Find $v_p\left(\binom{ap}{bp}-\binom{a}{b}\right)$, where $p>a>b>1$ and $p$ odd prime.

Question

Find $v_p\left(\binom{ap}{bp}-\binom{a}{b}\right)$, where $p>a>b>1$ and $p$ odd prime.

Here $v_p(k)$ denotes the largest $\alpha\in\mathbb Z_{\ge 0}$ s.t. $p^\alpha\mid k$.

We have $p\nmid\binom{ap}{bp}$ and $p\nmid \binom{a}{b}$. $$\binom{ap}{bp}-\binom{a}{b}=\frac{(ap)!b!(a-b)!-a!(bp)!(ap-bp)!}{(bp)!(ap-bp)!b!(a-b)!}$$

Answer 1

Let $k=v_2\Big (\tbinom {ap}{bp}-\tbinom{a}{b}\Big)$ :

If we use the well known Lucas's theorem we can easily conclude that $k\geq 1$. In fact, your question was studied by a lot of mathematicians, it's called Ljunggren’s and Jacobsthal’s binomial congruence and some results are:

• $k$ is greater then the power of $p$ dividing $p^3ab(a-b)$ with equality if $p$ does not divide the $p-3$th Bernoulli number $B_{p-3}$ (due to Jacobsthal)
• In $2008$ Helou and Terjanian proved that $k\geq v_2\Big(ab(a-b)\tbinom{a}{b}\Big)$

there are other refinement of this results, a good reference wold be Wolstenholme's theorem