Respuesta :
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.
Following are the solution to the given question:
Note that any prime number other than 2 is an odd number
[tex]P>3[/tex] is a prime number we divide by 3 the possible remainders are 1 or 2 ['0' is ruled out because p is not divisible by 3]
so, p has the form:
[tex]\to p=3(k+1) \ \ \ \ \ Or \ \ \ \ \ p=3q+2 =3(q+1)-1=3k-1\\\\[/tex]
So, p is of form [tex]3k+1 \ \ or\ \ 3k-1[/tex].
we have, 3, 5, 7 prime numbers, so 5 is prime no that takes part in two different prime pairs.
Now if p is a prime no. Such that [tex]p+2 \ and \ p-2[/tex] are also prime then
From a:
Case-I:
[tex]P=3k+1[/tex] then [tex]p-2=3k-1 \ \ and\ \ p\neq 2 =3k+3 = 3(K+1)[/tex]
so, [tex]P+2[/tex] is divisible by 3 that is [tex]P+2[/tex] is not a prime.
Case-II :
[tex]p=3K-1: p-2 = 3k-1-2=3 (K-1)[/tex]
if
[tex]K= 2, then \ 3(k-1) = 3\\\\P = 5[/tex]
When [tex]k>2[/tex] then [tex]3(k-1)[/tex] is divisible by 3, So it's not a prime
Hence, if [tex]p\neq 5[/tex] then p does not take part in two different twin primos
Note: Here divisible by 3 means for the numbers, not 3.
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