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Alecsey [184]
3 years ago
14

Twin Primes (a) Let p > 3 be a prime. Prove that p is of the form 3k +1 or 3k – 1 for some integer k. (b) Twin primes are pai

rs of prime numbers p and q that have a difference of 2. Use part (a) to prove that 5 is the only prime number that takes part in two different twin prime pairs.
Mathematics
1 answer:
Alecsey [184]3 years ago
8 0

Answer:

(a) Let us recall the division algorithm: given two positive integers n and p there exist other two positive integers k and r such that

n = pk+r where r and r is called the <em>remainder</em>.

So, given any positive integer n and 3 we can write

n=3k+r where r=0,1,2. Thus, every n can be written as

  • n=3k
  • n=3k+1
  • n=3k+2

Now, notice that n=3k+2 = 3k+3-1 = 3(k+1)-1 =3k'-1. Hence, every number can be written as n=3k, or n=3k+1 or n=3k-1.

A number p is prime if and only if its only factors are 1 and p itself. So, a number of the form n=3k cannot be prime. Therefore, every primer number is of the form n=3k+1 or n=3k-1.

(b) Assume that there are three prime numbers such that p, p+2 and p-2 are prime.

By the previous exercise p=3k+1 or p=3k-1. Let us analyze both cases separately.

<em>First case</em>: p=3k+1. Then p-2=3k-1 that can be prime, and p+2=3k+3 that is not prime. Hence, there are not such three primes with p=3k+1.

<em>Second case</em>: p=3k-1. Then, p+2=3k+1 that can be prime, and p-2=3k-3=3(k-1) that cannot be prime. Hence, there are not such three primes with p=3k-1.

Therefore, there are no three primes  of the form p, p+2 and p-2, except for 3, 5 and 7.

Notice that this is only possible because 5=2*3-1 and 2*3-3=3, that is the only ‘‘multiple’’ of 3 that is prime.

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Find the root(s) of f (x) = (x + 5)3(x - 9)2(x + 1). -5 with multiplicity 3 5 with multiplicity 3 -9
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Step-by-step explanation:

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