Subfield: Number Theory


Question: Prove that for any pair of positive integers $k$ and $n$ there exist $k$ positive integers $m_{1}, m_{2}, \ldots, m_{k}$ such that

$$
1+\frac{2^{k}-1}{n}=\left(1+\frac{1}{m_{1}}\right)\left(1+\frac{1}{m_{2}}\right) \cdots\left(1+\frac{1}{m_{k}}\right)
$$

Solution:
We proceed by induction on $k$. For $k=1$ the statement is trivial. Assuming we have proved it for $k=j-1$, we now prove it for $k=j$.

Case 1. $n=2 t-1$ for some positive integer $t$.

Observe that

$$
1+\frac{2^{j}-1}{2 t-1}=\frac{2\left(t+2^{j-1}-1\right)}{2 t} \cdot \frac{2 t}{2 t-1}=\left(1+\frac{2^{j-1}-1}{t}\right)\left(1+\frac{1}{2 t-1}\right) .
$$

By the induction hypothesis we can find $m_{1}, \ldots, m_{j-1}$ such that

$$
1+\frac{2^{j-1}-1}{t}=\left(1+\frac{1}{m_{1}}\right)\left(1+\frac{1}{m_{2}}\right) \cdots\left(1+\frac{1}{m_{j-1}}\right)
$$

so setting $m_{j}=2 t-1$ gives the desired expression.

Case 2. $n=2 t$ for some positive integer $t$.

Now we have

$$
1+\frac{2^{j}-1}{2 t}=\frac{2 t+2^{j}-1}{2 t+2^{j}-2} \cdot \frac{2 t+2^{j}-2}{2 t}=\left(1+\frac{1}{2 t+2^{j}-2}\right)\left(1+\frac{2^{j-1}-1}{t}\right),
$$

noting that $2 t+2^{j}-2>0$. Again, we use that

$$
1+\frac{2^{j-1}-1}{t}=\left(1+\frac{1}{m_{1}}\right)\left(1+\frac{1}{m_{2}}\right) \cdots\left(1+\frac{1}{m_{j-1}}\right)
$$

Setting $m_{j}=2 t+2^{j}-2$ then gives the desired expression.
Consider the base 2 expansions of the residues of $n-1$ and $-n$ modulo $2^{k}$ :

$$
\begin{aligned}
& n-1 \equiv 2^{a_{1}}+2^{a_{2}}+\cdots+2^{a_{r}}\left(\bmod 2^{k}\right) \quad \text { where } 0 \leqslant a_{1}<a_{2}<\ldots<a_{r} \leqslant k-1 \text {, } \\
& -n \equiv 2^{b_{1}}+2^{b_{2}}+\cdots+2^{b_{s}}\left(\bmod 2^{k}\right) \quad \text { where } 0 \leqslant b_{1}<b_{2}<\ldots<b_{s} \leqslant k-1
\end{aligned}
$$

Since $-1 \equiv 2^{0}+2^{1}+\cdots+2^{k-1}\left(\bmod 2^{k}\right)$, we have $\left\{a_{1}, \ldots, a_{r}\right\} \cup\left\{b_{1} \ldots, b_{s}\right\}=\{0,1, \ldots, k-1\}$ and $r+s=k$. Write

$$
\begin{aligned}
& S_{p}=2^{a_{p}}+2^{a_{p+1}}+\cdots+2^{a_{r}} \quad \text { for } \quad 1 \leqslant p \leqslant r \\
& T_{q}=2^{b_{1}}+2^{b_{2}}+\cdots+2^{b_{q}} \quad \text { for } \quad 1 \leqslant q \leqslant s .
\end{aligned}
$$



Also set $S_{r+1}=T_{0}=0$. Notice that $S_{1}+T_{s}=2^{k}-1$ and $n+T_{s} \equiv 0\left(\bmod 2^{k}\right)$. We have

$$
\begin{aligned}
1+\frac{2^{k}-1}{n} & =\frac{n+S_{1}+T_{s}}{n}=\frac{n+S_{1}+T_{s}}{n+T_{s}} \cdot \frac{n+T_{s}}{n} \\
& =\prod_{p=1}^{r} \frac{n+S_{p}+T_{s}}{n+S_{p+1}+T_{s}} \cdot \prod_{q=1}^{s} \frac{n+T_{q}}{n+T_{q-1}} \\
& =\prod_{p=1}^{r}\left(1+\frac{2^{a_{p}}}{n+S_{p+1}+T_{s}}\right) \cdot \prod_{q=1}^{s}\left(1+\frac{2^{b_{q}}}{n+T_{q-1}}\right),
\end{aligned}
$$

so if we define

$$
m_{p}=\frac{n+S_{p+1}+T_{s}}{2^{a_{p}}} \quad \text { for } 1 \leqslant p \leqslant r \quad \text { and } \quad m_{r+q}=\frac{n+T_{q-1}}{2^{b_{q}}} \quad \text { for } 1 \leqslant q \leqslant s
$$

the desired equality holds. It remains to check that every $m_{i}$ is an integer. For $1 \leqslant p \leqslant r$ we have

$$
n+S_{p+1}+T_{s} \equiv n+T_{s} \equiv 0 \quad\left(\bmod 2^{a_{p}}\right)
$$

and for $1 \leqslant q \leqslant r$ we have

$$
n+T_{q-1} \equiv n+T_{s} \equiv 0 \quad\left(\bmod 2^{b_{q}}\right)
$$

The desired result follows.