Question:
$$\Big({m\choose0}+{m\choose1}-{m\choose2}-{m\choose3}\Big) + \Big({m\choose4}+{m\choose5}-{m\choose6}-{m\choose7}\Big) + \cdots = 0$$ if and only if for some positive integer $k$, we have
(i) $m = 4k$.
(ii) $m = 4k + 1$.
(iii) $m = 4k - 1$.
(iv) $m = 4k + 2$.
My Attempt:
$$(1+x)^m={m\choose0}+{m\choose1}x+{m\choose2}x^2+{m\choose3}x^3 + \cdots +{m\choose m}x^m$$
Replacing $x$ by $i$,
$$(1+i)^m={m\choose0}+{m\choose1}i-{m\choose2}-{m\choose3}i+\cdots+{m\choose m}i^m\\=\Bigg({m\choose0}-{m\choose2}+{m\choose4}-{m\choose6}+ \cdots \Bigg)+i\Bigg({m\choose1}-{m\choose3}+{m\choose5}-{m\choose7}+ \cdots \Bigg)$$
Given that ${m\choose0}-{m\choose2}+{m\choose4}-{m\choose6}+\cdots=-\Big({m\choose1}-{m\choose3}+{m\choose5}-{m\choose7}+\cdots\Big):=p$, thus $(1+i)^m=p-ip=p(1-i)$. Taking mod on both sides,
$$(\sqrt2)^m=|p|\sqrt2$$
Can we conclude from this?
If we take $m$ to be $4k$, say $8$, then we'll have two groups $\Big({m\choose0}+{m\choose1}-{m\choose2}-{m\choose3}\Big)$ and $\Big({m\choose4}+{m\choose5}-{m\choose6}-{m\choose7}\Big)$, and $m\choose8$ will be left as it is.
If $m=4k-1$, say $7$, we'll have two groups and nothing will be left.
But don't know what to do with this observation.
I think it's easier to use the symmetry argument. $${m\choose n}={m\choose m -n}$$ so $$ {m\choose n}+{m\choose n+1}-{m\choose m -n}-{m\choose m -(n+1)}=0 $$ and the two terms at distance $n$ and $n+1$ from the beginning and the two terms at distance $n$ and $n+1$ from the end of the summation cancel each other, so this is true for $m=3, 7, ...=4k-1$.
For example $$ {3\choose 0}+{3\choose 1}-{3\choose 2}-{3\choose 3}=0 $$ but for $m=4$ $$ {4\choose 0}+{4\choose 1}-{4\choose 2}-{4\choose 3}+{4\choose4}\neq0 $$