A conjecture involving prime numbers and parallelograms
Clash Royale CLAN TAG#URR8PPP
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As already introduced in this post, given the series of prime numbers greater than $9$, let organize them in four rows, according to their last digit ($1,3,7$ or $9$). The column in which they are displayed is the ten to which they belong, as illustrated in the following scheme.
Within this scheme, and given the tens $N=0,3,6,9ldots$, we can uniquely define a parallelogram by means of the four points corresponding to the four integers $N+1$, $(N+10)+1$, $(N+40)+9$ and $(N+50)+9$, as easily illustrated below.
For instance, the parallelogram corresponding to the ten $N=3$ is defined by the integers $31,41,79$ and $89$, whereas the one corresponding to $N=6$ is defined by $61,71,109, 119$.
My conjecture is:
On the perimeter of each parallelogram there cannot be more than $7$ primes.
In the following picture, I denote with a red cross some of the missing primes, i.e. those integers that occupy one of the $8$ possible positions that the primes can occupy on the parallelograms, but that are not primes.
And here some more (sorry for the bad quality).
(This conjecture is motivated by the fact that, if true, it can perhaps be used to devise a method to determine which point will be missing on the parallelogram $N+1$, knowing which ones are missing on the previous $N$ parallelograms, but this is another problem!).
So far, I tried to use the strategies suggested in this post, but without much success.
I apologize in case this is a trivial question, and I will thank you for any suggestion and/or comment. Also, in case this question is not clear or not rigorous, please help me to improve it (I am not an expert of prime numbers). Thank you!
EDIT: A follow-up of this post can be found here, where I try to use this conjecture in order to locate the "missing primes" on the parallelograms...
number-theory prime-numbers euclidean-geometry quadrilateral
add a comment |Â
up vote
28
down vote
favorite
As already introduced in this post, given the series of prime numbers greater than $9$, let organize them in four rows, according to their last digit ($1,3,7$ or $9$). The column in which they are displayed is the ten to which they belong, as illustrated in the following scheme.
Within this scheme, and given the tens $N=0,3,6,9ldots$, we can uniquely define a parallelogram by means of the four points corresponding to the four integers $N+1$, $(N+10)+1$, $(N+40)+9$ and $(N+50)+9$, as easily illustrated below.
For instance, the parallelogram corresponding to the ten $N=3$ is defined by the integers $31,41,79$ and $89$, whereas the one corresponding to $N=6$ is defined by $61,71,109, 119$.
My conjecture is:
On the perimeter of each parallelogram there cannot be more than $7$ primes.
In the following picture, I denote with a red cross some of the missing primes, i.e. those integers that occupy one of the $8$ possible positions that the primes can occupy on the parallelograms, but that are not primes.
And here some more (sorry for the bad quality).
(This conjecture is motivated by the fact that, if true, it can perhaps be used to devise a method to determine which point will be missing on the parallelogram $N+1$, knowing which ones are missing on the previous $N$ parallelograms, but this is another problem!).
So far, I tried to use the strategies suggested in this post, but without much success.
I apologize in case this is a trivial question, and I will thank you for any suggestion and/or comment. Also, in case this question is not clear or not rigorous, please help me to improve it (I am not an expert of prime numbers). Thank you!
EDIT: A follow-up of this post can be found here, where I try to use this conjecture in order to locate the "missing primes" on the parallelograms...
number-theory prime-numbers euclidean-geometry quadrilateral
2
I think you will enjoy learning about wheel factorization.
â PM 2Ring
Aug 20 at 18:40
5
The person who downvoted this is an absolute swamp monster.
â The Count
Aug 21 at 2:05
add a comment |Â
up vote
28
down vote
favorite
up vote
28
down vote
favorite
As already introduced in this post, given the series of prime numbers greater than $9$, let organize them in four rows, according to their last digit ($1,3,7$ or $9$). The column in which they are displayed is the ten to which they belong, as illustrated in the following scheme.
Within this scheme, and given the tens $N=0,3,6,9ldots$, we can uniquely define a parallelogram by means of the four points corresponding to the four integers $N+1$, $(N+10)+1$, $(N+40)+9$ and $(N+50)+9$, as easily illustrated below.
For instance, the parallelogram corresponding to the ten $N=3$ is defined by the integers $31,41,79$ and $89$, whereas the one corresponding to $N=6$ is defined by $61,71,109, 119$.
