Solved on Jan 31, 2024

Prove that there are infinitely many primes of the form $3k+2$ , where $k$ is a natural number.

STEP 1

Assumptions
1. A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.
2. The form $3k+2$ represents numbers that are 2 more than a multiple of 3.
3. $\mathbb{N}$ represents the set of natural numbers.

STEP 2

We will use proof by contradiction to show that there are infinitely many primes of the form $3k+2$ . Assume, for the sake of contradiction, that there are only finitely many such primes. Let's denote them by $p_1, p_2, ..., p_n$ .

STEP 3

Consider the number $N$ which is defined as follows:
$N = 3(p_1p_2...p_n) + 2$

STEP 4

Note that $N$ is of the form $3k+2$ because it is 2 more than a multiple of 3.

STEP 5

Since $N$ is greater than any of the primes $p_1, p_2, ..., p_n$ , it is not equal to any of these primes.

STEP 6

If $N$ is prime, then it is a prime of the form $3k+2$ that is not in our list, which contradicts our assumption that we have listed all such primes.

STEP 7

If $N$ is not prime, then it must have a prime divisor. Let $p$ be a prime divisor of $N$ .

STEP 8

Since $p$ divides $N$ , it cannot divide $3$ (as $3$ is prime and $p$ is not necessarily $3$ ). Therefore, $p$ must be of the form $3m+1$ or $3m+2$ for some $m \in \mathbb{N}$ , because all integers are of the form $3m$ , $3m+1$ , or $3m+2$ .

STEP 9

$p$ cannot be of the form $3m+1$ because if it were, then $p$ would divide $N - 2 = 3(p_1p_2...p_n)$ , which is a multiple of $3$ . This would imply that $p$ divides $3$ , which is not possible as $p$ is a prime greater than $3$ .

STEP 10

Therefore, $p$ must be of the form $3m+2$ . Since $p$ divides $N$ , and $N$ is of the form $3k+2$ , it follows that $p$ is a prime of the form $3k+2$ .

STEP 11

If $p$ is one of the primes $p_1, p_2, ..., p_n$ , then $p$ would divide $N - 2 = 3(p_1p_2...p_n)$ , which is a contradiction because a prime cannot divide a product and leave a remainder of $2$ when it is one of the factors.

STEP 12

Therefore, $p$ must be a different prime of the form $3k+2$ that is not in the list $p_1, p_2, ..., p_n$ , which contradicts our initial assumption that there are only finitely many primes of this form.

STEP 13

Since our assumption led to a contradiction, we conclude that there must be infinitely many primes of the form $3k+2$ .
This completes the proof.

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Prove that there are infinitely many primes of the form $3k+2$ , where $k$ is a natural number.

STEP 1

Assumptions
1. A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.
2. The form $3k+2$ represents numbers that are 2 more than a multiple of 3.
3. $\mathbb{N}$ represents the set of natural numbers.

STEP 2

We will use proof by contradiction to show that there are infinitely many primes of the form $3k+2$ . Assume, for the sake of contradiction, that there are only finitely many such primes. Let's denote them by $p_1, p_2, ..., p_n$ .

STEP 3

Consider the number $N$ which is defined as follows:
$N = 3(p_1p_2...p_n) + 2$

STEP 4

Note that $N$ is of the form $3k+2$ because it is 2 more than a multiple of 3.

STEP 5

Since $N$ is greater than any of the primes $p_1, p_2, ..., p_n$ , it is not equal to any of these primes.

STEP 6

If $N$ is prime, then it is a prime of the form $3k+2$ that is not in our list, which contradicts our assumption that we have listed all such primes.

STEP 7

If $N$ is not prime, then it must have a prime divisor. Let $p$ be a prime divisor of $N$ .

STEP 8

Since $p$ divides $N$ , it cannot divide $3$ (as $3$ is prime and $p$ is not necessarily $3$ ). Therefore, $p$ must be of the form $3m+1$ or $3m+2$ for some $m \in \mathbb{N}$ , because all integers are of the form $3m$ , $3m+1$ , or $3m+2$ .

STEP 9

$p$ cannot be of the form $3m+1$ because if it were, then $p$ would divide $N - 2 = 3(p_1p_2...p_n)$ , which is a multiple of $3$ . This would imply that $p$ divides $3$ , which is not possible as $p$ is a prime greater than $3$ .

STEP 10

Therefore, $p$ must be of the form $3m+2$ . Since $p$ divides $N$ , and $N$ is of the form $3k+2$ , it follows that $p$ is a prime of the form $3k+2$ .

STEP 11

If $p$ is one of the primes $p_1, p_2, ..., p_n$ , then $p$ would divide $N - 2 = 3(p_1p_2...p_n)$ , which is a contradiction because a prime cannot divide a product and leave a remainder of $2$ when it is one of the factors.

STEP 12

Therefore, $p$ must be a different prime of the form $3k+2$ that is not in the list $p_1, p_2, ..., p_n$ , which contradicts our initial assumption that there are only finitely many primes of this form.

STEP 13

Since our assumption led to a contradiction, we conclude that there must be infinitely many primes of the form $3k+2$ .
This completes the proof.

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Prove that there are infinitely many primes of the form 3k+23k+23k+2, where kkk is a natural number.

