Subfield: Number Theory


Question: Let $n$ be a positive integer. Show that the numbers

$$
\left(\begin{array}{c}
2^{n}-1 \\
0
\end{array}\right), \quad\left(\begin{array}{c}
2^{n}-1 \\
1
\end{array}\right), \quad\left(\begin{array}{c}
2^{n}-1 \\
2
\end{array}\right), \quad \cdots, \quad\left(\begin{array}{c}
2^{n}-1 \\
2^{n-1}-1
\end{array}\right)
$$

are congruent modulo $2^{n}$ to $1,3,5, \ldots, 2^{n}-1$ in some order.

Solution:
It is well-known that all these numbers are odd. So the assertion that their remainders $\left(\bmod 2^{n}\right)$ make up a permutation of $\left\{1,3, \ldots, 2^{n}-1\right\}$ is equivalent just to saying that these remainders are all distinct. We begin by showing that

$$
\left(\begin{array}{c}
2^{n}-1 \\
2 k
\end{array}\right)+\left(\begin{array}{c}
2^{n}-1 \\
2 k+1
\end{array}\right) \equiv 0 \quad\left(\bmod 2^{n}\right) \quad \text { and } \quad\left(\begin{array}{c}
2^{n}-1 \\
2 k
\end{array}\right) \equiv(-1)^{k}\left(\begin{array}{c}
2^{n-1}-1 \\
k
\end{array}\right) \quad\left(\bmod 2^{n}\right)
\tag{1}
$$

The first relation is immediate, as the sum on the left is equal to $\left(\begin{array}{c}2^{n} \\ 2 k+1\end{array}\right)=\frac{2^{n}}{2 k+1}\left(\begin{array}{c}2^{n}-1 \\ 2 k\end{array}\right)$, hence is divisible by $2^{n}$. The second relation:

$$
\left(\begin{array}{c}
2^{n}-1 \\
2 k
\end{array}\right)=\prod_{j=1}^{2 k} \frac{2^{n}-j}{j}=\prod_{i=1}^{k} \frac{2^{n}-(2 i-1)}{2 i-1} \cdot \prod_{i=1}^{k} \frac{2^{n-1}-i}{i} \equiv(-1)^{k}\left(\begin{array}{c}
2^{n-1}-1 \\
k
\end{array}\right) \quad\left(\bmod 2^{n}\right) .
$$

This prepares ground for a proof of the required result by induction on $n$. The base case $n=1$ is obvious. Assume the assertion is true for $n-1$ and pass to $n$, denoting $a_{k}=\left(\begin{array}{c}2^{n-1}-1 \\ k\end{array}\right)$, $b_{m}=\left(\begin{array}{c}2^{n}-1 \\ m\end{array}\right)$. The induction hypothesis is that all the numbers $a_{k}\left(0 \leq k<2^{n-2}\right)$ are distinct $\left(\bmod 2^{n-1}\right)$; the claim is that all the numbers $b_{m}\left(0 \leq m<2^{n-1}\right)$ are distinct $\left(\bmod 2^{n}\right)$.

The congruence relations (1) are restated as

$$
b_{2 k} \equiv(-1)^{k} a_{k} \equiv-b_{2 k+1} \quad\left(\bmod 2^{n}\right)
\tag{2}
$$

Shifting the exponent in the first relation of (1) from $n$ to $n-1$ we also have the congruence $a_{2 i+1} \equiv-a_{2 i}\left(\bmod 2^{n-1}\right)$. We hence conclude:
$$
\text{If, for some j, }k<2^{n-2}, a_{k} \equiv-a_{j}\left(\bmod 2^{n-1}\right)\text{, then }\{j, k\}=\{2 i, 2 i+1\}\text{ for some i.}\tag{3}
$$

This is so because in the sequence $\left(a_{k}: k<2^{n-2}\right)$ each term $a_{j}$ is complemented to $0\left(\bmod 2^{n-1}\right)$ by only one other term $a_{k}$, according to the induction hypothesis.