My conjecture is:
On the perimeter of each parallelogram there cannot be more than $7$ primes.
In the following picture, I denote with a red cross some of the missing primes, i.e. those integers that occupy one of the $8$ possible positions that the primes can occupy on the parallelograms, but that are not primes.
And here some more (sorry for the bad quality).
(This conjecture is motivated by the fact that, if true, it can perhaps be used to devise a method to determine which point will be missing on the parallelogram $N+1$, knowing which ones are missing on the previous $N$ parallelograms, but this is another problem!).
So far, I tried to use the strategies suggested in this post, but without much success.
I apologize in case this is a trivial question, and I will thank you for any suggestion and/or comment. Also, in case this question is not clear or not rigorous, please help me to improve it (I am not an expert of prime numbers). Thank you!
EDIT: A follow-up of this post can be found here, where I try to use this conjecture in order to locate the "missing primes" on the parallelograms...
number-theory prime-numbers euclidean-geometry quadrilateral
As already introduced in this post, given the series of prime numbers greater than $9$, let organize them in four rows, according to their last digit ($1,3,7$ or $9$). The column in which they are displayed is the ten to which they belong, as illustrated in the following scheme.
Within this scheme, and given the tens $N=0,3,6,9ldots$, we can uniquely define a parallelogram by means of the four points corresponding to the four integers $N+1$, $(N+10)+1$, $(N+40)+9$ and $(N+50)+9$, as easily illustrated below.
For instance, the parallelogram corresponding to the ten $N=3$ is defined by the integers $31,41,79$ and $89$, whereas the one corresponding to $N=6$ is defined by $61,71,109, 119$.
My conjecture is:
On the perimeter of each parallelogram there cannot be more than $7$ primes.
In the following picture, I denote with a red cross some of the missing primes, i.e. those integers that occupy one of the $8$ possible positions that the primes can occupy on the parallelograms, but that are not primes.
And here some more (sorry for the bad quality).
(This conjecture is motivated by the fact that, if true, it can perhaps be used to devise a method to determine which point will be missing on the parallelogram $N+1$, knowing which ones are missing on the previous $N$ parallelograms, but this is another problem!).
So far, I tried to use the strategies suggested in this post, but without much success.
I apologize in case this is a trivial question, and I will thank you for any suggestion and/or comment. Also, in case this question is not clear or not rigorous, please help me to improve it (I am not an expert of prime numbers). Thank you!
EDIT: A follow-up of this post can be found here, where I try to use this conjecture in order to locate the "missing primes" on the parallelograms...
number-theory prime-numbers euclidean-geometry quadrilateral
edited Aug 21 at 20:52
asked Aug 20 at 8:08
Andrea Prunotto
1,284625
1,284625
2
I think you will enjoy learning about wheel factorization.
â PM 2Ring
Aug 20 at 18:40
5
The person who downvoted this is an absolute swamp monster.
â The Count
Aug 21 at 2:05
add a comment |Â
2
I think you will enjoy learning about wheel factorization.
â PM 2Ring
Aug 20 at 18:40
5
The person who downvoted this is an absolute swamp monster.
â The Count
Aug 21 at 2:05
2
2
I think you will enjoy learning about wheel factorization.
â PM 2Ring
Aug 20 at 18:40
I think you will enjoy learning about wheel factorization.
â PM 2Ring
Aug 20 at 18:40
5
5
The person who downvoted this is an absolute swamp monster.
â The Count
Aug 21 at 2:05
The person who downvoted this is an absolute swamp monster.
â The Count
Aug 21 at 2:05
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
35
down vote
accepted
The eight points on each parallelogram cover all residues mod 7.
1
Thanks! Please, can you expand your answer a bit more? I am not very familiar with this topic, and I did not get the connection between your answer and the problem. Thanks again!
â Andrea Prunotto
Aug 20 at 8:29
9
@Andrea The point is that at least one of the eight numbers on any parallelogram is divisible by 7 and therefore not a prime number.
â Marc Paul
Aug 20 at 8:53
@MarcPaul Thank you. I got it now! But of course, this does not help to predict which points will be missing on the $N$-th parallelogram, if we know which ones are missing in the previous $N-1$ parallelograms, right?
