfloor &= q+lfloor t/4rfloor\
lfloor 3n/4rfloor &= 3q+lfloor 3t/4rfloor
end{align}
$$
where we can make the following table:
$$
begin{array}{c|r|r|r|r}
t & -1 & 0 & 1 & 2\
hline
lfloor t/4rfloor & -1 & 0 & 0 & 0\
lfloor 3t/4rfloor & -1 & 0 & 0 & 1
end{array}
$$
and so we can cover each $t$-case based on that.
CASE $t=0$
Let us first reconsider the case $t=0$ so that $n=4q$, which was already covered above.
Then:
$$
xy=q(2mu+1)
$$
will yield "size 1.5" dots. If $q=ab$ is composite then:
$$
begin{align}
x&=a\
y&=b(2mu+1)
end{align}
$$
will yield a vertical line of points $2b$ apart that are all of "size 1.5". If on the other hand $q$ is prime, then:
$$
begin{align}
x&=q\
y&=2mu+1
end{align}
$$
will yield a very visible vertical line of points only $2$ apart.
CASE $t=1$
Suppose $t=1$. Then $n=4q+1$. For this case, having $xyequivlfloor n/4rfloor=q$ implies:
$$
begin{align}
xy &=q+mu n
end{align}
$$
Now, since $q,n$ are relatively prime, $mu$ must be a multiple of $q$ for two such situations to have any shared factor. So assume $mu=gamma q$. Then:
$$
xy=(gamma n+1)q
$$
So if $q=ab$ we have:
$$
begin{align}
x &= a\
y &= (gamma n+1)b
end{align}
$$
showing that no vertical lines can exist, since the $y$-values must be far more than $n$ units apart. On the other hand, plugging in $mu=0$ in the $y$-expression above shows why points tend to lie on the hyperbola:
$$
xy=q
$$
I think the other cases can be broken down in a similar fashion, so I will stop here.
$endgroup$
$begingroup$
Thanks for this answer! May I help with specific visualizations? Where and why are you "still a bit unsure"?
$endgroup$
– Hans-Peter Stricker
Feb 26 at 23:43
$begingroup$
@HansStricker: Thank you. Well, it is just that the "patternless" cases, ie. no horizontal or vertical lines, behave in a more chaotic-seeming manner regarding the 1.5 dots.
$endgroup$
– String
Feb 26 at 23:47
$begingroup$
@HansStricker: If you are able to draw more clear conclusions for those "chaotic" cases where $nneq 4q$ or to visualize them in order to see different principles at play, please let me know. I think the floor function is what throws in the "chaos" in the first place :o)
$endgroup$
– String
Feb 26 at 23:50
1
$begingroup$
@HansStricker: I would put my money on Robert Israel or the like for deeper insights. I merely took his work and explained some helpful details, I guess :o) I regret my comment about the floor function. The main players are the factorizations themselves. The floor function merely compensates for the unsuccesful division of $n$. Any rounding technique would be equally chaotic, I think.
$endgroup$
– String
Feb 27 at 0:00
1
$begingroup$
@HansStricker: I updated my answer. It has turned into more of a full distinct answer now. I hope it makes sense and sheds some light!
$endgroup$
– String
Feb 27 at 10:44
|
show 1 more comment
$begingroup$
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 $x=pm q,y=pm q$.
UPDATE: Some insights on the general case here. Write $n=4q+t$ for $tin{-1,0,1,2}$. Then we have:
$$
begin{align}
lfloor n/4rfloor &= q+lfloor t/4rfloor\
lfloor 3n/4rfloor &= 3q+lfloor 3t/4rfloor
end{align}
$$
where we can make the following table:
$$
begin{array}{c|r|r|r|r}
t & -1 & 0 & 1 & 2\
hline
lfloor t/4rfloor & -1 & 0 & 0 & 0\
lfloor 3t/4rfloor & -1 & 0 & 0 & 1
end{array}
$$
and so we can cover each $t$-case based on that.
CASE $t=0$
Let us first reconsider the case $t=0$ so that $n=4q$, which was already covered above.
Then:
$$
xy=q(2mu+1)
$$
will yield "size 1.5" dots. If $q=ab$ is composite then:
$$
begin{align}
x&=a\
y&=b(2mu+1)
end{align}
$$
will yield a vertical line of points $2b$ apart that are all of "size 1.5". If on the other hand $q$ is prime, then:
$$
begin{align}
x&=q\
y&=2mu+1
end{align}
$$
will yield a very visible vertical line of points only $2$ apart.
