# Prime Numbers in Linear Patterns

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 Revision as of 00:11, 8 December 2012 (edit)← Previous diff Revision as of 00:12, 8 December 2012 (edit) (undo)Next diff → Line 18: Line 18: Proof. Proof. - Given any prime number $p$, assume that $p$ is neither congruent to 1 (mod 30) nor $q$ (mod 30) for every prime $q$ less than 30. Then $p$ is congruent to $x$ (mod 30), where $x$ is some integer less than 30 that is not 1 and not a prime. Prime factorization of $x$ must contain one of 2, 3, and 5. (If the prime factorization of $x$ did not contain any of 2, 3, or 5, then the smallest possible value of $x$ will be $7 * 7 =49$, which is greater than 30). Thus, $x=2^a3^b5^c$, where $a,b,c >=0$, and at least one of $a,b,c$ is greater than 0. Since $p$ is congruent to $x$, we can write $p$ as $p=30*n+2^a3^b5^c$, where $n$ is an integer greater than or equal to 1. Then, $p=30*n+2^a3^b5^c=(2*3*5)n+2^a3^b5^c$. $p$ is then equal to one of $2(3*5*n+2^{a-1}3^b5^c$ or $3(2*5*n+2^a3^{b-1}5^c)$ or $5(2*3*n+2^a3^b5^{c-1})$, which contradicts $p$ being a prime number. Thus, $p \equiv 1 \pmod {30}$ or $p \equiv q \pmod {30}$. + Given any prime number $p$, assume that $p$ is neither congruent to 1 (mod 30) nor $q$ (mod 30) for every prime $q$ less than 30. Then $p$ is congruent to $x$ (mod 30), where $x$ is some integer less than 30 that is not 1 and not a prime. Prime factorization of $x$ must contain one of 2, 3, and 5. (If the prime factorization of $x$ did not contain any of 2, 3, or 5, then the smallest possible value of $x$ will be $7 * 7 =49$, which is greater than 30). Thus, $x=2^a3^b5^c$, where $a,b,c >=0$, and at least one of $a,b,c$ is greater than 0. Since $p$ is congruent to $x$, we can write $p$ as $p=30*n+2^a3^b5^c$, where $n$ is an integer greater than or equal to 1. Then, $p=30*n+2^a3^b5^c=(2*3*5)n+2^a3^b5^c$. $p$ is then equal to one of $2(3*5*n+2^{a-1}3^b5^c)$ or $3(2*5*n+2^a3^{b-1}5^c)$ or $5(2*3*n+2^a3^b5^{c-1})$, which contradicts $p$ being a prime number. Thus, $p \equiv 1 \pmod {30}$ or $p \equiv q \pmod {30}$. Line 39: Line 39: Let $p$ be a prime number less than 30 that is not equal to 2, 3, or 5. Let $q=30-p$. If $q$ were not a prime, then $q$ must have 2, 3, or 5 as a prime factor. Since $p=30-q$, $p$ will also be divisible by 2, 3, or 5, contradicting our condition that $p$ is a prime number. Let $p$ be a prime number less than 30 that is not equal to 2, 3, or 5. Let $q=30-p$. If $q$ were not a prime, then $q$ must have 2, 3, or 5 as a prime factor. Since $p=30-q$, $p$ will also be divisible by 2, 3, or 5, contradicting our condition that $p$ is a prime number. - Such observation triggers one's interest to see whether the above statement is true for any multiples of 30, or for any number that is a product of the first few primes. However, this turns out not to be the case. + Such observation triggers one's interest to see whether the above statement is true for any multiples of 30, or for any number that is a product of the first few primes. However, this turns out not to be the case. |AuthorName=Iris Yoon |AuthorName=Iris Yoon |Field=Algebra |Field=Algebra |InProgress=No |InProgress=No }} }}

## Revision as of 00:12, 8 December 2012

Prime numbers in table with 180 columns

Prime numbers marked in a table with 180 columns

# Basic Description

Arranging natural numbers in a particular way and marking the prime numbers can lead to interesting patterns. For example, consider a table with 180 columns and infinitely many rows. Write positive integers in increasing order from left to right, and top to bottom. If we mark all the prime numbers, we get a pattern shown in the figure. We can see that prime numbers show patterns of vertical line segments.

