Prove or disprove the following: If $n^3 − 5$ is an odd integer, then $n$ is even.

Question

Prove or disprove the following proposition: If $n^3 − 5$ is an odd integer, then $n$ is even.

I know that $n$ must be even in order for $n^3 - 5$ to be odd which means I have to prove the statement.. possibly with a contradiction? I have been able to successfully start the proof but I am unsure of where to go from here. Any help would be greatly appreciated!

Proof: Suppose $n$ is an odd integer which can be expressed as $n=2k+1$, and $n^3-5$ is also an odd integer. $$(2k+1)^3 - 5 = (2k+1)(2k+1)(2k+1) - 5 = 8k^3 + 12k^2 + 6k + 1 - 5...$$

Where should I go from here? Thanks!

Answer 1

You could say if $n=2k+1$ then $n^3-5=8k^3+12k^2+6k-4$ is even (being a sum of even numbers), which proves it by contrapositive, but I find it simpler to say if $n^3-5$ is odd, then $n^3$ is even, so $n$ is even.

Answer 2

Looks like you have an error multiplying out $(2k+1)^3.$ Anyway you are on the right track.

If $n^5 - 5$ is odd implies $n$ is even, the contra-positive would be "$n$ is odd implies $n^5 - 5$ is even."

Let odd $n = 2k + 1$ with integer $k.$

$(2k+1)^3 - 5 = (2k)^3 + 3(2k)^2 + 3(2k) + 1 - 5 = 8k^3 - 12k^2 + 6k - 4 = 2(4k^3 - 12k^2 + 3k - 2)$

$2$ divides $2(4k^3 - 12k^2 + 3k - 4)$

Answer 3

Since $$n^3-5=n^3-1-4$$ we can say that $ n^3-5 $ and $ n^3-1 $ have the same parity.

We will prove that if $ n $ is odd, then $ n^3-1 $ is even.

$$n^3-1=(n-1)(n^2+n+1)$$

So, if $ n=2k+1 $ then $$n^3-1=2k(n^2+n+1)$$ is even.

Answer 4

The above answers seem very convoluted and over-specific to me. If $n$ is any odd number, then all powers $n^i$ are odd, and, in particular, $n^3$ is odd, so $n^3 - 5$ is even. But we are given that $n^3 - 5$ is even, so $n$ cannot be odd, and hence $n$ is even.

Answer 5

Alternative proof using Euclid's lemma. Namely, $p|ab\implies p|a\lor p|b$.

Then since $2|n^3$, we get the result.