MATH SOLVE

3 months ago

Q:
# 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 pairs 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.

Accepted Solution

A:

Answer:(a) Let us recall the division algorithm: given two positive integers [tex]n[/tex] and [tex]p[/tex] there exist other two positive integers [tex]k[/tex] and [tex]r[/tex] such that[tex] n = pk+r[/tex] where [tex]r<p[/tex] and [tex]r[/tex] is called the remainder.So, given any positive integer [tex]n[/tex] and 3 we can write[tex]n=3k+r[/tex] where [tex]r=0,1,2[/tex]. Thus, every [tex]n[/tex] can be written as[tex]n=3k[/tex][tex]n=3k+1[/tex][tex]n=3k+2[/tex]Now, notice that [tex]n=3k+2 = 3k+3-1 = 3(k+1)-1 =3k'-1[/tex]. Hence, every number can be written as [tex]n=3k[/tex], or [tex]n=3k+1[/tex] or [tex]n=3k-1[/tex]. A number [tex]p[/tex] is prime if and only if its only factors are 1 and [tex]p[/tex] itself. So, a number of the form [tex]n=3k[/tex] cannot be prime. Therefore, every primer number is of the form [tex]n=3k+1[/tex] or [tex]n=3k-1[/tex].(b) Assume that there are three prime numbers such that [tex]p[/tex], [tex]p+2[/tex] and [tex]p-2[/tex] are prime. By the previous exercise [tex]p=3k+1[/tex] or [tex]p=3k-1[/tex]. Let us analyze both cases separately.First case: [tex]p=3k+1[/tex]. Then [tex]p-2=3k-1[/tex] that can be prime, and [tex]p+2=3k+3[/tex] that is not prime. Hence, there are not such three primes with [tex]p=3k+1[/tex].Second case: [tex]p=3k-1[/tex]. Then, [tex]p+2=3k+1[/tex] that can be prime, and [tex]p-2=3k-3=3(k-1)[/tex] that cannot be prime. Hence, there are not such three primes with [tex]p=3k-1[/tex].Therefore, there are no three primes of the form [tex]p[/tex], [tex]p+2[/tex] and [tex]p-2[/tex], 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.