From $(2)$ we see that $b_{4 i} \equiv a_{2 i}$ and $b_{4 i+3} \equiv a_{2 i+1}\left(\bmod 2^{n}\right)$. Let

$$
M=\left\{m: 0 \leq m<2^{n-1}, m \equiv 0 \text { or } 3(\bmod 4)\right\}, \quad L=\left\{l: 0 \leq l<2^{n-1}, l \equiv 1 \text { or } 2(\bmod 4)\right\}
$$

The last two congruences take on the unified form

$$
b_{m} \equiv a_{\lfloor m / 2\rfloor} \quad\left(\bmod 2^{n}\right) \quad \text { for all } \quad m \in M\tag{4}
$$

Thus all the numbers $b_{m}$ for $m \in M$ are distinct $\left(\bmod 2^{n}\right)$ because so are the numbers $a_{k}$ (they are distinct $\left(\bmod 2^{n-1}\right)$, hence also $\left(\bmod 2^{n}\right)$ ).

Every $l \in L$ is paired with a unique $m \in M$ into a pair of the form $\{2 k, 2 k+1\}$. So (2) implies that also all the $b_{l}$ for $l \in L$ are distinct $\left(\bmod 2^{n}\right)$. It remains to eliminate the possibility that $b_{m} \equiv b_{l}\left(\bmod 2^{n}\right)$ for some $m \in M, l \in L$.

Suppose that such a situation occurs. Let $m^{\prime} \in M$ be such that $\left\{m^{\prime}, l\right\}$ is a pair of the form $\{2 k, 2 k+1\}$, so that $(\operatorname{see}(2)) b_{m^{\prime}} \equiv-b_{l}\left(\bmod 2^{n}\right)$. Hence $b_{m^{\prime}} \equiv-b_{m}\left(\bmod 2^{n}\right)$. Since both $m^{\prime}$ and $m$ are in $M$, we have by (4) $b_{m^{\prime}} \equiv a_{j}, b_{m} \equiv a_{k}\left(\bmod 2^{n}\right)$ for $j=\left\lfloor m^{\prime} / 2\right\rfloor, k=\lfloor m / 2\rfloor$.

Then $a_{j} \equiv-a_{k}\left(\bmod 2^{n}\right)$. Thus, according to (3), $j=2 i, k=2 i+1$ for some $i$ (or vice versa). The equality $a_{2 i+1} \equiv-a_{2 i}\left(\bmod 2^{n}\right)$ now means that $\left(\begin{array}{c}2^{n-1}-1 \\ 2 i\end{array}\right)+\left(\begin{array}{c}2^{n-1}-1 \\ 2 i+1\end{array}\right) \equiv 0\left(\bmod 2^{n}\right)$. However, the sum on the left is equal to $\left(\begin{array}{l}2^{n-1} \\ 2 i+1\end{array}\right)$. A number of this form cannot be divisible by $2^{n}$. This is a contradiction which concludes the induction step and proves the result.
We again proceed by induction, writing for brevity $N=2^{n-1}$ and keeping notation $a_{k}=\left(\begin{array}{c}N-1 \\ k\end{array}\right), b_{m}=\left(\begin{array}{c}2 N-1 \\ m\end{array}\right)$. Assume that the result holds for the sequence $\left(a_{0}, a_{1}, a_{2}, \ldots, a_{N / 2-1}\right)$. In view of the symmetry $a_{N-1-k}=a_{k}$ this sequence is a permutation of $\left(a_{0}, a_{2}, a_{4}, \ldots, a_{N-2}\right)$. So the induction hypothesis says that this latter sequence, taken $(\bmod N)$, is a permutation of $(1,3,5, \ldots, N-1)$. Similarly, the induction claim is that $\left(b_{0}, b_{2}, b_{4}, \ldots, b_{2 N-2}\right)$, taken $(\bmod 2 N)$, is a permutation of $(1,3,5, \ldots, 2 N-1)$.

