Prove that 3n^3 + n + 5 is not divisible by 2 for all integers n ≥ 1?

3n^3 + n + 5

= n(3n^2 + 1) + 5

= n( 2n^2 + 2 + n^2 -1) + 5

= 2n(n^2 + 1) + n(n+1)(n-1) + 5

now 2n(n^2 + 1) is even as 2 is a factor

n(n+1)(n-1) product of 3 consecutive numbers is even

5 is odd

so sum and hence given expression is odd

Case 1: n is even.

Writing n = 2k for some positive integer k, we have

3n^3 + n + 5 = 3(2k)^3 + (2k) + 5

...................= 2 (12k^3 + k + 2) + 1.

Since 12k^3 + k + 2 is an integer, we see that 3n^3 + n + 5 is odd.

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Case 2: n is odd.

Writing n = 2k+1 for some non-negative integer k, we have

3n^3 + n + 5 = 3(2k+1)^3 + (2k+1) + 5

...................= 2(12k^3 + 18k^2 + 10k + 4) + 1

Since 12k^3 + 18k^2 + 10k + 4 is an integer, we see that 3n^3 + n + 5 is odd.

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Hence, 3n^3 + n + 5 is not divisible by 2 for all integers n ≥ 1.

I hope this helps!