â Andrea Prunotto
Aug 20 at 9:06
1
@AndreaPrunotto : you should ask another question if you have another precise request (this one was nice and fun, btw)
â Evargalo
Aug 20 at 13:25
1
@Evargalo Sure, you're definitely right. I will do it, after having edited the pictures! : )
â Andrea Prunotto
Aug 20 at 13:28
 |Â
show 1 more comment
up vote
12
down vote
This is an expansion of Michal's answer. We can find a multiple of 7 on each of the first seven parallelograms: 49, 77, 119, 91, 133, 161, 203.
Now every parallelogram can be obtained by taking one of these first seven parallelograms and adding 210 to each point. Since 210 is a multiple of 7, each point listed above will be translated onto missing points. This gives a way to compute a missing point on every parallelogram: if you find the remainder mod 210 of the points on the parallelogram, it must contain one of the seven values I listed above.
add a comment |Â
up vote
2
down vote
I think I have a closed formula for the missing points. These seem to be (conjeture, but should not be difficult to prove) the composite integers $ninmathbbN$ such that $n=pm 1 mod 6$. This sequence gives: 1,25,35,49,55,65,77,85,91,95,115,119,121...
1
Every positive integer, prime or composite (or 1!), is on one of the parallelograms if and only if it is congruent to $pm 1 mod 6$. The parallelograms were presumably constructed this way in the first place because except for 2 and 3, numbers with any other residues mod 6 will never be prime.
â aschepler
Aug 20 at 22:18
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
35
down vote
accepted
The eight points on each parallelogram cover all residues mod 7.
1
Thanks! Please, can you expand your answer a bit more? I am not very familiar with this topic, and I did not get the connection between your answer and the problem. Thanks again!
â Andrea Prunotto
Aug 20 at 8:29
9
@Andrea The point is that at least one of the eight numbers on any parallelogram is divisible by 7 and therefore not a prime number.
â Marc Paul
Aug 20 at 8:53
@MarcPaul Thank you. I got it now! But of course, this does not help to predict which points will be missing on the $N$-th parallelogram, if we know which ones are missing in the previous $N-1$ parallelograms, right?
â Andrea Prunotto
Aug 20 at 9:06
1
@AndreaPrunotto : you should ask another question if you have another precise request (this one was nice and fun, btw)
â Evargalo
Aug 20 at 13:25
1
@Evargalo Sure, you're definitely right. I will do it, after having edited the pictures! : )
â Andrea Prunotto
Aug 20 at 13:28
 |Â
show 1 more comment
up vote
35
down vote
accepted
The eight points on each parallelogram cover all residues mod 7.
1
Thanks! Please, can you expand your answer a bit more? I am not very familiar with this topic, and I did not get the connection between your answer and the problem. Thanks again!
â Andrea Prunotto
Aug 20 at 8:29
9
@Andrea The point is that at least one of the eight numbers on any parallelogram is divisible by 7 and therefore not a prime number.
â Marc Paul
Aug 20 at 8:53
@MarcPaul Thank you. I got it now! But of course, this does not help to predict which points will be missing on the $N$-th parallelogram, if we know which ones are missing in the previous $N-1$ parallelograms, right?
â Andrea Prunotto
Aug 20 at 9:06
1
@AndreaPrunotto : you should ask another question if you have another precise request (this one was nice and fun, btw)
â Evargalo
Aug 20 at 13:25
1
@Evargalo Sure, you're definitely right. I will do it, after having edited the pictures! : )
â Andrea Prunotto
Aug 20 at 13:28
 |Â
show 1 more comment
up vote
35
down vote
accepted
up vote
35
down vote
accepted
The eight points on each parallelogram cover all residues mod 7.
The eight points on each parallelogram cover all residues mod 7.
answered Aug 20 at 8:23
Michal Adamaszek
1,64948
1,64948
1
Thanks! Please, can you expand your answer a bit more? I am not very familiar with this topic, and I did not get the connection between your answer and the problem. Thanks again!
â Andrea Prunotto
Aug 20 at 8:29
9
@Andrea The point is that at least one of the eight numbers on any parallelogram is divisible by 7 and therefore not a prime number.
â Marc Paul
Aug 20 at 8:53
@MarcPaul Thank you. I got it now! But of course, this does not help to predict which points will be missing on the $N$-th parallelogram, if we know which ones are missing in the previous $N-1$ parallelograms, right?
