Subfield: Number Theory


Question: For every integer $k \geq 2$, prove that $2^{3 k}$ divides the number
$$
\left(\begin{array}{c}
2^{k+1} \\
2^{k}
\end{array}\right)-\left(\begin{array}{c}
2^{k} \\
2^{k-1}
\end{array}\right) 
\tag{1}
$$
but $2^{3 k+1}$ does not.

Solution:
We use the notation $(2 n-1) ! !=1 \cdot 3 \cdots(2 n-1)$ and $(2 n) ! !=2 \cdot 4 \cdots(2 n)=2^{n} n !$ for any positive integer $n$. Observe that $(2 n) !=(2 n) ! !(2 n-1) ! !=2^{n} n !(2 n-1) ! !$.

For any positive integer $n$ we have
$$
\begin{aligned}
& \left(\begin{array}{c}
4 n \\
2 n
\end{array}\right)=\frac{(4 n) !}{(2 n) !^{2}}=\frac{2^{2 n}(2 n) !(4 n-1) ! !}{(2 n) !^{2}}=\frac{2^{2 n}}{(2 n) !}(4 n-1) ! ! \\
& \left(\begin{array}{c}
2 n \\
n
\end{array}\right)=\frac{1}{(2 n) !}\left(\frac{(2 n) !}{n !}\right)^{2}=\frac{1}{(2 n) !}\left(2^{n}(2 n-1) ! !\right)^{2}=\frac{2^{2 n}}{(2 n) !}(2 n-1) ! !^{2} .
\end{aligned}
$$
Then expression (1) can be rewritten as follows:
$$
\begin{aligned}
\left(\begin{array}{c}
2^{k+1} \\
2^{k}
\end{array}\right) & -\left(\begin{array}{c}
2^{k} \\
2^{k-1}
\end{array}\right)=\frac{2^{2^{k}}}{\left(2^{k}\right) !}\left(2^{k+1}-1\right) ! !-\frac{2^{2^{k}}}{\left(2^{k}\right) !}\left(2^{k}-1\right) ! !^{2} \\
& =\frac{2^{2^{k}}\left(2^{k}-1\right) ! !}{\left(2^{k}\right) !} \cdot\left(\left(2^{k}+1\right)\left(2^{k}+3\right) \ldots\left(2^{k}+2^{k}-1\right)-\left(2^{k}-1\right)\left(2^{k}-3\right) \ldots\left(2^{k}-2^{k}+1\right)\right) .
\end{aligned} \tag{2}
$$
We compute the exponent of 2 in the prime decomposition of each factor (the first one is a rational number but not necessarily an integer; it is not important).

First, we show by induction on $n$ that the exponent of 2 in $\left(2^{n}\right)$ ! is $2^{n}-1$. The base case $n=1$ is trivial. Suppose that $\left(2^{n}\right) !=2^{2^{n}-1}(2 d+1)$ for some integer $d$. Then we have
$$
\left(2^{n+1}\right) !=2^{2^{n}}\left(2^{n}\right) !\left(2^{n+1}-1\right) ! !=2^{2^{n}} 2^{2^{n}-1} \cdot(2 d+1)\left(2^{n+1}-1\right) ! !=2^{2^{n+1}-1} \cdot(2 q+1)
$$
for some integer $q$. This finishes the induction step.

Hence, the exponent of 2 in the first factor in $(2)$ is $2^{k}-\left(2^{k}-1\right)=1$.

The second factor in (2) can be considered as the value of the polynomial
$$
P(x)=(x+1)(x+3) \ldots\left(x+2^{k}-1\right)-(x-1)(x-3) \ldots\left(x-2^{k}+1\right) .\tag{3}
$$
at $x=2^{k}$. Now we collect some information about $P(x)$.

Observe that $P(-x)=-P(x)$, since $k \geq 2$. So $P(x)$ is an odd function, and it has nonzero coefficients only at odd powers of $x$. Hence $P(x)=x^{3} Q(x)+c x$, where $Q(x)$ is a polynomial with integer coefficients.

Compute the exponent of 2 in $c$. We have
$$
\begin{aligned}
c & =2\left(2^{k}-1\right) ! ! \sum_{i=1}^{2^{k-1}} \frac{1}{2 i-1}=\left(2^{k}-1\right) ! ! \sum_{i=1}^{2^{k-1}}\left(\frac{1}{2 i-1}+\frac{1}{2^{k}-2 i+1}\right) \\
& =\left(2^{k}-1\right) ! ! \sum_{i=1}^{2^{k-1}} \frac{2^{k}}{(2 i-1)\left(2^{k}-2 i+1\right)}=2^{k} \sum_{i=1}^{2^{k-1}} \frac{\left(2^{k}-1\right) ! !}{(2 i-1)\left(2^{k}-2 i+1\right)}=2^{k} S .
\end{aligned}
$$

For any integer $i=1, \ldots, 2^{k-1}$, denote by $a_{2 i-1}$ the residue inverse to $2 i-1$ modulo $2^{k}$. Clearly, when $2 i-1$ runs through all odd residues, so does $a_{2 i-1}$, hence
$$
\begin{gathered}
S=\sum_{i=1}^{2^{k-1}} \frac{\left(2^{k}-1\right) ! !}{(2 i-1)\left(2^{k}-2 i+1\right)} \equiv-\sum_{i=1}^{2^{k-1}} \frac{\left(2^{k}-1\right) ! !}{(2 i-1)^{2}} \equiv-\sum_{i=1}^{2^{k-1}}\left(2^{k}-1\right) ! ! a_{2 i-1}^{2} \\
=-\left(2^{k}-1\right) ! ! \sum_{i=1}^{2^{k-1}}(2 i-1)^{2}=-\left(2^{k}-1\right) ! ! \frac{2^{k-1}\left(2^{2 k}-1\right)}{3} \quad\left(\bmod 2^{k}\right) .
\end{gathered}
$$
Therefore, the exponent of 2 in $S$ is $k-1$, so $c=2^{k} S=2^{2 k-1}(2 t+1)$ for some integer $t$.

Finally we obtain that
$$
P\left(2^{k}\right)=2^{3 k} Q\left(2^{k}\right)+2^{k} c=2^{3 k} Q\left(2^{k}\right)+2^{3 k-1}(2 t+1),
$$
which is divisible exactly by $2^{3 k-1}$. Thus, the exponent of 2 in $(2)$ is $1+(3 k-1)=3 k$.