8.9k views
4 votes
Show that there exists an integer n such that n, n + 2 and n + 4 are all primes.

User Marsad
by
8.1k points

1 Answer

5 votes

Final answer:

The question seeks to find a prime triplet, but after the set (3, 5, 7), there is no such triplet where n, n+2, and n+4 are all primes, because one of the numbers will be divisible by 3, thus it cannot be prime.

Step-by-step explanation:

The question asks to show that there exists an integer n such that n, n + 2 and n + 4 are all primes. However, the information provided does not directly relate to proving the existence of such a prime triplet. In general, a prime triplet is a set of three prime numbers in which the smallest and largest numbers differ by four.

The smallest and best-known example of a prime triplet is (3, 5, 7). After this triplet, the number of the form n, n + 2, and n + 4 cannot all be prime because one of them will be a multiple of 3. If n is not divisible by 3, then either n + 3 is, making n + 4 not prime, or n + 2 or n itself is divisible by 3.

To show that there exists an integer n such that n, n + 2, and n + 4 are all prime numbers, we can use the concept of prime numbers and the properties of addition.

Let's consider the number n. If n is an even number, then n + 2 will also be even, making it not prime. Therefore, n must be odd.

We can start by assuming n = 3, which is an odd number. Checking the values of n, n + 2, and n + 4:

n = 3, n + 2 = 5, n + 4 = 7.

All three numbers are prime numbers, proving that there exists an integer n (in this case, n = 3) where n, n + 2, and n + 4 are all prime numbers.

User LazyProphet
by
8.0k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories