Numbers with Square-Free Binomial Coefficients

Theorem
For every $n$ greater than $23$, there exists a binomial coefficient $\dbinom n k$ that is not square-free.

More specifically, the list of numbers $n$ such that $\dbinom n k$ are squarefree for all $k = 0, \dots, n$ is given by:


 * $1, 2, 3, 5, 7, 11, 23$

Case 1: $n \ge 25$
Consider the case $n \ge 25$.

for all $0 \le m \le n$, $\dbinom n m$ is square-free.

Then for any prime $p$:
 * $p^2 \nmid \dbinom n m$

for all $0 \le m \le n$.

In particular we have:
 * $2^2 \nmid \dbinom n m$

and:
 * $5^2 \nmid \dbinom n m$

There exists an integer $r$ where $2^r \le n < 2^{r + 1}$.

There also exists an integer $s$ where $5^s \le n < 5^{s + 1}$.

We have $r \ge 4$ and $s \ge 2$ from $n \ge 25 = 5^2 \ge 16 = 2^4$.

By the Lemma, we have:
 * $2^{r - 1} \divides \paren {n + 1}$

and:
 * $5^{s - 1} \divides \paren {n + 1}$

Because $2^{r - 1}$ and $5^{s - 1}$ are coprime, we have by Product of Coprime Factors:
 * $2^{r - 1} 5^{s - 1} \divides \paren {n + 1}$

Thus:

which is a contradiction.

Therefore $\dbinom n m$ is not square-free for any $0 \le m \le n$.

Case 2: $9 \le n \le 24$
Consider the case $9 \le n \le 24$.

Suppose for all $0 \le m \le n$, $\dbinom n m$ is square-free.

Then for any prime $p$:
 * $p^2 \nmid \dbinom n m$

for all $0 \le m \le n$.

In particular we have:
 * $2^2 \nmid \dbinom n m$

and:
 * $3^2 \nmid \dbinom n m$

There exists an integer $r$ where $2^r \le n < 2^{r + 1}$.

There also exists an integer $s$ where $3^s \le n < 3^{s + 1}$.

From:
 * $n \ge 9 = 3^2 \ge 8 = 2^3$

we have $r \ge 3$ and $s \ge 2$

By the Lemma, we have:
 * $2^{r - 1} \divides \paren {n + 1}$

and:
 * $3^{s - 1} \divides \paren {n + 1}$

So we have:
 * $2^2 \divides \paren {n + 1}$

and:
 * $3^1 \divides \paren {n + 1}$

Since $4$ and $3$ are coprime, we have by Product of Coprime Factors:
 * $12 \divides \paren {n + 1}$

In the range $9 \le n \le 24$, only $11$ and $23$ satisfy this condition.

Case 3: $4 \le n \le 8$
Consider the case $4 \le n \le 8$.

Suppose for all $0 \le m \le n$, $\dbinom n m$ is square-free.

Then for any prime $p$:
 * $p^2 \nmid \dbinom n m$

for all $0 \le m \le n$.

In particular we have:
 * $2^2 \nmid \dbinom n m$

There exists an integer $r$ where $2^r \le n < 2^{r + 1}$.

From $n \ge 4 = 2^2$:
 * $r \ge 2$

By the Lemma, we have:
 * $2^{r - 1} \divides \paren {n + 1}$

So we have:
 * $2^1 \divides \paren {n + 1}$

In the range $4 \le n \le 8$, only $5$ and $7$ satisfy this condition.

Manual Checking
We have narrowed our list of possible candidates to:
 * $1, 2, 3, 5, 7, 11, 23$

Manual checking each $n$ in the list shows that $\dbinom n m$ is square-free for each $0 \le m \le n$.