In place of the congruence relations (2) we now use the following ones,

$$
b_{4 i} \equiv a_{2 i} \quad(\bmod N) \quad \text { and } \quad b_{4 i+2} \equiv b_{4 i}+N \quad(\bmod 2 N)\tag{5}
$$

Given this, the conclusion is immediate: the first formula of (5) together with the induction hypothesis tells us that $\left(b_{0}, b_{4}, b_{8}, \ldots, b_{2 N-4}\right)(\bmod N)$ is a permutation of $(1,3,5, \ldots, N-1)$. Then the second formula of $(5)$ shows that $\left(b_{2}, b_{6}, b_{10}, \ldots, b_{2 N-2}\right)(\bmod N)$ is exactly the same permutation; moreover, this formula distinguishes $(\bmod 2 N)$ each $b_{4 i}$ from $b_{4 i+2}$.

Consequently, these two sequences combined represent $(\bmod 2 N)$ a permutation of the sequence $(1,3,5, \ldots, N-1, N+1, N+3, N+5, \ldots, N+N-1)$, and this is precisely the induction claim.

Now we prove formulas (5); we begin with the second one. Since $b_{m+1}=b_{m} \cdot \frac{2 N-m-1}{m+1}$,

$$
b_{4 i+2}=b_{4 i} \cdot \frac{2 N-4 i-1}{4 i+1} \cdot \frac{2 N-4 i-2}{4 i+2}=b_{4 i} \cdot \frac{2 N-4 i-1}{4 i+1} \cdot \frac{N-2 i-1}{2 i+1} .
$$

The desired congruence $b_{4 i+2} \equiv b_{4 i}+N$ may be multiplied by the odd number $(4 i+1)(2 i+1)$, giving rise to a chain of successively equivalent congruences:

$$
\begin{aligned}
b_{4 i}(2 N-4 i-1)(N-2 i-1) & \equiv\left(b_{4 i}+N\right)(4 i+1)(2 i+1) & & (\bmod 2 N) \\
b_{4 i}(2 i+1-N) & \equiv\left(b_{4 i}+N\right)(2 i+1) & & (\bmod 2 N) \\
\left(b_{4 i}+2 i+1\right) N & \equiv 0 & & (\bmod 2 N)
\end{aligned}
$$

and the last one is satisfied, as $b_{4 i}$ is odd. This settles the second relation in (5).

The first one is proved by induction on $i$. It holds for $i=0$. Assume $b_{4 i} \equiv a_{2 i}(\bmod 2 N)$ and consider $i+1$ :

$$
b_{4 i+4}=b_{4 i+2} \cdot \frac{2 N-4 i-3}{4 i+3} \cdot \frac{2 N-4 i-4}{4 i+4} ; \quad a_{2 i+2}=a_{2 i} \cdot \frac{N-2 i-1}{2 i+1} \cdot \frac{N-2 i-2}{2 i+2}
$$

Both expressions have the fraction $\frac{N-2 i-2}{2 i+2}$ as the last factor. Since $2 i+2<N=2^{n-1}$, this fraction reduces to $\ell / m$ with $\ell$ and $m$ odd. In showing that $b_{4 i+4} \equiv a_{2 i+2}(\bmod 2 N)$, we may ignore this common factor $\ell / m$. Clearing other odd denominators reduces the claim to

$$
b_{4 i+2}(2 N-4 i-3)(2 i+1) \equiv a_{2 i}(N-2 i-1)(4 i+3) \quad(\bmod 2 N)
$$

By the inductive assumption (saying that $b_{4 i} \equiv a_{2 i}(\bmod 2 N)$ ) and by the second relation of (5), this is equivalent to

$$
\left(b_{4 i}+N\right)(2 i+1) \equiv b_{4 i}(2 i+1-N) \quad(\bmod 2 N)
$$

a congruence which we have already met in the preceding proof a few lines above. This completes induction (on $i$ ) and the proof of (5), hence also the whole solution.