3943 Shares

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.
👍 286 | 👎 9
Did you like the answer? Discrete Math help please!: Show that there exists a prime number p such that p+2 and p+4 are also prime? Share with your friends

We found more questions related to the topic: Case ss number 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. Madyson
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.
👍 120 | 👎 8 Kerenhappuch
For all integers n, either n, n+1 or n+2 is a multiple of 3 If p > 3 is prime, then either p+1 or p+2 is a multiple of 3 If p+1 is a multiple of 3, then p+4 is also a multiple of 3 and therefore not prime If p+2 is a multiple of three, then p+2 is not prime. So for all prime numbers p>3, either p+2 or p+4 is a multiple of 3 and therefore there is no such p that p+2 and p+4 are prime. [edit: this is actually true for all number p that are not multiples of 3 (no prime besides 3 is a multiple of 3) ]
👍 112 | 👎 7 Originally Answered: Let G be a group with p^k elements where p is a prime number. Prove that G has a subgroup of order p.?
You're incorrect - G need not be isomorphic to Z_p^k. For example Z_2 x Z_2 is a group with four elements but has no element of order 4 and so cannot be isomorphic to Z_4. Let g be any nonidentity element of G. Since the order of G must divide the order of the group, |g| = p^l with 1 < l < or = k. Consider h = g^{p^{l-1}}. It is not hard to see that |h| = p, and so the subgroup generated by h has order p.

If you have your own answer to the question case ss number, then you can write your own version, using the form below for an extended answer.