Prove that a number is even, given the cube is even If it is known that $x^3$ is even, can we say that $x$ is even? It seems to be the case because an odd*odd*odd=odd (if we are dealing with natural numbers). But is there a proof?
 A: Yes it's a formal proof by contradiction and to be more accurate we write:
assume that $x$ is odd hence $x=2k+1$ for some $k$ and then
$$x^3=(2k+1)^3=8k^3+12k^2+6k+1=2(\underbrace{4k^3+6k^2+3k}_{=k'})+1=2k'+1$$
so $x^3$ is odd which's a contradiction.
A: Your proof is correct: if $n$ was odd then $n^3$ would be odd as well, so $n$ must be even.
A: Suppose $x$ is odd. Then 2 does not divide x. So the prime factorization of x does not contain $2$. What can we say about the prime factorization of $x^3$?
A: It sounds like you've more or less given a proof without realizing it.  Proof by contrapositive
A: Suppose $n^3$ is even.  Then $2$ is a prime factor of $n^3$ and therefore $n$. Hence, you actually have $8\mid n^3$ if $n^3$ is even. 
A: On top of the mathematical proofs, you can try this yourself with very high numbers with a bit of code.
from __future__ import division    

for i in range(0,1000):
    if type((i*i*i)/2)==int:
        print(i)
else:
    print("Done")

With any number you input, you can see it will never find any number that this statement does not hold true for.
http://labs.codecademy.com/
You can try the python code out here if you like.
