Writing the statement in english with proof Was wondering how to write this out in English and prove the statement. 
$$
\forall x \in \mathbb{N} (x > 1 \implies ∃k ∈ N∃m ∈ N (m=1 \pmod 2)^x = 2^k \cdot m))
$$
My answer so far: for all integers $x$, there are 2 natural numbers $k$ and $m$ such that.. 
I don't feel I have read that part right, I am just getting confused with the brackets. 

gt6989b has the right translation. Basically this is saying that for each natural number greater than 1, there exists two natural numbers $k$ and $m$, where $k$ will be a natural number that acts as a binary log value of an exponential whose base is 2 (e.g. $k$ is the exponent of $2^k$ and $m$ will be odd). We must prove that $x = 2^k * m$.
For the proof:
Assume $x > 1$. If the number $x$ is odd then let $k = 0$ and $m = x$. If the natural number $x$ is even then it is either a power of $2$ or by the fundamental theorem of arithmetic it can be written as a unique set of primes (up to ordering, there may be multiple primes as well, notice those $\alpha$ exponents) such that $x = p_1^{\alpha_1} \cdot p_2^{\alpha_2} \cdots p_c^{\alpha_c} \cdots p_n^{\alpha_n}$. If $x$ is a product of the powers of $2$ and contains no other prime factor then let $k = \log_2 x$ and $m = 1$. Otherwise, let the ordering of the primes be such that $p_1$ through $p_c$ are all of the even prime integers (the only even prime being 2). Thus let $k = c$, e.g. the power of $2^k$ is determined by the number of $2$'s in the factoring, and let $m = p_{c+1} \cdots p_n$. We have that $m$ must be odd, e.g. a product of odd primes (otherwise it would have a factor of $2$ which should have been in the first $p_1$ through $p_c$ factors). 
Thus $x = 2^k \cdot m$ and $m$ being odd implies $m \mod 2 = 1$.

Perhaps an easier way to think about this is that you're asked to prove $x = 2^k \cdot m$ with some contingencies. Let's rewrite the formula in mathjax:
$$\forall x \in \mathbb{N} (x > 1 \implies \exists k \in \mathbb{N},\ \ \exists m \in \mathbb{N}, \ \  ((m=1) \mod 2 \wedge x = 2^k * m))$$
Based on the statement of the formula, we first look outside and then progressively inward, applying restrictions: for all numbers x that are natural numbers, if $x$ is greater than $1$ then the inner part of the formula follows:
$$\exists k \in \mathbb{N},\ \ \exists m \in \mathbb{N}, \ \  ((m=1) \mod 2 \wedge x = 2^k * m$$
So assume $x$ greater than 1 and $x \in \mathbb{N}$. Then we have that there exists some $k$, a natural number and  there exists some $m$, a natural number, such that the remainder of the inner formula follows:
$$(m=1) \mod 2 \wedge x = 2^k * m$$
Taken together this means we can rule out $x$, $m$, and $k$ being negative or decimals for whatever we want to prove about the inner formula $(m=1) \mod 2 \wedge x = 2^k * m$. 
Now, based on all of that, we need to show that $x = 2^k * m$ and that for our choice of $m$ for every $x$ greater than 1, $(m = 1) \mod 2$.
So this requires us to think about how $x$ is written. Is $x$ odd? Well, that's a possibility. Is $x$ even? That too is a possibility. If $x$ is even, then can it be written as just a $2^k$ or is there some prime factor $p$ such that $x = 2^k * p$? Perhaps there are multiple prime factors? The proof above attempts to address each of these questions to say that "Yes the inner statement $(m=1) \mod 2 \wedge x = 2^k * m$ works whether or not $x$ is odd, or if $x$ is even with no odd prime factors or $x$ is even with many prime factors."
That's the reason the proof might be longer than you're expecting. Of course you can be terse, but the purpose of proof is to explain it to another human in a way that they believe the result and understand the steps you took to derive the result.
