Exercise 2.2.15* (Generalization of Problem 2.1.44)

Question

Show that $\binom{p^\alpha - 1}{k} \equiv (-1)^k \pmod p$ for $0 \leq k \leq p^\alpha - 1$.

Richard Ganaye · Accepted Answer

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We can generalize proof 1 of proof 2 of Problem 2.1.44.

\begin{proof} 
By Problem 14, $(1+x)^{p^\alpha} = 1 + x^{p^\alpha}$ is true in $\F_p[x]$.

Therefore the polynomial equality
$$(1+x)^{p^\alpha-1} = \frac{1+x^{p^\alpha}}{1+x} = \sum_{k=0}^{p^\alpha-1} (-1)^k x^k.$$
is true if $p$ is an odd prime, and remains true if $p = 2$: in $\F_2$, the equality  $-1 = 1$ implies 
$$(1 +x)^{2^\alpha -1} = (1 - x)^{2^\alpha -1} = \frac{1-x^{2^\alpha}}{1-x} = \sum_{k=0}^{2^\alpha-1} x^k=\sum_{k=0}^{2^\alpha-1} (-1)^k x^k.$$

By the binomial formula,
$$\sum_{k=0}^{p^\alpha-1} \binom{p^\alpha-1}{k} x^k = \sum_{k=0}^{p^\alpha-1} (-1)^k x^k.$$
This gives the equality in $\F_p$ for $0 \leq k \leq p^\alpha - 1$,
$$\left[\binom{p^\alpha-1}{k} ight]_p=\left [(-1)^kight]_p,$$
thus
$$\binom{p^\alpha-1}{k} \equiv (-1)^k \pmod p, \qquad 0 \leq k \leq p^\alpha - 1.$$
\end{proof}

Exercise 2.2.15* (Generalization of Problem 2.1.44)

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Exercise 2.2.15* (Generalization of Problem 2.1.44)

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