Is there a criterion to check if n choose k is divisible by 4? Is there a criterion to check if $4 | {{n} \choose {k}}$?
For example if $n = 4$ there are two values of $k$ when $4 | {{n} \choose {k}}$. When $n = 14$ or $n = 16$ there are also some $k$ where $4 | {{n} \choose {k}}$ is passed. However there is no such $k$ for $n = 15$ or $n = 127$. So is there a change to check it without calculating $ n \choose k$?
 A: It seems the comments have provided definitive resolutions. Here is a way to get to that fact. (Esp. the one about adding the numbers in binary)
Consider the fact that
$$\binom{n}{k} = \frac{n!}{k!(n-k)!}$$
For $4$ to divide this, examine $\nu_2(n!), \nu_2(k!), \nu_2((n-k)!)$. By Legendre's, the values of such are 
$$\sum_{i=0}^{\infty}\left\lfloor\frac{n}{2^i}\right\rfloor \\ \sum_{i=0}^{\infty}\left\lfloor\frac{k}{2^i}\right\rfloor \\ \sum_{i=0}^{\infty}\left\lfloor\frac{n-k}{2^i}\right\rfloor$$
We need 
$$\sum_{i=0}^{\infty}\left\lfloor\frac{n}{2^i}\right\rfloor - \left(\sum_{i=0}^{\infty}\left\lfloor\frac{k}{2^i}\right\rfloor + \sum_{i=0}^{\infty}\left\lfloor\frac{n-k}{2^i}\right\rfloor\right) > 1$$
Observe that if $a+b=c$, then $\lfloor c \rfloor \geq \lfloor a \rfloor + \lfloor b \rfloor$ (proven by simple casework). Thus, it suffices to show that for some $i$,
$$\left\lfloor\frac{n}{2^i}\right\rfloor > \left\lfloor\frac{k}{2^i}\right\rfloor + \left\lfloor\frac{n-k}{2^i}\right\rfloor$$
When this happens, it is equivalent to say that
$$\left\lbrace\frac{k}{2^i}\right\rbrace + \left\lbrace\frac{n-k}{2^i}\right\rbrace \geq 1$$
Multiplying by $2^i$, we need 
$$k \bmod 2^i + (n-k) \bmod 2^i \geq 2^i$$
Hopefully it is evident from the binary representations here.
Edit: To be clear, check if $k \text{ AND } (n-k) \neq 0$, or if $n = 2^i - 1$.
