I'm trying to prove a property involving the parity of a binomial coefficient using Lucas' theorem. For this, let $r=\lfloor \log_2(k_s)\rfloor+1$, with $k_s\geq2$. We know that if $0\leq i\leq 2^{r}-1$, then Lucas' theorem implies that $\binom{i+2^r}{k_s}\equiv\binom{i}{k_s}\binom{1}{0}\equiv\binom{i}{k_s}(\text{mod }2)$. The latter relation is clear. What I don't know is how do I prove that,
\begin{equation}
\binom{k_s+2^{r-1}}{k_s}\equiv0(\text{mod }2)
\end{equation}
using Lucas' theorem. I hope anyone can help me. I would appreciate any help that you can give me to clarify this.
Best Answer
Note that $r - 1 = \left\lfloor \log_2(k_s) \right\rfloor \; \rightarrow \; r - 1 \le \log_2(k_s) \lt r \; \rightarrow \; 2^{r-1} \le k_s \lt 2^{r}$. Thus, in the base $2$ expansions, $k_s$ has a $1$ coefficient for the $2^{r-1}$ term, but $k_s + 2^{r-1}$ has a carry of the $2^{r-1}$ term, meaning its coefficient is $0$. Thus, by Lucas's theorem, those coefficients give $\binom{0}{1} = 0$ for the $2^{r-1}$ term, meaning the congruence product, modulo $2$, is $0$.
FYI, Kummer's theorem can also be used to prove this since there's a carry when adding $k_s$ and $2^{r-1}$ in base $2$, as I explained earlier. In addition, since there's only one carry, this theorem shows that $\binom{k_s + 2^{r-1}}{k_s}$ has just one factor of $2$.