# A More Mathematical Explanation

Instead of studying a table with 180 columns, we will study a table with 30 columns, as shown in [[#1 [...]

Instead of studying a table with 180 columns, we will study a table with 30 columns, as shown in Image 1.

Image 1

## Construction

First, create a table with 30 columns and sufficiently many rows. Write all positive integers starting from 1 as one moves from left to right, and top to bottom. Then, each row will start with a multiple of 30 added by 1, such as 1, 31, 61, 91, 121, ... . If we mark the prime numbers in this table we get Image 2.

Image 2

## Theorem 1

All prime numbers appear on columns that have a 1 or a prime number on its top row. In other words, for every prime number $p$, either $p \equiv 1\pmod {30}$, or there exists a prime number $q$ less than 30 such that $p \equiv q \pmod {30}$.

Proof. Given any prime number $p$, assume that $p$ is neither congruent to 1 (mod 30) nor $q$ (mod 30) for every prime $q$ less than 30. Then $p$ is congruent to $x$ (mod 30), where $x$ is some integer less than 30 that is not 1 and not a prime. Prime factorization of $x$ must contain one of 2, 3, and 5. (If the prime factorization of $x$ did not contain any of 2, 3, or 5, then the smallest possible value of $x$ will be $7 * 7 =49$, which is greater than 30). Thus, $x=2^a3^b5^c$, where $a,b,c >=0$, and at least one of $a,b,c$ is greater than 0. Since $p$ is congruent to $x$, we can write $p$ as $p=30*n+2^a3^b5^c$, where $n$ is an integer greater than or equal to 1. Then, $p=30*n+2^a3^b5^c=(2*3*5)n+2^a3^b5^c$. $p$ is then equal to one of $2(3*5*n+2^{a-1}3^b5^c)$ or $3(2*5*n+2^a3^{b-1}5^c)$ or $5(2*3*n+2^a3^b5^{c-1})$, which contradicts $p$ being a prime number. Thus, $p \equiv 1 \pmod {30}$ or $p \equiv q \pmod {30}$.

However, the statement does not generalize to other integer modulo groups. For instance, when we look at a table with 60 columns, 49 is not a prime number, but the column containing 49 will contain other prime numbers, such as 109.

Moreover, not all integers that are congruent to 1 (mod 30) or $q$ (mod 30), where $q$ is a prime number less than 30, are prime numbers. For instance, 49, which is congruent to 19 (mod 30), is not a prime number, but 49 still appears on the same column as 19. Let's call the columns that have a 1 or a prime number on its top row as prime-concentrated columns. One can observe that all composite numbers that appear on these prime-concentrated columns, say 49,77,91,119,121,133,143,161,169,..., have prime factors that are greater than or equal to 7. In other words, these composite numbers do not have 2, 3, or 5 as a prime factor.

## Theorem 2

Composite numbers that appear on prime-concentrated columns do not have 2,3, or 5 as a prime factor.

Proof. Let $x$ be a composite number that appears on a prime-concentrated column, and assume that $x$ has at least one of 2, 3, or 5 as a prime factor. Since $x$ appears on the prime-concentrated column, $x$ can be written as $x=30n+k$, where $n$ is a positive integer and $k$ is a prime number smaller than 30. If $x$ had 2 as a prime factor, $k$ must also have 2 as a factor because 30 has 2 as a factor. This contradicts the fact that $k$ is a prime number. The same argument works for the case when $x$ has 3 or 5 as prime factors.

Another pattern to notice is that the prime-concentrated columns seem symmetric about the column that contains 15, which leads to the following observation.

## Theorem 3

If $p$ is a prime number less than 30 and if $p$ is not equal to 2,3, or 5, then 30-p is a prime number.

Proof. Let $p$ be a prime number less than 30 that is not equal to 2, 3, or 5. Let $q=30-p$. If $q$ were not a prime, then $q$ must have 2, 3, or 5 as a prime factor. Since $p=30-q$, $p$ will also be divisible by 2, 3, or 5, contradicting our condition that $p$ is a prime number.

Such observation triggers one's interest to see whether the above statement is true for any multiples of 30, or for any number that is a product of the first few primes. However, this turns out not to be the case.

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