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

$$
\left(a_{1}+a_{2}\right)\left(a_{2}+a_{3}\right) \cdots\left(a_{n-1}+a_{n}\right)\left(a_{n}+a_{1}\right)=2^{2 k-1} .
$$
Find all positive integers $n \geq 2$ for which there exists an expensive $n$-tuple.

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

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

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.
For odd $n$ the tuple $(1,1, \ldots, 1)$ is a solution.

Now consider $n$ even. Since the product $\prod\left(a_{i}+a_{i+1}\right)$ is a power of two, every factor needs to be a power of two. We are going to prove that for all tuples $\left(a_{1}, \ldots, a_{n}\right)$ such that $a_{i}+a_{i+1}$ is always a power of two , it is the case that $\prod\left(a_{i}+a_{i+1}\right)$ is equal to an even power of two. We are going to prove this with strong induction on $\sum a_{i}$. When all $a_{i}$ are equal to one this is certainly the case. Since $a_{i}+a_{i+1}>1$ it is even and we conclude that the $a_{i}$ are either all odd or all even. In the case they are all even, then consider the tuple $\left(b_{1}, \ldots, b_{n}\right)$ with $b_{i}=a_{i} / 2$. This tuple clearly satisfies the hypothesis as well and we have $\sum b_{i}<\sum a_{i}$. Furthermore we have $\prod\left(a_{i}+a_{i+1}\right)=2^{n} \prod\left(b_{i}+b_{i+1}\right)$ and since $n$ is even we are done in this case.

Now all $a_{i}$ are odd. Suppose none of are one, then consider the tuple $\left(b_{1}, \ldots, b_{n}\right)$ with $b_{i}=\left(a_{i}+(-1)^{i}\right) / 2$. Since all $a_{i}$ are odd and strictly larger than one, the $b_{i}$ are positive integers and satisfy $b_{i}+b_{i+1}=\left(a_{i}+a_{i+1}\right) / 2$, a power of two. Since $\sum b_{i}<\sum a_{i}$ and $\prod\left(a_{i}+a_{i+1}\right)=2^{n} \prod\left(b_{i}+b_{i+1}\right)$ we are done in this case again. Now there is at least one $a_{i}$ being one. We may assume $i=1$, because the condition is cyclic. Moreover we may also assume that $a_{2}>1$ since not all of the $a_{i}$ are equal to one. Let now $k$ be the smallest index larger than one such that $a_{k}$ is equal to one. We are not excluding the case $k=n+1$, yet. Now for $i=1, \ldots, k-1$ we have $a_{i}+a_{i+1}>2$ and thus divisible by four. By induction it easily follows that $a_{i} \equiv(-1)^{i+1} \bmod (4)$ for $i=1, \ldots, k-1$. In particular, since $a_{k}=1$ we find that $k$ is odd and at least three. Now consider the tuple $\left(b_{1}, \ldots, b_{n}\right)$ with $b_{i}=\left(a_{i}-(-1)^{i}\right) / 2$ for $i=1, \ldots, k$ and $b_{i}=a_{i}$ otherwise. This is again a tuple that satisfies the hypothesis, since $b_{1}=a_{1}=1=b_{k}=a_{k}$. Moreover $b_{2}<a_{2}$ and thus $\sum b_{i}<\sum a_{i}$. Finally we have $\prod\left(a_{i}+a_{i+1}\right)=2^{k-1} \prod\left(b_{i}+b_{i+1}\right)$ and since $k$ is odd we conclude the proof.

Final answer: odd integers $n>2$