<s>[INST] Primes dividing sum of binomial coefficients. Given that $p>3$ is a prime, we have $k=\lfloor \frac {2p}{3} \rfloor$ then prove or disprove that $\sum_{i=1}^k\binom{p}{i}\equiv 0\pmod{p^2}$.
Its easy to see how $p$ divides each of the binomial coefficients, but I tried using maximal powers of $p$ in each coefficient but was stuck can someone help.
 [/INST] Ahmad showed the problem is equivalent to


*

*$H_{2k}-H_k \equiv 0 \pmod p$ for $p \equiv 1 \pmod 3$

*$H_{2k}-H_k+(2k+1)^{-1}  \equiv 0 \pmod p$ for $p \equiv 2 \pmod 3$


1:
$$H_{2k}-H_k \equiv \sum_{i=1}^{2k} i^{-1} - \sum_{i=1}^{k} i^{-1} \equiv \sum_{i=k+1}^{2k} i^{-1} \equiv
\sum_{i=k+1}^{\frac{3}{2}k} i^{-1} + \sum_{i=\frac{3}{2}k+1}^{2k} i^{-1} \equiv
\sum_{i=k+1}^{\frac{3}{2}k} i^{-1} + \sum_{i=-\frac{3}{2}k-1}^{-2k} (-i)^{-1} \equiv
\sum_{i=k+1}^{\frac{3}{2}k} i^{-1} + \sum_{i=3k+1-\frac{3}{2}k-1}^{3k+1-2k} (-i)^{-1} \equiv
\sum_{i=k+1}^{\frac{3}{2}k} i^{-1} + \sum_{i=k+1}^{\frac{3}{2}k} (-i)^{-1} \equiv
\sum_{i=k+1}^{\frac{3}{2}k} i^{-1} + \sum_{i=k+1}^{\frac{3}{2}k} (-1)^{-1}(i)^{-1}\equiv
\sum_{i=k+1}^{\frac{3}{2}k} i^{-1} + -i^{-1}
\equiv
\sum_{i=k+1}^{\frac{3}{2}k} 0
\equiv
0
\pmod p$$
The idea is that for each inverse the opposite inverse is also in the sum and therefore the sum cancelled out. You need to show $k$ is even so no inverse has no companion. If $k$ is odd, $3k+1$ is divisible by $2$ and therefore not a prime.
2:
Same idea just with odd number of terms.
$$\begin{align}
H_{2k}-H_k+(2k+1)^{-1}  \equiv \\ &
\equiv
(2k)^{-1}+(2k-1)^{-1}+\cdots+(1)^{-1} - ((k)^{-1}+(k-1)^{-1}+\cdots+(1)^{-1}) + (2k+1)^{-1} \\ &
\equiv
(2k)^{-1}+(2k-1)^{-1}+\cdots+(k+1)^{-1} +(2k+1)^{-1} \\ &
\equiv
(2k)^{-1}+(2k-1)^{-1}+\cdots+(k+1)^{-1} + (2k+1)^{-1} \\ &
\equiv
-(k+2)^{-1}+-(k+3)^{-1}+\cdots+(k+1)^{-1} + -(k+1)^{-1} \\ &
\equiv
0
\pmod p \end{align}$$
Again we need to show $k$ is always odd. If $k$ is even $3k+2$ is divisible by $2$.
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