Prove or disprove: If $n^3$ is odd then $n$ is odd.

Question

If $n^3$ is odd, then $n$ is odd.

I need to prove or disprove by means of counterexample why this is true or false.

$\forall x P(x) = x^3$, $x = 1,3,5,7,9$

I am having a very difficult time finding a way to prove this by counterexample!

Answer 1

Suppose $n^3$ is odd, but $n$ is not odd. Then $n$ must be even. The product of any two even numbers is even, so $n^2$ is even. Then $n(n^2)=n^3$ is the product of two even numbers again: and so $n(n^2)=n^3$ is even. This contradicts the assumption that $n^3$ is odd, and so $n$ is not even. So $n$ must be odd.

Answer 2

• if $n$ is even, $n^3$ is even.
• if $n$ is odd, $n^3$ is odd.

Therefore $n$ is odd iff $n^3$ is odd.

Answer 3

If $n^3$ is odd, then $n^3 = 2m+1$ for some $m$.

Therefore $2m = n^3-1 =(n-1)(n^2+n+1) $.

$n^2+n+1$ is odd, since $n^2+n = n(n+1)$ is even, being the product of two consecutive integers.

But $2m$ is even, so $n-1$ is even, since the product of two odd numbers is odd.

Therefore $n$ is odd.

Answer 4

You can never prove anything by showing a counter-example. If you can find a counter example then the statement is not true.

Now this is VERY VERY VERY important. WHAT is the exact wording of the problem? Does it, or does it not, specifically state that $n$ must be a natural number?

The answer are COMPLETELY different depending on whether $n$ is assumed to be a natural number or not.

1) $n$ need not be a natural number:

This is tricky because if $n$ is odd, then $n^3$ is odd. But the converse, if $n^3$ is odd implies $n$ is odd, is only true if $n$ is an integer... or in other words if $n^3$ is not a perfect cube then $n$ is not an integer and $n$ is neither even nor odd.

So counter examples are $n^3 = 3,5,7,9,11,13,,15,...25,29,.... $ etc.

Now... IF the problem specifies that $n$ must be a natural number. Then those are not counter examples because $n^3 = 3 \iff n = \sqrt[3]3 \not \in \mathbb N$.

2))

$n \in \mathbb N$

Either $n$ is odd and $n = 2k + 1$ for some $k \in \mathbb N \cup \{0\}$. Or $n$ is even and $n=2k$ for $k \in \mathbb N \cup \{0\}$. If $n = 2k$ is even, then $n^3 = (2k)^3 = 8k^3 = 2*(4k^3)$ is even. So if $n^3$ is odd, $n$ can not be even. So $n$ must be odd.