Binomial coefficients mod p I want to find the following sum mod $p$ (prime number): Let $i\geq \frac{p-1}{2}$,
$
\sum_{k=i}^{p-1} \binom{k}{i}\binom{k}{p-i-1} \pmod{p}
$
OK, I succeeded in simplyfying this argument to the following, I reversed $i$ to be $\leq \frac{p-1}{2}$:
$\sum_{j=0}^i (-1)^j\binom{i}{j}\binom{i+j}{j}\equiv \pm 1\pmod{p}$
 A: Consider the projective space of dimension $i$ and let $V$ be a vector space of dimension $i$ with $p\in \mathbb{P}^i$ (over say complex numbers). I will, for ease of typing, write $\wedge^kV(l)$ for $\wedge^kV\otimes \mathcal{O}(l)$. Then we have the Koszul resolution, $$0\to \wedge^iV(-i)\to \wedge^{i-1} V(-i+1)\to\cdots \to V(-1)\to\mathcal{O}\to k(p)\to 0.$$ For any $l<0$ (you can also do for other $l$, but we don't need it for your question) twisting by $l$ we see that $\sum(-1)^j\chi(\wedge^jV(-j+l))=\pm\chi(k(p))=\pm 1$. But $\chi(\wedge^jV(-j+l))=h^i(\wedge^jV(-j+l))=h^0(\wedge^jV(j-l-i-1))$ (Canonically there are duals involved, but we are just computing dimensions). Taking $l=-i-1$ and noting that $h^0(\mathcal{O}(l)=\binom{i+l}{l}$, you get what you want.
A: We have:
$$\sum_{k=0}^\infty (-1)^k{n\choose k} x^k= (1-x)^n$$
$$\sum_{k=0}^\infty {{n+k} \choose n} x^k = (1-x)^{-n-1} $$
Multiplying and comparing coefficients gives us:
$$\sum_{j=0}^n (-1)^j {n\choose j} {{n+j}\choose j} = (-1)^n\sum_{j=0}^n (-1)^{n-j} {n\choose {n-j}} {{n+j}\choose n} =$$
$$(-1)^n[x^n] \frac{1}{1-x} = (-1)^n$$
No reduction modulo $p$ is required.
