$\binom{2p-1}{p-1}\equiv 1\pmod{\! p^2}$ implies $\binom{ap}{bp}\equiv\binom{a}{b}\pmod{\! p^2}$; where $p>3$ is a prime?

Question

From $\binom{2p-1}{p-1}\equiv 1\pmod{\! p^2}$ how does one get $\binom{ap}{bp}\equiv\binom{a}{b}\pmod{\! p^2},\,\forall a,b \in \mathbb N,\, a>b$; where $p>3$ is a prime ?

Answer 1

We can begin with Wilson's Theorem $(p-1)!=-1\mod p$. This allows $\dfrac{(kp-1)!}{(k-1)p!}=-1\mod p$, for example with $p=5$ and $k=2$, $9.8.7.6=3024\equiv-1\mod5$, because we are looking at $(kp-1)(kp-2)\dots ((k-1)p+1)$ in which the terms without a $p$ coefficient relate to Wilson's Theorem.

So we have $\binom{2p-1}{p-1}\equiv1\mod p^2$, using Wolstenholme's Theorem.

Similarly, $\binom{ap-1}{bp-1}\equiv 1\mod p^2$, and by dividing through by appropriate $p$'s we get to the result.

Answer 2

We have two congruences $$ (1)\hspace{1cm}{2p-1\choose p-1}\equiv 1\!\!\!\mod p^2,\hspace{10mm}\text{and}\hspace{10mm} {ap\choose bp}\equiv {a\choose b}\!\!\! \mod p^2,\,\, a,b,\in\mathbb{N}.\hspace{1.1cm}(2) $$ We would like to obtain $(2)$ from $(1)$. Both congruences actually hold modulo $p^3$.

For $k$ an integer, define $ c_k=(kp-1)(kp-2)\ldots(kp-p+1) $. We can write our binomial coefficients in terms of the $c_k$: $$ {2p-1\choose p-1}=\frac{c_2}{c_1},\;\;\;\;\;{ap\choose bp}={a\choose b}\frac{c_a c_{a-1}\ldots c_{a-b+1}}{c_bc_{b-1}\ldots c_1}. $$

Let $H_n:=1+\frac{1}{2}+\ldots+\frac{1}{n}$ be the $n$-th harmonic number. We can compute \begin{align*} \frac{c_k}{c_1}&=\left(1-\frac{kp}{1}\right)\left(1-\frac{kp}{2}\right)\ldots\left(1-\frac{kp}{p-1}\right)\\ &\equiv 1-k\,p H_{p-1}\mod p^2,\tag{3} \end{align*} where for $r$, $s$ rational numbers we write $r\equiv s\!\!\mod p^n$ to mean $p^n$ divides the numerator of $r-s$. Thus $(1)$ is equivalent to the statement $H_{p-1}\equiv 0\mod p$. Now $(3)$ and $H_{p-1}\equiv 0\!\!\!\mod p$ imply $c_k/c_1\equiv 1\!\!\!\mod p^2$ for all $k$ (and so $c_i/c_j\equiv 1\!\!\!\mod p^2$ for all $i$, $j$), so we get $(2)$.

---

If instead we are interested in $(1)$ and $(2)$ mod $p^3$, we can strengthen $(3)$: $$ \frac{c_k}{c_1}\equiv1-k\,pH_{p-1}+k^2\,p^2H_{p-1}(1,1)\mod p^3, $$ where $H_{p-1}(1,1)$ is the value of the second elementary symmetric polynomial evaluated at $1,\frac{1}{2},\ldots,\frac{1}{p-1}$. The final term vanishes (it can be checked that $H_{p-1}(1,1)\equiv 0\!\!\mod p$), so the $p^3$ strengthening of $(1)$ is equivalent to $H_{p-1}\equiv 0\!\!\!\mod p^2$, which implies the strengthening of $(2)$.