### Theorem of Euclid’s on prime number : There are an infinite number of primes.

Proof: Euclid’s proof is by contradiction let p_{1 }= 2, p_{2} = 3, p_{3} = 5, p_{4} = 7, … be the primes in according order and suppose that there is a least prime called p_{n}. Now consider the positive integer P = p_{1} p_{2}p_{3}…p_{n} + 1.

Since P 1, we conclude that P is a divisible by some prime p. But p_{1} p_{2}p_{3}…p_{n} are the only prime numbers, so that p must be equal to one of p_{1} p_{2}p_{3}…p_{n} combining the relation, p\p_{1} p_{2}p_{3}…p_{n} with p\ P, we arrive p \ P – p_{1} p_{2}p_{3}…p_{n} or, equivalently, p\1. The only positive of the integer 1 is 1 itself, and because p > 1, a contradiction arises. Thus, no finite list of primes is complete, whence the number of primes is infinite. (Proved)

**Problem:** *The only prime of the form n ^{3} – 1 is 7.*

**Proof:** Given that, n^{3} – 1= n^{3} – 1^{3} = (n – 1) (n^{2} + n + 1)

We know if n is an integer, n – 1 is an integer and n^{2} + n + 1 is also an integer.

Thus (n – 1) (n^{2} + n + 1) = ab for integers and a/b.

If ab is prime, either a or b must be 1(or else we would have more than one factors).

Thus either n -1 = 1 or (n^{2} + n + 1) = 1. If the first is true then n = 2

∴ n^{3} – 1 = 7

If the 2^{nd} is true then n = 0 and n = – 1. So, we test for n = 0, n = – 1, and n = 2.

When n = 0, then n^{3} – 1 = – 1 is not prime.

When n = – 1 then n^{3} – 1 = (– 1)^{3} – 1 = – 1 – 1 = – 2, is not prime.

When n = 2 then n^{3} – 1 = 8 – 1 = 7.

∴ The only prime of the form n^{3} – 1 is 7. (Proved).

**Theorem:** If p_{n} is the nth prime number, then p_{n} ≤ 2^{2^n-1}

Proof: Let us proceed by induction on n, the asserted inequality being clearly true when n = 1. As the hypothesis of the induction, we assume that n > 1 and that the result holds for all integers up to n. Then

p_{n +1} ≤ p_{1}p_{2}…p_{n} + 1

≤ 2.2^{2}…(2^{2^(n-1)})+1 = 2^{1 + 2 + 2^2+…+2^(n-1)}+1

Recalling the identity 1 + 2 + 2 ^{2} + …+ 2^{n-1} = 2^{n }– 1 , we obtain

p_{n +1} ≤ 2^{2^n -1}+1

However, 1 ≤ 2^{2^n-1} for all n, whence

P_{n+1} ≤ 2^{2^n-1}+2^{2^n-1} = 2.2^{2^n-1} = 2^{2^n}.

Hence, Completing the induction step and the argument.** (Proved)**