Let $n$ be an integer. Prove that there exist $k, m$ integers such that $n=2^km$

Question

Let $n$ be an integer. Prove that there exist $k, m$ integers such that $n=2^km$.

I am stuck with this proof. I thought at first that I need to use contradiction but that did not work. What do you suggest? I know I have to break them into cases basically when $n=1$ then $1=2^0 \cdot 1$ but what about the other case?

I suspect there are some missing criteria, otherwise one may choose $k = 0$ and $m = n$, since $n = 2^0 \cdot n$. — 2017-02-13
Related http://math.stackexchange.com/questions/1998419/how-to-prove-partition/ — 2017-02-13
If you don't say what kind of integers are you considering, you can easily generalized to $n=p^km$ for all prime $p$ — 2017-02-13

user id:34692 12k · Accepted Answer

Induction argument hint. Do the basis step. Then for the inductive step, assume it holds for $1, \dots, n$. Now take two cases: If $m$ is odd, you are done (why?). If $n$ is even consider $(n+1)/2$.... you fill in the rest.

user id:202857 127k · Accepted Answer

Hint:

Consider the factorisation of $n$ as a product of primes: $$n=2^k\prod_{\substack{p_i\;\text{odd}\\\text{prime}}} p_i^{k_i}.$$

Let $n$ be an integer. Prove that there exist $k, m$ integers such that $n=2^km$

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