â Andrea Prunotto
Aug 20 at 9:06
1
@AndreaPrunotto : you should ask another question if you have another precise request (this one was nice and fun, btw)
â Evargalo
Aug 20 at 13:25
1
@Evargalo Sure, you're definitely right. I will do it, after having edited the pictures! : )
â Andrea Prunotto
Aug 20 at 13:28
 |Â
show 1 more comment
1
Thanks! Please, can you expand your answer a bit more? I am not very familiar with this topic, and I did not get the connection between your answer and the problem. Thanks again!
â Andrea Prunotto
Aug 20 at 8:29
9
@Andrea The point is that at least one of the eight numbers on any parallelogram is divisible by 7 and therefore not a prime number.
â Marc Paul
Aug 20 at 8:53
@MarcPaul Thank you. I got it now! But of course, this does not help to predict which points will be missing on the $N$-th parallelogram, if we know which ones are missing in the previous $N-1$ parallelograms, right?
â Andrea Prunotto
Aug 20 at 9:06
1
@AndreaPrunotto : you should ask another question if you have another precise request (this one was nice and fun, btw)
â Evargalo
Aug 20 at 13:25
1
@Evargalo Sure, you're definitely right. I will do it, after having edited the pictures! : )
â Andrea Prunotto
Aug 20 at 13:28
1
1
Thanks! Please, can you expand your answer a bit more? I am not very familiar with this topic, and I did not get the connection between your answer and the problem. Thanks again!
â Andrea Prunotto
Aug 20 at 8:29
Thanks! Please, can you expand your answer a bit more? I am not very familiar with this topic, and I did not get the connection between your answer and the problem. Thanks again!
â Andrea Prunotto
Aug 20 at 8:29
9
9
@Andrea The point is that at least one of the eight numbers on any parallelogram is divisible by 7 and therefore not a prime number.
â Marc Paul
Aug 20 at 8:53
@Andrea The point is that at least one of the eight numbers on any parallelogram is divisible by 7 and therefore not a prime number.
â Marc Paul
Aug 20 at 8:53
@MarcPaul Thank you. I got it now! But of course, this does not help to predict which points will be missing on the $N$-th parallelogram, if we know which ones are missing in the previous $N-1$ parallelograms, right?
â Andrea Prunotto
Aug 20 at 9:06
@MarcPaul Thank you. I got it now! But of course, this does not help to predict which points will be missing on the $N$-th parallelogram, if we know which ones are missing in the previous $N-1$ parallelograms, right?
â Andrea Prunotto
Aug 20 at 9:06
1
1
@AndreaPrunotto : you should ask another question if you have another precise request (this one was nice and fun, btw)
â Evargalo
Aug 20 at 13:25
@AndreaPrunotto : you should ask another question if you have another precise request (this one was nice and fun, btw)
â Evargalo
Aug 20 at 13:25
1
1
@Evargalo Sure, you're definitely right. I will do it, after having edited the pictures! : )
â Andrea Prunotto
Aug 20 at 13:28
@Evargalo Sure, you're definitely right. I will do it, after having edited the pictures! : )
â Andrea Prunotto
Aug 20 at 13:28
 |Â
show 1 more comment
up vote
12
down vote
This is an expansion of Michal's answer. We can find a multiple of 7 on each of the first seven parallelograms: 49, 77, 119, 91, 133, 161, 203.
Now every parallelogram can be obtained by taking one of these first seven parallelograms and adding 210 to each point. Since 210 is a multiple of 7, each point listed above will be translated onto missing points. This gives a way to compute a missing point on every parallelogram: if you find the remainder mod 210 of the points on the parallelogram, it must contain one of the seven values I listed above.
add a comment |Â
up vote
12
down vote
This is an expansion of Michal's answer. We can find a multiple of 7 on each of the first seven parallelograms: 49, 77, 119, 91, 133, 161, 203.
Now every parallelogram can be obtained by taking one of these first seven parallelograms and adding 210 to each point. Since 210 is a multiple of 7, each point listed above will be translated onto missing points. This gives a way to compute a missing point on every parallelogram: if you find the remainder mod 210 of the points on the parallelogram, it must contain one of the seven values I listed above.
add a comment |Â
up vote
12
down vote
up vote
12
down vote
This is an expansion of Michal's answer. We can find a multiple of 7 on each of the first seven parallelograms: 49, 77, 119, 91, 133, 161, 203.
