Find all positive integers 
n
n such that 
n
n, 
n
+
2
n+2, and 
n
+
4
n+4 are all prime.

We are tasked with finding all positive integers \( n \) such that \( n \), \( n+2 \), and \( n+4 \) are all prime numbers.

### Step 1: Consider small values of \( n \)

We start by checking small values of \( n \), since these numbers need to be primes, and prime numbers tend to grow sparse as \( n \) increases.

- **For \( n = 3 \):**
  - \( n = 3 \) (prime),
  - \( n+2 = 5 \) (prime),
  - \( n+4 = 7 \) (prime).

  So, \( n = 3 \) works.

- **For \( n = 5 \):**
  - \( n = 5 \) (prime),
  - \( n+2 = 7 \) (prime),
  - \( n+4 = 9 \) (not prime).

  So, \( n = 5 \) does not work.

- **For \( n = 7 \):**
  - \( n = 7 \) (prime),
  - \( n+2 = 9 \) (not prime),
  - \( n+4 = 11 \) (prime).

  So, \( n = 7 \) does not work.

- **For \( n = 11 \):**
  - \( n = 11 \) (prime),
  - \( n+2 = 13 \) (prime),
  - \( n+4 = 15 \) (not prime).

  So, \( n = 11 \) does not work.

- **For \( n = 13 \):**
  - \( n = 13 \) (prime),
  - \( n+2 = 15 \) (not prime),
  - \( n+4 = 17 \) (prime).

  So, \( n = 13 \) does not work.

### Step 2: General analysis

Let’s now analyze the problem more generally. We want \( n \), \( n+2 \), and \( n+4 \) to all be prime numbers.

- **Modulo 3 reasoning**: Since any integer \( n \) is congruent to either 0, 1, or 2 modulo 3, we will check each case to see if it leads to a contradiction.

  - If \( n \equiv 0 \mod 3 \), then \( n \) is divisible by 3. The only prime divisible by 3 is 3 itself. Hence, \( n = 3 \) is a solution.
  
  - If \( n \equiv 1 \mod 3 \), then \( n+2 \equiv 0 \mod 3 \), meaning \( n+2 \) is divisible by 3. Since the only prime divisible by 3 is 3 itself, we must have \( n+2 = 3 \), which leads to \( n = 1 \). But \( n = 1 \) is not prime, so there is no solution in this case.
  
  - If \( n \equiv 2 \mod 3 \), then \( n+4 \equiv 0 \mod 3 \), meaning \( n+4 \) is divisible by 3. As before, \( n+4 \) must be 3, which gives \( n = -1 \), but \( n \) must be a positive integer. Hence, there is no solution in this case.

### Conclusion

The only solution is \( n = 3 \), where \( n = 3 \), \( n+2 = 5 \), and \( n+4 = 7 \) are all prime numbers.

### Would you like more details or have any questions?

Here are 5 related questions to further explore the topic:

1. What are twin primes, and how are they related to this problem?
2. How sparse are prime triplets as \( n \) increases?
3. Are there other prime triplets of the form \( n, n+2, n+4 \)?
4. How can modular arithmetic help in solving problems involving primes?
5. Why do primes become less frequent as numbers get larger?

**Tip:** When solving problems involving primes, checking small cases first can give helpful insights into patterns or constraints that apply to larger cases.

Find all positive integers n such that n, n+2, and n+4 are all prime.

Explore how to find all positive integers n such that n, n+2, and n+4 are prime numbers. This detailed solution uses number theory, properties of prime numbers, and modular arithmetic to demonstrate that n = 3 is the only solution. Suitable for high school students in Grades 9-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation