$\binom{mp^n}{p^n}\cong m$ (mod $p$)

Question

I proved that $\binom{mp^n}{p^n}\cong m$ (mod $p$) for every prime number $p$ and positive integer $m$. But in my book $m$ is assumed to be coprime to $p$, while my proof did not make use of that assumption. Is the identity without that assumption really correct?

Proof sketch: We first expand $(1+x)^{mp^n}$ in the field $\mathbb F_p$ using the binomial theorem. Then we expand it in another way: since $(a+b)^p=a^p+b^p$ in $\mathbb F_p$, we have $(1+x)^{mp^n}=(1+x^{p^n})^m$. Now equating the coefficients of $x^{p^n}$, we get the result.

What is your proof? If we had that, we could tell you either that you're correct, or pinpoint for you where you've made a mistake. — Arthur, Jan 29 '23 at 15:50
See this post for a general proof (Lucas' theorem). And here, why coprime is not needed. — Dietrich Burde, Jan 29 '23 at 15:56
Does this answer your question? A problem regarding the proof of ${p^nk\choose p^n}\equiv k\mod p$, where $p\nmid k$. — Dietrich Burde, Jan 29 '23 at 15:58

Robert Israel · Accepted Answer · 2023-01-29T21:39:19.267

1

Yes, it is true. It follows from Lucas's theorem $$ {a \choose b} \equiv \prod_k {a_k \choose b_k} \mod p $$ where $a = \sum_{k} a_k p^k$ and $ b = \sum_k b_k p^k$ are the base-$p$ representations of $a$ and $b$.

edited Jan 29 '23 at 21:39

answered Jan 29 '23 at 15:55

Robert Israel

448,999

Right, exactly the above link. – Dietrich Burde Jan 29 '23 at 15:56
Thanks. I didn’t know the Lucas theorem – Jan 29 '23 at 16:02

$\binom{mp^n}{p^n}\cong m$ (mod $p$)

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