Let the numbers $c_{i}$ be defined by the power series identity $$\frac{1+x+x^{2}+\ldots+x^{p-1}}{(1-x)^{p-1}}= 1+c_{1}x+c_{2}x^{2}+\ldots$$ Show that $c_{i}\equiv 0\pmod{p}$ for all $i\geq 1$.

$\textbf{Solution.}$ First, from the Binomial Theorem we have $$(1-x)^{-n} = 1 + nx + \frac{n(n+1)}{2!} x^{2} + \frac{n(n+1)(n+2)}{3!}x^{3} + \cdots$$ \begin{align*} (1-x)^{-(p-1)} &= 1 + (p-1)x + \frac{(p-1)p}{2!} x^{2} + \frac{(p-1)p(p+1)}{3!}x^{3} + \cdots \\ &= 1 + \binom{p-1}{1} x + \binom{p}{2}x^{2} + \binom{p+1}{3}x^{3} + \binom{p+2}{4}x^{4} + \cdots \\ &= 1+\binom{p-1}{1}x + \binom{p}{2}x^{2} + \left[\binom{p}{2}+\binom{p}{3}\right]x^{3}+\left[\binom{p+1}{3}+\binom{p+1}{4}\right]x^{4} +...\\ &=1+\binom{p-1}{1}x + \binom{p}{2}x^{2} + \left[\binom{p}{2}+\binom{p}{3}\right]x^{3}+\left[\binom{p}{2}+2\binom{p}{3}+\binom{p}{4}\right]x^{4} +... \end{align*} where $\binom{p+1}{3}x^{3}=\Bigl\{\binom{p}{2}+\binom{p}{3}\Bigr\}x^{3}$ and $\binom{p}{4}x^{4}=\Bigl\{\binom{p}{2}+2\binom{p}{3}+\binom{p}{4}\Bigr\}x^{4}$ follows from using $\textit{Pascal's Identity.}$ Now we have $$\frac{1+x+x^{2}+\ldots+x^{p-1}}{(1-x)^{p-1}}=(1+x+x^{2}+\ldots)\cdot (1-x)^{-(p-1)} =1+c_{1}x+c_{2}x^{2}+\ldots$$ Using the binomial expansion of $(1-x)^{-(p-1)}$ and comparing the coefficients, we get \begin{align*} c_{1} &= 1 + \binom{p-1}{1} =p\\ c_{2} &= \binom{p}{2} + \binom{p-1}{1} + 1 = \binom{p}{2} + c_{1}\\ c_{3} &= \binom{p+1}{3} + \binom{p}{2} + \binom{p-1}{1} + 1 =\binom{p}{2}+\binom{p}{3}+c_{2}\\ c_{4} &=\binom{p+2}{4}+c_{3} = \binom{p}{2}+2\binom{p}{3}+\binom{p}{4}+c_{3} \end{align*} In general $c_{i}=\binom{p+i-2}{i}+c_{i-1}$. Now by we know that $\binom{p}{i}\equiv 0\pmod{p}$ whenever $0<i<p$.

  • So I am able to observe that $c_{1},c_{2},c_{3},c_{4}$ are $0\pmod{p}$. But still my argument is not correct. Since $c_{i}\neq\binom{p+i-2}{p}+c_{i-1}$ if $i\geq p$. A nice solution would be helpful.
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    $\begingroup$ Taking $x\ne1$ $$\frac{1+x+x^{2}+\ldots+x^{p-1}}{(1-x)^{p-1}}=\dfrac{1-x^p}{(1-x)^p}$$ Fermat's Little Theorem says $$x^p\equiv x,(1-x)^p\equiv1-x\pmod p$$ $\endgroup$ – lab bhattacharjee Jun 29 '16 at 15:32
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    $\begingroup$ @labbhattacharjee You can't say $x^p\equiv x\pmod{p}$. Note that $x$ is a variable (i.e., a transcendental element with respect to $\mathbb{F}_p$). $\endgroup$ – Batominovski Jun 29 '16 at 15:43
  • $\begingroup$ @SC, Expand $$(1-x)^{-p}$$ Observe that each power$(\ge1)$ of $x$ has coefficient divisible by prime $p$ $\endgroup$ – lab bhattacharjee Jun 29 '16 at 15:48

In $R:=\mathbb{F}_p[\![x]\!]$, $(1-x)^p=1-x^p$. Because $1-x$ is an invertible element of $R$ (with inverse $1+x+x^2+\ldots$), $$\frac{1-x^p}{(1-x)^p}=1$$ in $R$. Therefore, [...]. The rest is up to you.

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    $\begingroup$ Yes it's true, but it's implicitly what the OP want's to prove. So, I don't think that it's really a solution of the problem. $\endgroup$ – Surb Jun 29 '16 at 15:44
  • $\begingroup$ @Surb I don't know what you are talking about. This is almost a complete solution to the OP's problem. I only left two easy steps to him to verify. $\endgroup$ – Batominovski Jun 29 '16 at 15:46
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    $\begingroup$ It's of course a complete answer if he know's this formula. But I think it's what the OP implicitly has to prove (that $(1+x)^p=1+x^p$ in a field of characteristic $p$), therefore, I don't think that he can use this. $\endgroup$ – Surb Jun 29 '16 at 15:50
  • $\begingroup$ @Surb That is one of the two missing steps I intentionally left out. The other step is to finish the proof. $\endgroup$ – Batominovski Jun 29 '16 at 15:51

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