To find all odd integers $n>1$ such that $2n \choose r$ , where $1 \le r \le n$ , is odd only for $r=2$

For which odd integers $n>1$ is it true that $2n \choose r$ where $1 \le r \le n$ is odd only for $r=2$ ? I know that $2n \choose 2$ is odd if $n$ is odd but I want to find those odd $n$ for which the only value of $r$ between $1$ and $n$ such that $2n \choose r$ is odd , is $2$ . I am wanting to know this as for such an $n$ , if $a^{2n}=a , \forall a \in R $ where $R$ is a ring then $R$ is commutative .Indeed $-a=(-a)^{2n}=a^{2n}=a$ so $2a=0 , \forall a \in R$ ; then since $a^2,a$ commute , so in the expansion of $(a^2+a)^{2n}$ , since $2n \choose r$ is even for all $r$ between $3$ and $2n-3$ , and also ${2n \choose 1 }={2n \choose 2n-1}=2n$ is even , these terms vanish due to $2a=0 , \forall a \in R$ , so

$(a^2+a)^{2n}=(a^2)^{2n}+ {2n \choose 2}(a^2)^{2n-2}.a^2+{2n \choose 2n-2}(a^2)^2.a^{2n-2}+a^{2n}=a^2+ a^{4n-2}+a^{2n+2}+a=a^2+a^{2n-1}+a^3+a$

so $0=(a^2+a)^{2n}-(a^2+a)=a^{2n-1}+a^3=a^{2n-1}-a^3$ that is $a^{2n-1}=a^3 , \forall a \in R$ , so

$a^4=a.a^3=a.a^{2n-1}=a^{2n}=a , \forall a \in R$ , thus R is commutative ; so please help to find all odd

integers $n>1$ such that $2n \choose r$ , where $1 \le r \le n$ , is odd only for $r=2$ . Thanks in advance

Solutions Collecting From Web of "To find all odd integers $n>1$ such that $2n \choose r$ , where $1 \le r \le n$ , is odd only for $r=2$"

Let $2 n = n_{0} + n_{1} 2 + \dots + n_{k} 2^{k}$ be the base $2$ expansion of $2 n$, so all $n_{i} \in \{0, 1\}$. We have $n_{0} = 0$, $n_{1} = 1$, and may assume $n_{k} = 1$, that is, $2^{k-1} \le n < 2^{k}$.

Similarly, let $1 \le r = r_{0} + r_{1} 2 + \dots + r_{k-1} 2^{k-1} \le n$, with $n_{i} \in \{0, 1\}$.

Then by Lucas’ Theorem
$$
\binom{2n}{r}
\equiv
\prod_{i=0}^{k} \binom{n_{i}}{r_{i}}
\equiv
\prod_{i=0}^{k-1} \binom{n_{i}}{r_{i}}\pmod{2}.
$$
So the left hand side is even when there is an $i$ such that $n_{i} = 0$ and $r_{i} = 1$. You want this to happen for all $2 < r \le n$.

If there is $1 < i < k$ such that $n_{i} = 1$, then $r = 2^{i} \le 2^{k-1} \le n$, and
$$
\binom{2n}{r} \equiv 1 \pmod{2}.
$$

So the required condition is that for all $1 < i < k$ we have $n_{i} = 0$, that is, $2 n = 2 + 2^{k}$, that is, $n = 1 + 2^{k-1}$ for some $k > 1$.