Prove that if n is an integer then 2n^2+n+1 is not divisible by 3

Given

2n2 + 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 = 2n2 + n + 1

Taking MOD 3 on both sides

S % 3 = ( 2n2 + 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 = 2n2 + n + 1

Taking MOD 3 on both sides

S % 3 = ( 2n2 + 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 = 2n2 + n + 1

Taking MOD 3 on both sides

S % 3 = ( 2n2 + 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 2n2 + n + 1 is not divisible by 3 ∀ n ∈ Z. Hence Proved.