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I was proving that $p(x) = 1 + x + \text{ ... } + x^{p-1}$, where $p$ is a prime number, is irreducible over the rationals, by using the translation $1 + (1 + x) + \text{ ... } + (1 + x)^{p-1}$. I managed to do it, but I would like to get a closed expression for this as a polynomial.

So far, I have: \begin{equation} \sum_{k = 0}^{p-1}{(1+x)^k} = \sum_{k = 0}^{p-1}{(\sum_{i = 0}^k{\binom{k}{i}x^i)}} \end{equation}

By inspection, it looks like the final result will be:

\begin{equation} \sum_{k = 0}^{p-1}{(1+x)^k} = \sum_{k = 0}^{p-1}{(\sum_{i=k}^{p-1}{\binom{i}{k}}} )x^k \end{equation}

I.E.

\begin{equation} \sum_{i=0}^{p-1}{\binom{i}{0}} + \sum_{i=1}^{p-1}{\binom{i}{1}} x + \text{ ... } + \sum_{i=p-1}^{p-1}{\binom{i}{p-1}}x^{p-1} \end{equation}

First of all, is this correct? If not, what would be the correct expression and why? And if it is correct, how can I get from the first equation to the second?

Thanks in advance!

Gauss
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2 Answers2

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Assuming that I have not misinterpreted your query:

A case can be made that the following hint-answer is defective, because it avoids examining your work. Through no fault of your own, you went down a natural but bad path.

I actually don't know how to prove that any polynomial is irreducible, but I can give you a hint about how to very easily find a closed form expression for the requested polynomial.

Hint-1:
Given the geometric series,
$f(t) = 1 + t + t^2 + \cdots + t^n$, what is the result of
$f(t) \times (1-t)$?

Hint-2:
What happens if you set $t = (x+1)$?

user2661923
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If $p\in\Bbb P$ then for $0\le k\le p-1$ the $x^k$ coefficient in $\sum_{j=0}^{p-1}(1+x)^j$ is $\sum_{j=k}^{p-1}\binom{j}{k}=\binom{p}{k+1}$ by the hockey-stick identity. This is presumably how you proved irreducibility by Eisenstein's criterion, since the non-leading coefficients will be multiples of $p$. This implies the closed form$$\sum_{k=0}^{p-1}\binom{p}{k+1}x^k=\frac1x\sum_{\ell=1}^p\binom{p}{\ell}x^\ell=\frac{(1+x)^p-1}{x},$$but that's more easily obtained by @user2661923's strategy. Both approaches make obvious that this closed form is true even for composite $p$.

J.G.
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