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Discrete Math help please!: Show that there exists a prime number p such that p+2 and p+4 are also prime?

Topic: Case ss number
June 20, 2019 / By Sequoia
Question: There's also a Part 2 that says: Show that there is only one prime number p such that p+2 and p+4 are also prime. In other words, show that for every prime number p other than the number you used to justify part (a), p+2 or p+4 is not prime. I need to use proofs to show this, I can't just say that the prime number 3 would satisfy part a. Any help would be greatly appreciated!!!

Best Answers: Discrete Math help please!: Show that there exists a prime number p such that p+2 and p+4 are also prime?

Ollie | 9 days ago
Since the second part refers to "the number you used to justify part (a)", it sounds like it's perfectly acceptable to simply demonstrate that 3 satisfies a. In any case, the best you could do would be to "formalize" the proof as follows: Let p = 3; then p + 2 = 5 and p + 4 = 7. 3, 5, and 7 are prime; therefore, p, p+2 and p+4 are prime. Therefore, there exists a prime number p such that p+2 and p+4 are also prime. Part b is the harder part. Let p = a prime number other than 3. Every integer is either divisible by 3, one greater than a number that is divisible by 3, or two greater than a number that is divisible by 3. So, p must be equal to 3k, 3k+1, or 3k+2 for some integer k. If p = 3k, it is divisible by 3, so is not prime. If p = 3k+1, p+2 = 3k+3; p+2 is divisible by 3, so it is not prime. If p = 3k+2, p+4 = 3k + 6; p+4 is divisible by 3, so it is not prime. Therefore, there is no number p that satisfies the conditions.
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Originally Answered: Discrete Math help please!: Show that there exists a prime number p such that p+2 and p+4 are also prime?
Since the second part refers to "the number you used to justify part (a)", it sounds like it's perfectly acceptable to simply demonstrate that 3 satisfies a. In any case, the best you could do would be to "formalize" the proof as follows: Let p = 3; then p + 2 = 5 and p + 4 = 7. 3, 5, and 7 are prime; therefore, p, p+2 and p+4 are prime. Therefore, there exists a prime number p such that p+2 and p+4 are also prime. Part b is the harder part. Let p = a prime number other than 3. Every integer is either divisible by 3, one greater than a number that is divisible by 3, or two greater than a number that is divisible by 3. So, p must be equal to 3k, 3k+1, or 3k+2 for some integer k. If p = 3k, it is divisible by 3, so is not prime. If p = 3k+1, p+2 = 3k+3; p+2 is divisible by 3, so it is not prime. If p = 3k+2, p+4 = 3k + 6; p+4 is divisible by 3, so it is not prime. Therefore, there is no number p that satisfies the conditions.
Originally Answered: Discrete Math help please!: Show that there exists a prime number p such that p+2 and p+4 are also prime?
Actually, you can just say that 3 would satisfy part (a). 3, 5, and 7 are all prime, so 3 works. The second part requires proof. You need to show that at least one of the p, p+2, p+4 is divisible by 3. One of three cases follows: (1) p=3, in which case we have the prime from (a). (2) p=1, which isn't prime (3) p=-1, which also isn't prime.