Non-Divisbility of Binomial Coefficients of n by Prime

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Theorem

Let $n \in \Z_{\ge 0}$ be a positive integer.

Let $k \in \Z_{\ge 0}$ such that $0 \le k \le n$.

Let $p$ be a prime number.


Then:

$\dbinom n k$ is not divisible by $p$

if and only if:

$n = a p^m - 1$ where $1 \le a < p$

for some $m \in \Z_{\ge 0}$.


Proof

The statement:

$\dbinom n k$ is not divisible by $p$

is equivalent to:

$\dbinom n k \not \equiv 0 \pmod p$


The corollary to Lucas' Theorem gives:

$\displaystyle \dbinom n k \equiv \prod_{j \mathop = 0}^r \dbinom {a_j} {b_j} \pmod p$

where:

$n, k \in \Z_{\ge 0}$ and $p$ is prime
the representations of $n$ and $k$ to the base $p$ are given by:
$n = a_r p^r + \cdots + a_1 p + a_0$
$k = b_r p^r + \cdots + b_1 p + b_0$


Consider the form of $n = a p^m - 1$ when represented to the base $p$:

\(\displaystyle n\) \(=\) \(\displaystyle a p^m - 1\)
\(\displaystyle \) \(=\) \(\displaystyle a \left({p^m - 1}\right) + \left({a - 1}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle a \left({p - 1}\right) \sum_{j \mathop = 0}^{m - 1} p^i + \left({a - 1}\right)\) Sum of Geometric Progression: Corollary 1
\(\displaystyle \) \(=\) \(\displaystyle \left({a - 1}\right) p^m + \sum_{j \mathop = 0}^{m - 1} \left({p - 1}\right) p^i\) after algebra

That is, all of the digits of $n$ are $p - 1$ except perhaps the first one.

For $k$ such that $0 \le k \le n$, the digits of $k$ will all range over $0$ to $p - 1$.


Necessary Condition

Let $n = a p^m - 1$.

Then all the binomial coefficients $\dbinom {a_j} {b_j}$ (except perhaps the first) are of the form $\dbinom {p - 1} k$ for $0 \le k < p$.


By Binomial Coefficient of Prime Minus One Modulo Prime:

$\dbinom {p - 1} k \equiv \left({-1}\right)^k \pmod p$

and so:

$\dbinom {p - 1} k \equiv \left({-1}\right)^k \pmod p$



and so:

$\displaystyle \prod_{j \mathop = 0}^r \dbinom {a_j} {b_j} \not \equiv 0 \pmod p$


Sufficient Condition

Let $\dbinom n k$ not be divisible by $p$.

Aiming for a contradiction, suppose $n \ne a p^m - 1$.

Then one of the digits $a_j$ will be less than $p - 1$.

But the digits of all possible $k$ will range over $0$ to $p - 1$.

Therefore there must be at least one $\dbinom {a_j} {b_j}$ such that $b_j > a_j$.

Such a condition leads to $\dbinom {a_j} {b_j} = 0$.

That is:

$\displaystyle \prod_{j \mathop = 0}^r \dbinom {a_j} {b_j} \equiv 0 \pmod p$

which means $\dbinom n k$ not be divisible by $p$.

From this contradiction it follows that $n = a p^m - 1$ for some $a \in \Z, 0 < a < p$ and $m \in \Z_{>0}$.

$\blacksquare$


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