STEP 1

Assumptions1. A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.2. The form 3k+23k+23k+2 represents numbers that are 2 more than a multiple of 3.3. N\mathbb{N}N represents the set of natural numbers.

STEP 2

We will use proof by contradiction to show that there are infinitely many primes of the form 3k+23k+23k+2. Assume, for the sake of contradiction, that there are only finitely many such primes. Let's denote them by p1,p2,...,pnp_1, p_2, ..., p_np1​,p2​,...,pn​.

STEP 3

Consider the number NNN which is defined as follows:N=3(p1p2...pn)+2N = 3(p_1p_2...p_n) + 2N=3(p1​p2​...pn​)+2

STEP 4

Note that NNN is of the form 3k+23k+23k+2 because it is 2 more than a multiple of 3.

STEP 5

Since NNN is greater than any of the primes p1,p2,...,pnp_1, p_2, ..., p_np1​,p2​,...,pn​, it is not equal to any of these primes.

STEP 6

If NNN is prime, then it is a prime of the form 3k+23k+23k+2 that is not in our list, which contradicts our assumption that we have listed all such primes.

STEP 7

If NNN is not prime, then it must have a prime divisor. Let ppp be a prime divisor of NNN.

STEP 8

Since ppp divides NNN, it cannot divide 333 (as 333 is prime and ppp is not necessarily 333). Therefore, ppp must be of the form 3m+13m+13m+1 or 3m+23m+23m+2 for some m∈Nm \in \mathbb{N}m∈N, because all integers are of the form 3m3m3m, 3m+13m+13m+1, or 3m+23m+23m+2.

STEP 9

ppp cannot be of the form 3m+13m+13m+1 because if it were, then ppp would divide N−2=3(p1p2...pn)N - 2 = 3(p_1p_2...p_n)N−2=3(p1​p2​...pn​), which is a multiple of 333. This would imply that ppp divides 333, which is not possible as ppp is a prime greater than 333.

STEP 10

Therefore, ppp must be of the form 3m+23m+23m+2. Since ppp divides NNN, and NNN is of the form 3k+23k+23k+2, it follows that ppp is a prime of the form 3k+23k+23k+2.

STEP 11

If ppp is one of the primes p1,p2,...,pnp_1, p_2, ..., p_np1​,p2​,...,pn​, then ppp would divide N−2=3(p1p2...pn)N - 2 = 3(p_1p_2...p_n)N−2=3(p1​p2​...pn​), which is a contradiction because a prime cannot divide a product and leave a remainder of 222 when it is one of the factors.

STEP 12

Therefore, ppp must be a different prime of the form 3k+23k+23k+2 that is not in the list p1,p2,...,pnp_1, p_2, ..., p_np1​,p2​,...,pn​, which contradicts our initial assumption that there are only finitely many primes of this form.

STEP 13

Since our assumption led to a contradiction, we conclude that there must be infinitely many primes of the form 3k+23k+23k+2.This completes the proof.

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Prove that there are infinitely many primes of the form $3k+2$ , where $k$ is a natural number.

Assumptions
1. A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.
2. The form $3k+2$ represents numbers that are 2 more than a multiple of 3.
3. $\mathbb{N}$ represents the set of natural numbers.

We will use proof by contradiction to show that there are infinitely many primes of the form $3k+2$ . Assume, for the sake of contradiction, that there are only finitely many such primes. Let's denote them by $p_1, p_2, ..., p_n$ .

Consider the number $N$ which is defined as follows:
$N = 3(p_1p_2...p_n) + 2$

Note that $N$ is of the form $3k+2$ because it is 2 more than a multiple of 3.

Since $N$ is greater than any of the primes $p_1, p_2, ..., p_n$ , it is not equal to any of these primes.

If $N$ is prime, then it is a prime of the form $3k+2$ that is not in our list, which contradicts our assumption that we have listed all such primes.

If $N$ is not prime, then it must have a prime divisor. Let $p$ be a prime divisor of $N$ .

Since $p$ divides $N$ , it cannot divide $3$ (as $3$ is prime and $p$ is not necessarily $3$ ). Therefore, $p$ must be of the form $3m+1$ or $3m+2$ for some $m \in \mathbb{N}$ , because all integers are of the form $3m$ , $3m+1$ , or $3m+2$ .

$p$ cannot be of the form $3m+1$ because if it were, then $p$ would divide $N - 2 = 3(p_1p_2...p_n)$ , which is a multiple of $3$ . This would imply that $p$ divides $3$ , which is not possible as $p$ is a prime greater than $3$ .

Therefore, $p$ must be of the form $3m+2$ . Since $p$ divides $N$ , and $N$ is of the form $3k+2$ , it follows that $p$ is a prime of the form $3k+2$ .

If $p$ is one of the primes $p_1, p_2, ..., p_n$ , then $p$ would divide $N - 2 = 3(p_1p_2...p_n)$ , which is a contradiction because a prime cannot divide a product and leave a remainder of $2$ when it is one of the factors.

Therefore, $p$ must be a different prime of the form $3k+2$ that is not in the list $p_1, p_2, ..., p_n$ , which contradicts our initial assumption that there are only finitely many primes of this form.

Since our assumption led to a contradiction, we conclude that there must be infinitely many primes of the form $3k+2$ .
This completes the proof.