**Given**

2n^{2} + n + 1 is not divisible by 3 ∀ n ∈ Z

**Proof**

We will prove the given statement by direct method.

For each value n, we have three choices

**CASE 1 : n % 3 = 0**

S = 2n^{2} + n + 1

Taking MOD 3 on both sides

S % 3 = ( 2n^{2} + n + 1 ) % 3

= ( 2 ( n % 3 ) ( n % 3 ) + ( n % 3 ) + ( 1 % 3 ) ) % 3

= ( 2(0)(0) + (0) + 1 ) % 3

= 1 % 3

= 1

Therefore, the given statement is not divisible by 3 if n % 3 = 0.

**CASE 1 : n % 3 = 1**

S = 2n^{2} + n + 1

Taking MOD 3 on both sides

S % 3 = ( 2n^{2} + n + 1 ) % 3

= ( 2 ( n % 3 ) ( n % 3 ) + ( n % 3 ) + ( 1 % 3 ) ) % 3

= ( 2(1)(1) + (1) + 1 ) % 3

= 4 % 3

= 1

Therefore, the given statement is not divisible by 3 if n % 3 = 1.

**CASE 1 : n % 3 = 2**

S = 2n^{2} + n + 1

Taking MOD 3 on both sides

S % 3 = ( 2n^{2} + n + 1 ) % 3

= ( 2 ( n % 3 ) ( n % 3 ) + ( n % 3 ) + ( 1 % 3 ) ) % 3

= ( 2(2)(2) + (2) + 1 ) % 3

= 11 % 3

= 2

Therefore, the given statement is not divisible by 3 if n % 3 = 2.

From the above 3 cases, we can conclude that the given statement 2n^{2} + n + 1 is not divisible by 3 ∀ n ∈ Z. Hence Proved.