CASE $t=1$
Suppose $t=1$. Then $n=4q+1$. For this case, having $xyequivlfloor n/4rfloor=q$ implies:
$$
begin{align}
xy &=q+mu n
end{align}
$$
Now, since $q,n$ are relatively prime, $mu$ must be a multiple of $q$ for two such situations to have any shared factor. So assume $mu=gamma q$. Then:
$$
xy=(gamma n+1)q
$$
So if $q=ab$ we have:
$$
begin{align}
x &= a\
y &= (gamma n+1)b
end{align}
$$
showing that no vertical lines can exist, since the $y$-values must be far more than $n$ units apart. On the other hand, plugging in $mu=0$ in the $y$-expression above shows why points tend to lie on the hyperbola:
$$
xy=q
$$
I think the other cases can be broken down in a similar fashion, so I will stop here.
$endgroup$
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 $x=pm q,y=pm q$.
UPDATE: Some insights on the general case here. Write $n=4q+t$ for $tin{-1,0,1,2}$. Then we have:
$$
begin{align}
lfloor n/4rfloor &= q+lfloor t/4rfloor\
lfloor 3n/4rfloor &= 3q+lfloor 3t/4rfloor
end{align}
$$
where we can make the following table:
$$
begin{array}{c|r|r|r|r}
t & -1 & 0 & 1 & 2\
hline
lfloor t/4rfloor & -1 & 0 & 0 & 0\
lfloor 3t/4rfloor & -1 & 0 & 0 & 1
end{array}
$$
and so we can cover each $t$-case based on that.
CASE $t=0$
Let us first reconsider the case $t=0$ so that $n=4q$, which was already covered above.
Then:
$$
xy=q(2mu+1)
$$
will yield "size 1.5" dots. If $q=ab$ is composite then:
$$
begin{align}
x&=a\
y&=b(2mu+1)
end{align}
$$
will yield a vertical line of points $2b$ apart that are all of "size 1.5". If on the other hand $q$ is prime, then:
$$
begin{align}
x&=q\
y&=2mu+1
end{align}
$$
will yield a very visible vertical line of points only $2$ apart.
CASE $t=1$
Suppose $t=1$. Then $n=4q+1$. For this case, having $xyequivlfloor n/4rfloor=q$ implies:
$$
begin{align}
xy &=q+mu n
end{align}
$$
Now, since $q,n$ are relatively prime, $mu$ must be a multiple of $q$ for two such situations to have any shared factor. So assume $mu=gamma q$. Then:
$$
xy=(gamma n+1)q
$$
So if $q=ab$ we have:
$$
begin{align}
x &= a\
y &= (gamma n+1)b
end{align}
$$
showing that no vertical lines can exist, since the $y$-values must be far more than $n$ units apart. On the other hand, plugging in $mu=0$ in the $y$-expression above shows why points tend to lie on the hyperbola:
$$
xy=q
$$
I think the other cases can be broken down in a similar fashion, so I will stop here.
edited Feb 27 at 11:38
answered Feb 26 at 23:37
StringString
13.8k32756
13.8k32756
$begingroup$
Thanks for this answer! May I help with specific visualizations? Where and why are you "still a bit unsure"?
$endgroup$
– Hans-Peter Stricker
Feb 26 at 23:43
$begingroup$
@HansStricker: Thank you. Well, it is just that the "patternless" cases, ie. no horizontal or vertical lines, behave in a more chaotic-seeming manner regarding the 1.5 dots.
$endgroup$
– String
Feb 26 at 23:47
$begingroup$
@HansStricker: If you are able to draw more clear conclusions for those "chaotic" cases where $nneq 4q$ or to visualize them in order to see different principles at play, please let me know. I think the floor function is what throws in the "chaos" in the first place :o)
$endgroup$
– String
Feb 26 at 23:50
1
$begingroup$
@HansStricker: I would put my money on Robert Israel or the like for deeper insights. I merely took his work and explained some helpful details, I guess :o) I regret my comment about the floor function. The main players are the factorizations themselves. The floor function merely compensates for the unsuccesful division of $n$. Any rounding technique would be equally chaotic, I think.
$endgroup$
– String
Feb 27 at 0:00
1
$begingroup$
@HansStricker: I updated my answer. It has turned into more of a full distinct answer now. I hope it makes sense and sheds some light!
$endgroup$
– String
Feb 27 at 10:44
|
show 1 more comment
$begingroup$
Thanks for this answer! May I help with specific visualizations? Where and why are you "still a bit unsure"?
$endgroup$
– Hans-Peter Stricker
Feb 26 at 23:43
$begingroup$
@HansStricker: Thank you. Well, it is just that the "patternless" cases, ie. no horizontal or vertical lines, behave in a more chaotic-seeming manner regarding the 1.5 dots.
$endgroup$
– String
Feb 26 at 23:47
$begingroup$
@HansStricker: If you are able to draw more clear conclusions for those "chaotic" cases where $nneq 4q$ or to visualize them in order to see different principles at play, please let me know. I think the floor function is what throws in the "chaos" in the first place :o)
$endgroup$
– String
Feb 26 at 23:50
1
$begingroup$
@HansStricker: I would put my money on Robert Israel or the like for deeper insights. I merely took his work and explained some helpful details, I guess :o) I regret my comment about the floor function. The main players are the factorizations themselves. The floor function merely compensates for the unsuccesful division of $n$. Any rounding technique would be equally chaotic, I think.
