Problem. Let $p$ be an odd prime number, $m\in\mathbb{N}_+$ and $(p-1)|m$, prove that
$$\sum_{i=0}^{m/(p-1)}\binom m{(p-1)i}\equiv2+p(1+m)\pmod{p^2}$$
where $\binom ab=\dfrac{a!}{b!(a-b)!}$ is the binomial coefficient.
The following conclusions may help :
(a) Let $g$ be the primitive root of $p^k$, then $1+(g^t)+(g^t)^2+...+(g^t)^{\varphi(p^k)-1}=\varphi(p^k)[\varphi(p^k)|t]$, where $[A]$ is the Iverson Bracket ($[A]$ is $1$ when proposition $A$ holds, and $0$ otherwise).
(b) By applying the binomial theorem, we can handle this type of combinatorial summing neatly. For a example, $\sum_{j=0}^{k-1}(1+g^{6ji})^{3k}=\sum_{i=0}^{3k}\binom{3k}i\sum_{j=0}^{k-1}g^{6ji}\equiv(\binom{3k}0+\binom{3k}k+\binom{3k}{2k}+\binom{3k}{3k})k\pmod p$, where $p=6k+1$ is a prime.
Can anyone help me?
Let $r=\dfrac m{p-1}$, then $S_r=\sum_{i=1}^r\binom{r(p-1)}{i(p-1)}$. Let $g$ be an primitive root of $p$, $h=g^p$. According to Euler's Theorem, $h^{p-1}\equiv g^{p(p-1)}\equiv g^{\varphi(p^2)}\equiv1\pmod {p^2}$. If $h^k\not\equiv1\pmod p$, then $\sum_{j=1}^{p-2}h^{jk}\equiv0\pmod{p^2}$, otherwise, $\sum_{j=1}^{p-2}h^{jk}\equiv p-1\pmod{p^2}$. Note that $h^k\equiv g^k\pmod p$, hence $h^k\equiv1\pmod p\Longleftrightarrow g^k\equiv1\pmod p\Longleftrightarrow p-1|k$. We have $$ \sum_{j=0}^{p-1}h^{jk}\equiv(p-1)[p-1|k]\pmod{p^2} $$ where $[A]$ is the Iverson Bracket. According to the binomial theorem, $$ \sum_{j=0}^{p-2}(1+h^j)^{r(p-1)}=\sum_{k=0}^{r(p-1)}\binom{r(p-1)}k\sum_{j=0}^{p-2}h^{jk}\equiv(p-1)S_r\pmod{p^2} $$ According to Fermat's little Theorem, $(1+h_j)^{p-1}=1+pa_j$, where $a_j\in\mathbb{N}_+$.
By the properties of primitive roots, $h^j\equiv-1\pmod p(j=0,1,...,p-2)\Longleftrightarrow j=\dfrac{p-1}2$. Then $$ (p-1)S_r\equiv\sum_{j=0}^{p-2}(1+h_j)^{r(p-1)}\equiv\sum_{0\le j\le p-2\\j\neq\frac{p-1}2}(1+pa_j)^r\equiv p-2+prA\pmod{p^2} $$ where $A=\sum_{j=0}^{p-2}a_j$. Considering that $S_1=2$, we have $A\equiv-1\pmod p$, hence $(p-1)S_r\equiv p-2-pr\pmod{p^2}$. And then $$ S_r=-(p+1)(p-1)S_r=-(p+1)(p-2-pr)\equiv2+p+pr\equiv2+p(1+m)\pmod{p^2} $$ QED.