Now every parallelogram can be obtained by taking one of these first seven parallelograms and adding 210 to each point. Since 210 is a multiple of 7, each point listed above will be translated onto missing points. This gives a way to compute a missing point on every parallelogram: if you find the remainder mod 210 of the points on the parallelogram, it must contain one of the seven values I listed above.
This is an expansion of Michal's answer. We can find a multiple of 7 on each of the first seven parallelograms: 49, 77, 119, 91, 133, 161, 203.
Now every parallelogram can be obtained by taking one of these first seven parallelograms and adding 210 to each point. Since 210 is a multiple of 7, each point listed above will be translated onto missing points. This gives a way to compute a missing point on every parallelogram: if you find the remainder mod 210 of the points on the parallelogram, it must contain one of the seven values I listed above.
answered Aug 20 at 16:39
Jonathan Love
1693
1693
add a comment |Â
add a comment |Â
up vote
2
down vote
I think I have a closed formula for the missing points. These seem to be (conjeture, but should not be difficult to prove) the composite integers $ninmathbbN$ such that $n=pm 1 mod 6$. This sequence gives: 1,25,35,49,55,65,77,85,91,95,115,119,121...
1
Every positive integer, prime or composite (or 1!), is on one of the parallelograms if and only if it is congruent to $pm 1 mod 6$. The parallelograms were presumably constructed this way in the first place because except for 2 and 3, numbers with any other residues mod 6 will never be prime.
â aschepler
Aug 20 at 22:18
add a comment |Â
up vote
2
down vote
I think I have a closed formula for the missing points. These seem to be (conjeture, but should not be difficult to prove) the composite integers $ninmathbbN$ such that $n=pm 1 mod 6$. This sequence gives: 1,25,35,49,55,65,77,85,91,95,115,119,121...
1
Every positive integer, prime or composite (or 1!), is on one of the parallelograms if and only if it is congruent to $pm 1 mod 6$. The parallelograms were presumably constructed this way in the first place because except for 2 and 3, numbers with any other residues mod 6 will never be prime.
â aschepler
Aug 20 at 22:18
add a comment |Â
up vote
2
down vote
up vote
2
down vote
I think I have a closed formula for the missing points. These seem to be (conjeture, but should not be difficult to prove) the composite integers $ninmathbbN$ such that $n=pm 1 mod 6$. This sequence gives: 1,25,35,49,55,65,77,85,91,95,115,119,121...
I think I have a closed formula for the missing points. These seem to be (conjeture, but should not be difficult to prove) the composite integers $ninmathbbN$ such that $n=pm 1 mod 6$. This sequence gives: 1,25,35,49,55,65,77,85,91,95,115,119,121...
edited Aug 20 at 16:48
answered Aug 20 at 10:39
galois1989
508
508
1
Every positive integer, prime or composite (or 1!), is on one of the parallelograms if and only if it is congruent to $pm 1 mod 6$. The parallelograms were presumably constructed this way in the first place because except for 2 and 3, numbers with any other residues mod 6 will never be prime.
â aschepler
Aug 20 at 22:18
add a comment |Â
1
Every positive integer, prime or composite (or 1!), is on one of the parallelograms if and only if it is congruent to $pm 1 mod 6$. The parallelograms were presumably constructed this way in the first place because except for 2 and 3, numbers with any other residues mod 6 will never be prime.
â aschepler
Aug 20 at 22:18
1
1
Every positive integer, prime or composite (or 1!), is on one of the parallelograms if and only if it is congruent to $pm 1 mod 6$. The parallelograms were presumably constructed this way in the first place because except for 2 and 3, numbers with any other residues mod 6 will never be prime.
â aschepler
Aug 20 at 22:18
Every positive integer, prime or composite (or 1!), is on one of the parallelograms if and only if it is congruent to $pm 1 mod 6$. The parallelograms were presumably constructed this way in the first place because except for 2 and 3, numbers with any other residues mod 6 will never be prime.
â aschepler
Aug 20 at 22:18
add a comment |Â
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2
I think you will enjoy learning about wheel factorization.
â PM 2Ring
Aug 20 at 18:40
5
The person who downvoted this is an absolute swamp monster.
â The Count
Aug 21 at 2:05