I'm trying to prove that ${4a \choose 2a} - {2a \choose a}$ is divisible by $4$, and I've currently tried using induction, though I have been unsuccessful. I'm not really sure where to go to for now, any help would be appreciated.
This is what I have currently done; For $a=1$ we have that ${4a \choose 2a} - {2a \choose a} = {4 \choose 2} - {2 \choose 1} = 4$, which is divisible by $4$ and hence the statement is true for $a=1$.
Assume true for $a =n$ (where $n \in \mathbb{N}$).
Required to prove true for $a = n+1$. We have that; $${4(a+1) \choose 2(a+1)} - {2(a+1) \choose a+1} = {4a+4 \choose 2a+2} - {2a+2 \choose a+1} $$ Though $${2a+2 \choose a+1} = {2a+1 \choose a} + {2a+1 \choose a+1} = 2 {2a+1 \choose a}$$ However, applying the same binomial identity we get that ${2a+1 \choose a} = {2a \choose a - 1} + {2a \choose a}$. Hence our original expression can be written as; $${4a+4 \choose 2a+2} - {2a+2 \choose a+1} = {4a+4 \choose 2a+2} - 2 {2a+1 \choose a} $$
After here I'm not sure where to go?
Beware: I am going for the overkill. By Legendre's theorem exponent of the largest power of $2$ dividing $n!$ is $\sum_{k\geq 1}\left\lfloor\frac{n}{2^k}\right\rfloor$, hence $$ \nu_2\binom{2n}{n} = \sum_{k\geq 1}\left(\left\lfloor\frac{2n}{2^k}\right\rfloor-2\left\lfloor\frac{n}{2^k}\right\rfloor\right)\tag{1}$$ and the LHS is given by the number of "$1$"s in the binary representation of $n$. If we assume that $n$ has $k$ "1"s in its binary representation then both $\binom{2n}{n}$ and $\binom{4n}{2n}$ are numbers of the form $2^k\cdot\text{odd}$, hence their difference is a multiple of $2^{k+1}$. Since $k\geq 1$, we are done.