What is the minimum distance between vertices on an integer grid with the form $(m(m+2), 0)p + (m, 1)q$?
$begingroup$
Suppose, for given $m > 0$, we have a set of points of the form $v = (m(m+2), 0)p + (m, 1)q$, with $p, q, m$ integers. What is the minimum distance between two (distinct ones) of them?
Here is the square distance between any point and the origin:
$|v|^2 = (pm^2 + 2pm + qm)^2 + q^2,$
we need to find the $p, q$ that minimizes this distance given $m > 0$.
In practice it is easy enough to check a few points, and find the optimum by trial and error, but I am wondering if there is a closed form solution.
Background: this came up while working on this question: How many colors is necessary so that a rectangle always covers no color more than once?. See the second last conjecture.
(Note: I am not sure if there are more appropriate tags.)
integer-programming
asked Dec 14 '18 at 22:36
Herman TullekenHerman Tulleken
968620
$endgroup$
add a comment |
$begingroup$
Suppose, for given $m > 0$, we have a set of points of the form $v = (m(m+2), 0)p + (m, 1)q$, with $p, q, m$ integers. What is the minimum distance between two (distinct ones) of them?
Here is the square distance between any point and the origin:
$|v|^2 = (pm^2 + 2pm + qm)^2 + q^2,$
we need to find the $p, q$ that minimizes this distance given $m > 0$.
In practice it is easy enough to check a few points, and find the optimum by trial and error, but I am wondering if there is a closed form solution.
Background: this came up while working on this question: How many colors is necessary so that a rectangle always covers no color more than once?. See the second last conjecture.
(Note: I am not sure if there are more appropriate tags.)
integer-programming
asked Dec 14 '18 at 22:36
Herman TullekenHerman Tulleken
968620
$endgroup$
add a comment |
$begingroup$
Suppose, for given $m > 0$, we have a set of points of the form $v = (m(m+2), 0)p + (m, 1)q$, with $p, q, m$ integers. What is the minimum distance between two (distinct ones) of them?
Here is the square distance between any point and the origin:
$|v|^2 = (pm^2 + 2pm + qm)^2 + q^2,$
we need to find the $p, q$ that minimizes this distance given $m > 0$.
In practice it is easy enough to check a few points, and find the optimum by trial and error, but I am wondering if there is a closed form solution.
Background: this came up while working on this question: How many colors is necessary so that a rectangle always covers no color more than once?. See the second last conjecture.
(Note: I am not sure if there are more appropriate tags.)
integer-programming
asked Dec 14 '18 at 22:36
Herman TullekenHerman Tulleken
968620
$endgroup$
Suppose, for given $m > 0$, we have a set of points of the form $v = (m(m+2), 0)p + (m, 1)q$, with $p, q, m$ integers. What is the minimum distance between two (distinct ones) of them?
Here is the square distance between any point and the origin:
$|v|^2 = (pm^2 + 2pm + qm)^2 + q^2,$
we need to find the $p, q$ that minimizes this distance given $m > 0$.
In practice it is easy enough to check a few points, and find the optimum by trial and error, but I am wondering if there is a closed form solution.
Background: this came up while working on this question: How many colors is necessary so that a rectangle always covers no color more than once?. See the second last conjecture.
(Note: I am not sure if there are more appropriate tags.)
integer-programming
integer-programming
asked Dec 14 '18 at 22:36
Herman TullekenHerman Tulleken
968620
asked Dec 14 '18 at 22:36
Herman TullekenHerman Tulleken
968620
edited Dec 14 '18 at 22:57
Herman Tulleken
asked Dec 14 '18 at 22:36
Herman TullekenHerman Tulleken
968620
asked Dec 14 '18 at 22:36
Herman TullekenHerman Tulleken
968620
asked Dec 14 '18 at 22:36
Herman TullekenHerman Tulleken
968620
968620
add a comment |
add a comment |
1 Answer
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$begingroup$
Since $v^2$ is a sum of squares of integers, and it's not $0$, then it must be $1$ (if we can find such a solution). There are two choices:
Case 1: $q=0$ and $pm^2+2pm+qm=pm1$. I can easily guess a solution for
$pm^2+2pm=-1$: $p=1$ and $m=-1$
Case 2: $q=pm 1$ and $pm^2+2pm+qm=0$. Same $p$ and $m$ as above also verify this case.
answered Dec 14 '18 at 22:51
AndreiAndrei
12.6k21128
$endgroup$
$begingroup$
I realized now my question did not specify to find $p, q$ given $m$. I will update my question.
