I've been grappling with this problem for a while but haven't solved it.
Given integers $a \ge b > 0$ and a prime number $p$, prove that ${pa \choose pb} \equiv {a \choose b} \mod p$.
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NP. I think this can be solved just using the definition – Squirtle Jan 27 '14 at 05:20
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1You may be interested in the Proof section in the Wikipedia article on Lucas' theorem. – anon Jan 27 '14 at 05:34
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In fact $\displaystyle\binom{pa}{pb}\equiv\binom{a}{b}\pmod{p^3}$ – Grobber Jan 27 '14 at 06:17
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@Grobber $\binom{2\cdot 3}{2 \cdot 1} = \binom{6}{2} = 15$ and $\binom{3}{1} = 3$, but $15 \not\equiv 3 \bmod 8$. – J.-E. Pin Jan 27 '14 at 08:15
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J.E. Pin, I am afraid that I had forgotten to mention the above holds for $p\ge 5$. – Grobber Jan 27 '14 at 08:58
2 Answers
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The following is a combinatorial argument.
Draw $a$ concentric circles and divide them radially into $p$ parts each. There are then a total of $pa$ regions. There are $\binom{pa}{pb}$ ways to select $pb$ of these regions. Consider the action of rotation by $2\pi/p$ on these selections. There are $\binom{a}{b}$ selections which are fixed by the rotation: these are the selections that consist of $b$ complete annuli. All others fall into orbits of size $p$. The desired conclusion $$\binom{pa}{pb} \equiv \binom{a}{b}\, (\text{mod}\,p)$$ follows.
Ted
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8Nice proof! You can easily extend this to $p^2$ by rotating each block individually and considering the equivalence classes formed, which have order $p^n$ where $n$ is the number of partially filled annuli. At least two annuli are partially filled for every valid coloring, so $n\ge 2$ and $p^2$ divides the expression desired. – tc1729 Jul 17 '14 at 00:06
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$$(1+x)^{pa}= \sum_{n=0}^{pa} {pa \choose n} x^{n}$$
\begin{align} (1+x)^{pa}=\left ((1+x)^{p} \right )^a=\left (\sum_{k=0}^{p}{p \choose k}x^{k} \right )^{a} \Rightarrow\\ \end{align}
\begin{align} (1+x)^{pa}&\equiv\left (\sum_{k=0}^{p}{p \choose k}x^{k} \right )^{a} \mod p\\ &\equiv(1+x^p)^{a} \mod p\\ &\equiv \sum_{i=0}^{a} {a \choose i}x^{pi} \mod p \end{align}
Equating coefficients $\mod p$ it yields $$ { pa \choose pb} \equiv {a\choose b} \mod p$$
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2While reading your proof I think I came up with a variant...or maybe this is the same as your proof: Suppose I use the identity $(a+b+c+...)^p \equiv (a^p+b^p+c^p+...) \mod p$ to show that $((1+x)^a)^p = (\sum {a \choose b}x^b)^p \equiv \sum {a \choose b}^px^{pb}$. Then since $(1+x)^{pa} = \sum {pa \choose k}x^k$ by comparing coefficients we get ${a \choose b}^p \equiv {pa \choose pb} \mod p$, so by Fermat's Little Theorem ${a \choose b} \equiv {pa \choose pb} \mod p$ since $p$ is prime. Is this the same as your proof? (Sorry if this is a stupid question) – user1299784 Jan 27 '14 at 06:16
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@Ted Could you please explain in this proof why the coefficients can be equated? I believe that you are correct but do not see why. Relevant question here: http://math.stackexchange.com/questions/1538321/equating-coefficients-of-binomial-expansion-modulo-p – yeons Nov 25 '15 at 15:57