There are certain points to be noted about prime and composite numbers.

1. The number of prime numbers are infinite
2. Every prime number is odd except 2.
3. Every prime number which is more than 3, can be represented as \(6k + 1\) or \(6k – 1\)
4. Every prime number which is more than 2, can be written as \(4k + 1\) or  \(4k – 1\).
5. Every prime number which is more than 3, can be represented as \(8k \pm 1\) or \(8k \pm 3\)

6. There in no known algebraic formula which can represent prime numbers, many attempt have been made to make a formula to generate a formula for prime numbers for example the number \({n^2} + n + 41\) is always prime for all values of \(n\) from 1 to 39 but it fails when \(n = 40\) or any multiple of 41. 

   Similarly another number of this type is  

   \({n^2} + n + 17\) is prime for all values from 1 to 15.

7. If \(p\) is a prime number more than 3 then \({p^2} - 1\) is always multiple of 24.
8. If \(p\) denotes prime numbers and \(c\) denotes composite numbers then \({p^c},\,\,{c^p},\,\,{p^p},\,\,p \times c,\,\,c \times c\) and \(p \times p\) are always composite. \(p + c, p + p\) may not be a prime.

Example 1: If \(p, p + 4, p + 14\) all three numbers are prime. How many different values can \(p\) take?

Solution: By simple observation we see that  \(p\) is 3, now suppose \(p > 3\), since \(p\) is a prime number it can be only \(6k + 1\) or \(6k – 1\). 

(a) \(p = 6k + 1\), 

\(p + 4 = 6k + 5\) and

\(p +14\) will be \(6k + 15\), now \(6k + 15\) can never be a prime number as it is multiple of 3.

(b) \(p = 6k – 1\), 

\(p + 4 = 6k + 3\)

\(p +14\) will be \(6k + 13\)

Since \(p + 4 = 6k + 3\) which can never be a prime number as it is multiple of 3.

So \(p, p + 4\) and \(p + 14\) can never be prime simultaneously for a value of \(p\) other than 3.