Even number sum proof Prove that every even number $k\geqslant 8$ can be represented as $m,n\in \mathbb{N}\setminus \{1\}$
$$k=m+n$$
and $\gcd(m,n)=1$.
I was able to do it, if $k$ is an odd number but not if it is even. I could use some hints.

If $k = 4m: m \in \Bbb{N}$ then $k = (2m-1) + (2m+1)$ and both of those terms are odd and differ by exactly $2$, so one step of Euclid's algorithm shows that $$\gcd(2m+1,2,-1) = \gcd(2m-1,2) = 1.$$
Having taken care of all multiples of $4$, now consider $k = 4m+2$. In that case, $k = (2m-3) + (2m+1)$ and 
$$\gcd(2m+1,2,2m-3) = \gcd(2m-3,4) = 1$$ since $2m-3$ contains no powers of $2$.
