South African Mathematics Olympiad Third Round

We prove the statement by induction on $m$. When $m=k$, we have $$L_i\left[\left(\genfrac{}{}{0ex}{}{k}{i}\right)-\left(\genfrac{}{}{0ex}{}{0}{i}\right)\right]=L_i\left(\genfrac{}{}{0ex}{}{k}{i}\right)=L_i\cdot \frac{k!}{i!(k-i)!}=k\cdot \frac{L_i}{i}\cdot \frac{(k-1)!}{(i-1)!(k-i)!}=k\cdot \frac{L_i}{i}\cdot \left(\genfrac{}{}{0ex}{}{k-1}{i-1}\right).$$ Since $\frac{L_i}{i}$ is an integer (by definition of $L_i$), and $\left(\genfrac{}{}{0ex}{}{k-1}{i-1}\right)$ is a binomial coefficient and thus also an integer, we see that $k$ is indeed a divisor.

For the induction step, assume that the statement holds for a specific value of $m$. We use the recursion $\left(\genfrac{}{}{0ex}{}{m+1}{i}\right)=\left(\genfrac{}{}{0ex}{}{m}{i}\right)+\left(\genfrac{}{}{0ex}{}{m}{i-1}\right)$ to show that it holds for $m+1$ as well: $$\begin{array}{rl}{L}_i\left[\left(\genfrac{}{}{0ex}{}{m+1}{i}\right)-\left(\genfrac{}{}{0ex}{}{m+1-k}{i}\right)\right]& =L_i\left[\left(\genfrac{}{}{0ex}{}{m}{i}\right)-\left(\genfrac{}{}{0ex}{}{m-k}{i}\right)\right]+L_i\left[\left(\genfrac{}{}{0ex}{}{m}{i-1}\right)-\left(\genfrac{}{}{0ex}{}{m-k}{i-1}\right)\right]\\ & =L_i\left[\left(\genfrac{}{}{0ex}{}{m}{i}\right)-\left(\genfrac{}{}{0ex}{}{m-k}{i}\right)\right]+\frac{L_i}{L_{i-1}}\cdot L_{i-1}\left[\left(\genfrac{}{}{0ex}{}{m}{i-1}\right)-\left(\genfrac{}{}{0ex}{}{m-k}{i-1}\right)\right].\end{array}$$ Note that $k$ divides both $L_i[(\genfrac{}{}{0ex}{}{m}{i})-(\genfrac{}{}{0ex}{}{m-k}{i})]$ and $L_{i-1}[(\genfrac{}{}{0ex}{}{m}{i-1})-(\genfrac{}{}{0ex}{}{m-k}{i-1})]$ by the induction hypothesis (if $i=1$, the latter term is simply zero), and $L_{i-1}$ (the least common multiple of $1,2,\dots ,i-1$) divides $L_i$ (the least common multiple of $1,2,\dots ,i$, or equivalently the least common multiple of $L_{i-1}$ and $i$). It follows that $k$ is also a divisor of $L_i[(\genfrac{}{}{0ex}{}{m+1}{i})-(\genfrac{}{}{0ex}{}{m+1-k}{i})]$, which completes the proof.