Tag Archives: prime numbers

Prove It by Velleman, Introduction, Problem 6

Problem

The sequence 3, 5, 7 is a list of three prime numbers such that each pair of adjacent numbers in the list differ by two. Are there any more such “triplet primes”?

Solution

No. Any triplet would have to be three consecutive odd numbers since all even numbers are not prime. The possibilities for the last digits of the triplets are: (1, 3, 5), (3, 5, 7), (5, 7, 9). In each case, there is a 5 in the last digit of one of the numbers. Any such number will be divisible by 5, thus not prime. Therefore, once you get pass 7, there are no more triplets.

Prove It by Velleman, Introduction, Problem 3

Problem

The proof of Theorem 3 gives a method for finding a prime number different from any in a given list of prime numbers.

Problem a

Use this method to find a prime different from 2, 3, 5, 7.

Solution

The method in Theorem 3 is as follows: Let p_{1}, p_{2}, p_{3}, … , p_{n} be a list of all prime numbers. Let m = p_{1} p_{2} ... p_{n} + 1. m will be a prime or a product of primes.

2 \cdot 3 \cdot 5 \cdot 7 + 1 = 211

211 is prime.

Problem b

Use this method to find  a prime different from 2, 5, 11.

Solution

The section on Theorem 3 also discusses Marsenne Primes – if n is prime, then 2^{n} - 1 might be prime.

2^{2} -1 = 3

That is one possible solution.

The answer key lists 3 and 37 and solutions, but it’s unclear how to calculate the 37.

Prove It by Velleman, Introduction, Problem 2

Problem

Make some conjectures about the value of n for which 3{n} - 1 is a prime or the value of n for which 3{n} - 2 is a prime. (You might start by making a table similar to Figure 1.)

Solution

 

n Prime 3^{n}-1 Prime 3^{n}-2 Prime
2 Yes 8 No 7 Yes
3 Yes 26 No 25 No
4 No 80 No 79 Yes
5 Yes 242 No 241 Yes
6 No 728 No 727 Yes
7 Yes 2186 No 2185 No
8 No 6560 No 6559 No
9 No 19682 No 19681 Yes
10 No 59048 No 59047 No

 

Conjecture: 3^{n} - 1 is never prime.

Conjecture: 3^{n} - 2 may be prime.