Subfield: Algebra


Question: Let $n \geq 2$ be an integer. An $n$-tuple $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ of positive integers is expensive if there exists a positive integer $k$ such that

$$
\left(a_{1}+a_{2}\right)\left(a_{2}+a_{3}\right) \cdots\left(a_{n-1}+a_{n}\right)\left(a_{n}+a_{1}\right)=2^{2 k-1} .
$$
Prove that for every positive integer $m$ there exists an integer $n \geq 2$ such that $m$ belongs to an expensive $n$-tuple.

There are exactly $n$ factors in the product on the left hand side.

Solution:
Notice that for odd integers $n>2$, the tuple $(1,1, \ldots, 1)$ is expensive. We will prove that there are no expensive $n$-tuples for even $n$.

Lemma 0.1. If an expensive $n$-tuple exists for some $n \geq 4$, then also an expensive $n-2$-tuple.

Proof. In what follows all indices are considered modulo $n$. Let $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ be an expensive $n$-tuple and $a_{t}$ the largest element of the tuple. We have the inequalities

$$
\begin{gathered}
a_{t-1}+a_{t} \leq 2 a_{t}<2\left(a_{t}+a_{t+1}\right) \\
a_{t}+a_{t+1} \leq 2 a_{t}<2\left(a_{t-1}+a_{t}\right) .
\end{gathered}
$$

Since both $a_{t-1}+a_{t}$ and $a_{t}+a_{t+1}$ are powers of 2 (they are divisors of a power of 2), we deduce from (1) and $(2)$

$$
a_{t-1}+a_{t}=a_{t}+a_{t+1}=2^{r}
$$

for some positive integer $r$, and in particular $a_{t-1}=a_{t+1}$.

Consider now the $n-2$-tuple $\left(b_{1}, \ldots, b_{n-2}\right)$ obtained by removing $a_{t}$ and $a_{t+1}$ from $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$. By what we just said we have

$$
\prod_{i=1}^{n-2}\left(b_{i}+b_{i+1}\right)=\frac{\prod_{i=1}^{n}\left(a_{i}+a_{i+1}\right)}{\left(a_{t-1}+a_{t}\right)\left(a_{t}+a_{t+1}\right)}=2^{2(k-r)-1}
$$

and hence $\left(b_{1}, \ldots, b_{n-2}\right)$ is again expensive.

From the lemma we now conclude that if there exists an expensive $n$-tuple for some even $n$, then also an expensive 2 -tuple i.e.

$$
\left(a_{1}+a_{2}\right)^{2}=2^{2 k-1}
$$

for some positive integers $a_{1}, a_{2}$, which is impossible since the right hand side is not a square.

We prove this by induction. From above we saw that 1 belongs to an expensive $n$-tuple. Assume now that all odd positive integers less that $2^{k}$ belong to an expensive $n$-tuple, for some $k \geq 1$. Hence for any odd $r<2^{k}$ there is an integer $n$ and an expensive $n$-tuple $\left(a_{1}, \ldots, r, \ldots, a_{n}\right)$. We notice that then also $\left(a_{1}, \ldots, r, 2^{k+1}-\right.$ $\left.r, r, \ldots, a_{n}\right)$ is expensive. Since $2^{k+1}-r$ can take all odd values between $2^{k}$ and $2^{k+1}$ the induction step is complete.
Consider the operators $T_{ \pm}(n)=2 n \pm 1$. We claim that for every odd integer $m$ there is an integer $r$ and signs $\epsilon_{i} \in\{+,-\}$ for $i=1, \ldots r$ such that $T_{\epsilon_{r}} \circ \cdots \circ T_{\epsilon_{1}}(1)=m$. This is certainly true for $m=1$ and for $m>1$ we find that $m=T_{-}((m+1) / 2)$ if $m \equiv 1 \bmod (4)$ and $m=T_{+}((m-1) / 2)$ if $m \equiv 3 \bmod (4)$. Note that both $(m+1) / 2$ and $(m-1) / 2$ are odd integers in their respective cases and $(m-1) / 2 \leq(m+1) / 2<m$ for $m>1$. Therefore iterating the procedure will eventually terminate in one.

For the construction it is most convenient to set $n=2 l+1$ and label the tuple $\left(a_{-l}, a_{-l+1}, \ldots, a_{l}\right)$. For $m=1$ we have the expensive tuple $(1,1,1)$. For $m>1$ we will define operators $T_{ \pm}$on expensive tuples with the condition $a_{-l}=a_{l}=1$ that give rise to a new expensive tuple $\left(b_{-l^{\prime}}, \ldots, b_{l^{\prime}}\right)$ with $b_{-l^{\prime}}=b_{l^{\prime}}=1$ and $b_{0}=T_{ \pm} a_{0}$. It is then clear that $T_{\epsilon_{r}} \circ \cdots \circ T_{\epsilon_{1}}((1,1,1))$ is an expensive tuple containing $m$. We define $T_{ \pm}$as follows: set $l^{\prime}=l+1$ and $b_{-l^{\prime}}=b_{l^{\prime}}=1$ and $b_{i}=T_{ \pm(-1)^{i}}\left(a_{i}\right)$ for $i=-l, \ldots, l$. Here we identify + with +1 and - with -1 . We are left to prove that the new tuple is indeed expensive. If $\pm(-1)^{l}=-1$, then $\prod\left(b_{i}+b_{i+1}\right)=4 \cdot 2^{2 l} \prod\left(a_{i}+a_{i+1}\right)$, and if $\pm(-1)^{l}=+1$, then $\prod\left(b_{i}+b_{i+1}\right)=4 \cdot 2^{2 l+2} \prod\left(a_{i}+a_{i+1}\right)$. In both cases we end up with an expensive tuple again.