$endgroup$
– Herman Tulleken
Dec 14 '18 at 22:55
add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
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oldest
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active
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active
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votes
$begingroup$
Since $v^2$ is a sum of squares of integers, and it's not $0$, then it must be $1$ (if we can find such a solution). There are two choices:
Case 1: $q=0$ and $pm^2+2pm+qm=pm1$. I can easily guess a solution for
$pm^2+2pm=-1$: $p=1$ and $m=-1$
Case 2: $q=pm 1$ and $pm^2+2pm+qm=0$. Same $p$ and $m$ as above also verify this case.
answered Dec 14 '18 at 22:51
AndreiAndrei
12.6k21128
$endgroup$
$begingroup$
I realized now my question did not specify to find $p, q$ given $m$. I will update my question.
$endgroup$
– Herman Tulleken
Dec 14 '18 at 22:55
add a comment |
$begingroup$
Since $v^2$ is a sum of squares of integers, and it's not $0$, then it must be $1$ (if we can find such a solution). There are two choices:
Case 1: $q=0$ and $pm^2+2pm+qm=pm1$. I can easily guess a solution for
$pm^2+2pm=-1$: $p=1$ and $m=-1$
Case 2: $q=pm 1$ and $pm^2+2pm+qm=0$. Same $p$ and $m$ as above also verify this case.
answered Dec 14 '18 at 22:51
AndreiAndrei
12.6k21128
$endgroup$
$begingroup$
I realized now my question did not specify to find $p, q$ given $m$. I will update my question.
$endgroup$
– Herman Tulleken
Dec 14 '18 at 22:55
add a comment |
$begingroup$
Since $v^2$ is a sum of squares of integers, and it's not $0$, then it must be $1$ (if we can find such a solution). There are two choices:
Case 1: $q=0$ and $pm^2+2pm+qm=pm1$. I can easily guess a solution for
$pm^2+2pm=-1$: $p=1$ and $m=-1$
Case 2: $q=pm 1$ and $pm^2+2pm+qm=0$. Same $p$ and $m$ as above also verify this case.
answered Dec 14 '18 at 22:51
AndreiAndrei
12.6k21128
$endgroup$
Since $v^2$ is a sum of squares of integers, and it's not $0$, then it must be $1$ (if we can find such a solution). There are two choices:
Case 1: $q=0$ and $pm^2+2pm+qm=pm1$. I can easily guess a solution for
$pm^2+2pm=-1$: $p=1$ and $m=-1$
Case 2: $q=pm 1$ and $pm^2+2pm+qm=0$. Same $p$ and $m$ as above also verify this case.
answered Dec 14 '18 at 22:51
AndreiAndrei
12.6k21128
answered Dec 14 '18 at 22:51
AndreiAndrei
12.6k21128
answered Dec 14 '18 at 22:51
AndreiAndrei
12.6k21128
answered Dec 14 '18 at 22:51
AndreiAndrei
12.6k21128
12.6k21128
$begingroup$
I realized now my question did not specify to find $p, q$ given $m$. I will update my question.
$endgroup$
– Herman Tulleken
Dec 14 '18 at 22:55
add a comment |
$begingroup$
I realized now my question did not specify to find $p, q$ given $m$. I will update my question.
$endgroup$
– Herman Tulleken
Dec 14 '18 at 22:55
$begingroup$
I realized now my question did not specify to find $p, q$ given $m$. I will update my question.
$endgroup$
– Herman Tulleken
Dec 14 '18 at 22:55
$begingroup$
I realized now my question did not specify to find $p, q$ given $m$. I will update my question.
$endgroup$
– Herman Tulleken
Dec 14 '18 at 22:55
add a comment |
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