TITLE: binomial congruence $\sum_{i=1}^{\frac{p-1}{2}}\binom{2i}{i}\equiv 0~or (-2)\pmod p$
QUESTION [11 upvotes]: Let $p\ge 5$ be a prime number. Show that $$\sum_{i=1}^{\frac{p-1}{2}}\binom{2i}{i}\equiv 0 \text{ or } -2\pmod p .$$

Examples:
If $p=5$, then
$$f=\sum_{i=1}^{\frac{p-1}{2}}\binom{2i}{i}=\binom{2}{1}+\binom{4}{2}=8\equiv -2\pmod 5 .$$
If $p=7$, then
$$f=\sum_{i=1}^{\frac{p-1}{2}}\binom{2i}{i}=\binom{2}{1}+\binom{4}{2}+\binom{6}{3}=28\equiv 0\pmod 7 .$$
If $p=11$, then
$$f=\sum_{i=1}^{\frac{p-1}{2}}\binom{2i}{i}=\binom{2}{1}+\binom{4}{2}+\binom{6}{3}+\binom{8}{4}+\binom{10}{5}=350\equiv -2 \pmod{11} .$$

REPLY [2 votes]: My solution:\begin{align*}&\sum_{i=0}^{\frac{p-1}{2}}\binom{2i}{i}=\sum_{i=0}^{\frac{p-1}{2}}\binom{2i}{i}\left(\dfrac{1}{2}+\frac{1}{2}\right)^{i}
=\sum_{i=0}^{\frac{p-1}{2}}\binom{2i}{i}\sum_{j=0}^{i}\binom{i}{j}\dfrac{1}{2^i}\\
&=\sum_{i=0}^{\frac{p-1}{2}}\sum_{j=0}^{i}\dfrac{(2i)!}{2^i\cdot i!\cdot j!(i-j)!}\\
&=\sum_{i=0}^{\frac{p-1}{2}}\sum_{j=0}^{i}\dfrac{(2i-1)!!}{j!(i-j)!}\\
&\equiv\sum_{i=0}^{\frac{p-1}{2}}\sum_{j=0}^{i}\dfrac{(-1)^i(p-1)(p-3)(p-5)\cdots(p-2i+1)}{j!(i-j)!}\\
&=\sum_{i=0}^{\frac{p-1}{2}}\sum_{j=0}^{i}\dfrac{(-1)^i2^i\left(\frac{p-1}{2}\right)!}{\left(\frac{p-1-2i}{2}\right)!j!(i-j)!}\\
&=\sum_{i=0}^{\frac{p-1}{2}}\sum_{j=0}^{i}(-2)^i\binom{\frac{p-1}{2}}{i}\binom{i}{j}\\
&=\sum_{i=0}^{\frac{p-1}{2}}(-4)^i\binom{\frac{p-1}{2}}{i}\\
&=(1-4)^{\frac{p-1}{2}}\\
&=(-3)^{\frac{p-1}{2}}\\
&\equiv \pm 1\pmod p
\end{align*}