Question

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.

Answer 1

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 $r<p$ and $r$ is called the remainder.

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.

First case: $p=3k+1$. 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 $p=3k+1$.

Second case: $p=3k-1$. 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.

Answer 2

Following are the solution to the given question:

Note that any prime number other than 2 is an odd number

$P>3$ 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:

$\to p=3(k+1) \ \ \ \ \ Or \ \ \ \ \ p=3q+2 =3(q+1)-1=3k-1$

So, p is of form $3k+1 \ \ or\ \ 3k-1$.

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 $p+2 \ and \ p-2$ are also prime then

From a:

Case-I:

$P=3k+1$ then $p-2=3k-1 \ \ and\ \ p\neq 2 =3k+3 = 3(K+1)$

so, $P+2$ is divisible by 3 that is  $P+2$ is not a prime.

Case-II :

$p=3K-1: p-2 = 3k-1-2=3 (K-1)$

if

$K= 2, then \ 3(k-1) = 3\\\\P = 5$

When $k>2$ then $3(k-1)$ is divisible by 3, So it's not a prime

Hence, if $p\neq 5$ 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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