Thai Mathematical Olympiad

Observe that $-2008\equiv 6\equiv 5^2(\mathrm{mod}19)$. Thus, $$\begin{array}{rl}2\sum _{k=0}^n\left(\genfrac{}{}{0ex}{}{2n+1}{2k+1}\right)(-2008{)}^k& \equiv 2\sum _{k=0}^n\left(\genfrac{}{}{0ex}{}{2n+1}{2k+1}\right)5^{2k}(\mathrm{mod}19)\\ & \equiv (1+5{)}^{2n+1}-(1-5{)}^{2n+1}(\mathrm{mod}19)\\ & \equiv 6^{2n+1}+4^{2n+1}\\ & \equiv 2^{2n+1}(3^{2n+1}+2^{2n+1})(\mathrm{mod}19).\end{array}$$ Since $$3^{2n+1}+2^{2n+1}\equiv (-16{)}^{2n+1}+2^{2n+1}\equiv 2^{2n+1}(1-2^{6n+3})(\mathrm{mod}19)$$ and $2^{18}\equiv 1(\mathrm{mod}19)$. We can see that $$2^{6(n+3)+3}\equiv 2^{6n+3}(\mathrm{mod}19)$$ for each $n=0,1,2,\dots$. Therefore, it suffices to consider the divisibility of $3^{2n+1}+2^{2n+1}$ by $19$ when $n=0,1,2$. We now verify that $$3^1+2^1\equiv 5(\mathrm{mod}19)$$ $$3^3+2^3\equiv 35\equiv 16(\mathrm{mod}19)$$ $$3^5+2^5\equiv 275\equiv 9(\mathrm{mod}19)$$ Hence, $19\nmid {\sum }_{k=0}^n\left(\genfrac{}{}{0ex}{}{2n+1}{2k+1}\right)(-2008{)}^k$.