February, 2017

Solution:

We use the following theorem of Lucas:
Theorem. Given a prime $p$ and nonnegative integers $a, b$ written in base $p$ as $a=\{\overline{a_{n}} a_{n-1} \ldots a_{0}\}_{p}$ and $b=\overline{b_{n} b_{n-1} \ldots b_{0}}$ respectively, where $0 \leq a_{i}, b_{i} \leq p-1$ for $0 \leq i \leq n$, we have
$$\binom{a}{b}=\prod_{i=0}^{n}\binom{a_{i}}{b_{i}} \quad(\bmod p)$$
Now, let $n=(p-1) \times p+(p-2)=p^{2}-2$. For $k=pq+r$ with $0 \leq q, r \leq p-1$, applying Lucas's theorem gives
$$\binom{n}{k} \equiv \binom{p-1}{q}\binom{p-2}{r} \quad(\bmod p)$$
Note that
$$\binom{p-1}{q}=\prod_{i=1}^{q} \frac{p-i}{i} \equiv (-1)^{q} \quad(\bmod p)$$
and
$$\binom{p-2}{r}=\prod_{i=1}^{r} \frac{p-1-i}{i} \equiv (-1)^{r} \frac{(r+1)!}{r!}=(-1)^{r}(r+1) \quad(\bmod p)$$
So for $2 \leq i \leq p$ we can take $k=(p+1)(i-1)$ and obtain $\binom{n}{k} \equiv i\ (\bmod p)$, while for $i=1$ we can take $k=0$. Thus this row satisfies the desired property.