How can I prove this in a systematic manner?

Question

I have to prove the following claim.

For all $n \in \mathbb{N}, 2$ divides $3n^{3} + 13n^{2} + 18n + 8.$

I want to have a systematic proof or even just a hint, to start.

Answer 1

You can use Proof by division into cases.

For all n in $N$ 2 divides $3n^3 + 13n^2 + 18n + 8$

You can take two cases, where n is odd and where n is Even.

Note that these two cases indeed exhaust all the possible cases in Natural numbers.

First we can rewrite the claimed polynomial as:

$n^2(3n+13) + 18n + 8$

Case, n is odd, meaning n = $2k + 1$, where k is any arbitrary Natural number

You can substitute n in the rewritten polynomial to get

$(2k+1)^2(3(2k+1)+13) + 18(2k+1) + 8$

Solving it a bit, you get.

$(2k+1)^2(6k + 3+13) + 18(2k+1) + 8$

Or more simply

$2 [(2k+1)^2(3k+8) + 9(2k+1) + 4] $

This shows that it can be written in the form of 2D where D is any arbitrary Natural number.

Note that, we can write the polynomial between $[ ]$ because, k is also a Natural.

This proves that if n is odd, 2 divides $3n^3 + 13n^2 + 18n + 8$.

Case, n is even, meaning n = $2k$, where k is any arbitrary Natural number.

Substituting the value of n, in the polynomial we get

$(2k)^2(3(2k)+13) + 18(2k) + 8$

Solving it a bit we can write it as:

$4k^2(6k +13) + 36k + 8$

Or even as,

$2 [2k^2(6k+13) + 18k + 4] $

This shows that it can be written in the form of 2D where D is any arbitrary Natural number.

Note that, we can write the polynomial between $[ ]$ because, k is also a Natural.

This proves that if n is even, 2 divides $3n^3 + 13n^2 + 18n + 8$.

Thus, we prove the claim.

QED.

Answer 2

Use that $$3n^3+13n^2+18n+8=(n+1)(n+2)(3n+4)$$

Answer 3

For fun:

$3n^3+3n^2+10n^2 +18n +8=$

$3n^2(n+1) + 2(5n^2+9n+4).$

The first term is divisible by $2$ (why?)