Show that for $m\in \mathbb{Z}^{+}$ and $k\in \lbrace 1,2,\cdots , p^{m-1} \rbrace$ then $p\mid {p^m \choose k}$

Asked Sep 17 '21 at 18:07

Active Sep 17 '21 at 21:52

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Show that for $m\in \mathbb{Z}^{+}$ and $k\in \lbrace 1,2,\cdots , p^{m-1} \rbrace$ then $p\mid{p^m \choose k}$ where $p$ is a prime number.

I try induction over $m$, the case $m=1$ implies $p\mid{p\choose 1}$ which is true. Now I suppose that the proposition is true for $m=l-1$ it is I can use $p\mid{p^{l-1} \choose k}$ and I must show that $p\mid{p^l \choose k}$.

Unfortunelly I expand both binomial expressions and I´don´t know how apply a trick to use my induction hypothesis, also I try use a property of binomial coefficient to reduce the power of $p$, but I´cant find a useful property.

I think that make induction don´t is good idea, but I don´t know other approach of this fact. Any hint was useful Thanks in advice.

edited Sep 17 '21 at 18:36

Michael Hardy

asked Sep 17 '21 at 18:07

Alan Jr

1

Cf. Lucas's theorem. – John Omielan Sep 17 '21 at 18:28
I´m gonna try use the theorem – Alan Jr Sep 17 '21 at 18:53
yes it answer my question – Alan Jr Sep 17 '21 at 19:54
2

Note that all the linked answers are proving the stronger statement for all $k \in {1, 2, \ldots, p^m-1}$. You might think this makes it harder, but on the contrary, this is why Daniel Schepler's inductive proof here works so easily. The statement becomes stronger, but it also strengthens the inductive hypothesis, making the inductive step much easier. – Brian Moehring Sep 17 '21 at 20:21

Show that for $m\in \mathbb{Z}^{+}$ and $k\in \lbrace 1,2,\cdots , p^{m-1} \rbrace$ then $p\mid {p^m \choose k}$

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