$endgroup$
– String
Feb 27 at 0:00
1
$begingroup$
@HansStricker: I updated my answer. It has turned into more of a full distinct answer now. I hope it makes sense and sheds some light!
$endgroup$
– String
Feb 27 at 10:44
$begingroup$
Thanks for this answer! May I help with specific visualizations? Where and why are you "still a bit unsure"?
$endgroup$
– Hans-Peter Stricker
Feb 26 at 23:43
$begingroup$
Thanks for this answer! May I help with specific visualizations? Where and why are you "still a bit unsure"?
$endgroup$
– Hans-Peter Stricker
Feb 26 at 23:43
$begingroup$
@HansStricker: Thank you. Well, it is just that the "patternless" cases, ie. no horizontal or vertical lines, behave in a more chaotic-seeming manner regarding the 1.5 dots.
$endgroup$
– String
Feb 26 at 23:47
$begingroup$
@HansStricker: Thank you. Well, it is just that the "patternless" cases, ie. no horizontal or vertical lines, behave in a more chaotic-seeming manner regarding the 1.5 dots.
$endgroup$
– String
Feb 26 at 23:47
$begingroup$
@HansStricker: If you are able to draw more clear conclusions for those "chaotic" cases where $nneq 4q$ or to visualize them in order to see different principles at play, please let me know. I think the floor function is what throws in the "chaos" in the first place :o)
$endgroup$
– String
Feb 26 at 23:50
$begingroup$
@HansStricker: If you are able to draw more clear conclusions for those "chaotic" cases where $nneq 4q$ or to visualize them in order to see different principles at play, please let me know. I think the floor function is what throws in the "chaos" in the first place :o)
$endgroup$
– String
Feb 26 at 23:50
1
1
$begingroup$
@HansStricker: I would put my money on Robert Israel or the like for deeper insights. I merely took his work and explained some helpful details, I guess :o) I regret my comment about the floor function. The main players are the factorizations themselves. The floor function merely compensates for the unsuccesful division of $n$. Any rounding technique would be equally chaotic, I think.
$endgroup$
– String
Feb 27 at 0:00
$begingroup$
@HansStricker: I would put my money on Robert Israel or the like for deeper insights. I merely took his work and explained some helpful details, I guess :o) I regret my comment about the floor function. The main players are the factorizations themselves. The floor function merely compensates for the unsuccesful division of $n$. Any rounding technique would be equally chaotic, I think.
$endgroup$
– String
Feb 27 at 0:00
1
1
$begingroup$
@HansStricker: I updated my answer. It has turned into more of a full distinct answer now. I hope it makes sense and sheds some light!
$endgroup$
– String
Feb 27 at 10:44
$begingroup$
@HansStricker: I updated my answer. It has turned into more of a full distinct answer now. I hope it makes sense and sheds some light!
$endgroup$
– String
Feb 27 at 10:44
|
show 1 more comment
$begingroup$
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 $y = 3p$.
$endgroup$
1
$begingroup$
Where and how does the primeness of $p$ come into play?
$endgroup$
– Hans-Peter Stricker
Feb 26 at 22:14
add a comment |
$begingroup$
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 $y = 3p$.
$endgroup$
1
$begingroup$
Where and how does the primeness of $p$ come into play?
$endgroup$
– Hans-Peter Stricker
Feb 26 at 22:14
add a comment |
$begingroup$
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 $y = 3p$.
$endgroup$
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 $y = 3p$.
answered Feb 26 at 22:04
Robert IsraelRobert Israel
330k23219473
330k23219473
1
$begingroup$
Where and how does the primeness of $p$ come into play?
$endgroup$
– Hans-Peter Stricker
Feb 26 at 22:14
add a comment |
1
$begingroup$
Where and how does the primeness of $p$ come into play?
$endgroup$
– Hans-Peter Stricker
Feb 26 at 22:14
1
1
$begingroup$
Where and how does the primeness of $p$ come into play?
$endgroup$
– Hans-Peter Stricker
Feb 26 at 22:14
$begingroup$
Where and how does the primeness of $p$ come into play?
$endgroup$
– Hans-Peter Stricker
Feb 26 at 22:14
add a comment |
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7
$begingroup$
What do they look like when $n$ is not $4$ times a prime?
$endgroup$
– Robert Israel
Feb 26 at 21:53
$begingroup$
Could you include an example of a not so regular table?
$endgroup$
– Servaes
Feb 26 at 21:54
$begingroup$
To me the $nne 4p$ examples look just about as regular as the $n=4p$ ones, with the exception of the locations of the "size $1.5$" dots.
$endgroup$
– Robert Israel
Feb 26 at 22:02
$begingroup$
@RobertIsrael: To me, too. My question is just about the "size 1.5" dots.
$endgroup$
– Hans-Peter Stricker
Feb 26 at 22:04