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This is a question in Niven's An Introduction to the Theory of Numbers.

I believe a result from the previous exercise

If $p\geq 5$ and $m$ is a positive integer then $\binom{mp-1}{p-1} \equiv 1 \pmod {p^{3}}$

would help, and I had tried to apply Hensel's Lemma to it. But I failed to construct a function that can link the two questions.

Another approach that I had come up with is by induction on the positive integer $m$.

First for the base case where $m=1$, the result is trivial.

And for the induction step I assume that the proposition holds for $k=m-1$. Then for $k=m$, note that $\frac{(mp)!}{(p)![(m-1)p]!}=\binom{mp}{p}$ , and by the induction hypothesis we have $[(m-1)p]! \equiv (m-1)!p!^{m-1} \pmod{p^{m+2}}$, but I cannot deduce its remainder modulo $p^{m+3}$ from the hypothesis. So the method may comes to a dead end.

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1 Answer 1

up vote 2 down vote accepted

It is a consequence of Wolstenholme's theorem, in the form: $$ \binom{mp}{p}\equiv m\pmod{p^3}. $$ [There is a quite neat proof of this fact, using the orbit formula and the Chu-Vandermonde identity, please ask if needed.] From the last identity, we get: $$ (mp)! = (kp^3+m)\; p!\; ((m-1)p)! = K\, p^{m+3} + m\, p!\; ((m-1)p)!.$$ Now, $m\, p!\; ((m-1)p)!$ can be rewritten as: $$ m\, p!\; ((m-1)p)! = m\,(p!)^m \prod_{j=1}^{m-1}\binom{jp}{p},$$ so, by applying Wolstenholme's theorem again, we have done.

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Thank You @Jack D'Aurizio, this helps a lot! –  Ethan Wang Oct 23 '12 at 18:41
    
Inspired by your prove, I think it would be feasible using induction on $m$. –  Ethan Wang Oct 24 '12 at 0:18
    
Here is my proof: for The base case where $m=1$, is just trivial. Now for the induction step, assume that $[(m-1)p]! \equiv (m-1)!p!^{m-1} \pmod{p^{m+2}}$ for $k=m-1$. Then set $[(m-1)p]!=t_{m-1}p^{m+2}+(m-1)!p!^{m-1}$. Then for $k=m$ we have $(mp)!=[tp^{m+2}+(m-1)!p!^{m-1}](mp)(mp-1)(mp-2)\cdots (mp-p+1)$ That is, $(mp)!=t_{m}p^{m+3}+m!p!^{m-1}\cdot p(mp-1)(mp-2) \cdots (mp-p+1)$. And by the result given by a previous exercise in the book(Given in the question above), namely $\binom{mp-1}{p-1} \equiv 1\pmod{p^3}$, we have $(mp)!=m!p!^{m}$. –  Ethan Wang Oct 24 '12 at 0:35

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