Subfield: Number Theory


Question: Let $n$ be an integer with $n \geqslant 2$. Does there exist a sequence $\left(a_{1}, \ldots, a_{n}\right)$ of positive integers with not all terms being equal such that the arithmetic mean of every two terms is equal to the geometric mean of some (one or more) terms in this sequence?

Solution:
Suppose that $a_{1}, \ldots, a_{n}$ satisfy the required properties. Let $d=\operatorname{gcd}\left(a_{1} \ldots, a_{n}\right)$. If $d>1$ then replace the numbers $a_{1}, \ldots, a_{n}$ by $\frac{a_{1}}{d}, \ldots, \frac{a_{n}}{d}$; all arithmetic and all geometric means will be divided by $d$, so we obtain another sequence satisfying the condition. Hence, without loss of generality, we can assume that $\operatorname{gcd}\left(a_{1} \ldots, a_{n}\right)=1$.

We show two numbers, $a_{m}$ and $a_{k}$ such that their arithmetic mean, $\frac{a_{m}+a_{k}}{2}$ is different from the geometric mean of any (nonempty) subsequence of $a_{1} \ldots, a_{n}$. That proves that there cannot exist such a sequence.

Choose the index $m \in\{1, \ldots, n\}$ such that $a_{m}=\max \left(a_{1}, \ldots, a_{n}\right)$. Note that $a_{m} \geqslant 2$, because $a_{1}, \ldots, a_{n}$ are not all equal. Let $p$ be a prime divisor of $a_{m}$.

Let $k \in\{1, \ldots, n\}$ be an index such that $a_{k}=\max \left\{a_{i}: p \nmid a_{i}\right\}$. Due to $\operatorname{gcd}\left(a_{1} \ldots, a_{n}\right)=1$, not all $a_{i}$ are divisible by $p$, so such a $k$ exists. Note that $a_{m}>a_{k}$ because $a_{m} \geqslant a_{k}, p \mid a_{m}$ and $p \nmid a_{k}$.

Let $b=\frac{a_{m}+a_{k}}{2}$; we will show that $b$ cannot be the geometric mean of any subsequence of $a_{1}, \ldots, a_{n}$.

Consider the geometric mean, $g=\sqrt[t]{a_{i_{1}} \cdot \ldots \cdot a_{i_{t}}}$ of an arbitrary subsequence of $a_{1}, \ldots, a_{n}$. If none of $a_{i_{1}}, \ldots, a_{i_{t}}$ is divisible by $p$, then they are not greater than $a_{k}$, so

$$
g=\sqrt[t]{a_{i_{1}} \cdot \ldots \cdot a_{i_{t}}} \leqslant a_{k}<\frac{a_{m}+a_{k}}{2}=b
$$

and therefore $g \neq b$.

Otherwise, if at least one of $a_{i_{1}}, \ldots, a_{i_{t}}$ is divisible by $p$, then $2 g=2 \sqrt[t]{a_{i_{1}} \cdot \ldots \cdot a_{i_{t}}}$ is either not an integer or is divisible by $p$, while $2 b=a_{m}+a_{k}$ is an integer not divisible by $p$, so $g \neq b$ again.
Like in the previous solution, we assume that the numbers $a_{1}, \ldots, a_{n}$ have no common divisor greater than 1 . The arithmetic mean of any two numbers in the sequence is half of an integer; on the other hand, it is a (some integer order) root of an integer. This means each pair's mean is an integer, so all terms in the sequence must be of the same parity; hence they all are odd. Let $d=\min \left\{\operatorname{gcd}\left(a_{i}, a_{j}\right): a_{i} \neq a_{j}\right\}$. By reordering the sequence we can assume that $\operatorname{gcd}\left(a_{1}, a_{2}\right)=d$, the sum $a_{1}+a_{2}$ is maximal among such pairs, and $a_{1}>a_{2}$.

We will show that $\frac{a_{1}+a_{2}}{2}$ cannot be the geometric mean of any subsequence of $a_{1} \ldots, a_{n}$.

Let $a_{1}=x d$ and $a_{2}=y d$ where $x, y$ are coprime, and suppose that there exist some $b_{1}, \ldots, b_{t} \in\left\{a_{1}, \ldots, a_{n}\right\}$ whose geometric mean is $\frac{a_{1}+a_{2}}{2}$. Let $d_{i}=\operatorname{gcd}\left(a_{1}, b_{i}\right)$ for $i=1,2, \ldots, t$ and let $D=d_{1} d_{2} \cdot \ldots \cdot d_{t}$. Then

$$
D=d_{1} d_{2} \cdot \ldots \cdot d_{t} \mid b_{1} b_{2} \cdot \ldots \cdot b_{t}=\left(\frac{a_{1}+a_{2}}{2}\right)^{t}=\left(\frac{x+y}{2}\right)^{t} d^{t}
$$

We claim that $D \mid d^{t}$. Consider an arbitrary prime divisor $p$ of $D$. Let $\nu_{p}(x)$ denote the exponent of $p$ in the prime factorization of $x$. If $p \mid \frac{x+y}{2}$, then $p \nmid x, y$, so $p$ is coprime with $x$; hence, $\nu_{p}\left(d_{i}\right) \leqslant \nu_{p}\left(a_{1}\right)=\nu_{p}(x d)=\nu_{p}(d)$ for every $1 \leqslant i \leqslant t$, therefore $\nu_{p}(D)=\sum_{i} \nu_{p}\left(d_{i}\right) \leqslant$ $t \nu_{p}(d)=\nu_{p}\left(d^{t}\right)$. Otherwise, if $p$ is coprime to $\frac{x+y}{2}$, we have $\nu_{p}(D) \leqslant \nu_{p}\left(d^{t}\right)$ trivially. The claim has been proved.



Notice that $d_{i}=\operatorname{gcd}\left(b_{i}, a_{1}\right) \geqslant d$ for $1 \leqslant i \leqslant t$ : if $b_{i} \neq a_{1}$ then this follows from the definition of $d$; otherwise we have $b_{i}=a_{1}$, so $d_{i}=a_{1} \geqslant d$. Hence, $D=d_{1} \cdot \ldots \cdot d_{t} \geqslant d^{t}$, and the claim forces $d_{1}=\ldots=d_{t}=d$.

Finally, by $\frac{a_{1}+a_{2}}{2}>a_{2}$ there must be some $b_{k}$ which is greater than $a_{2}$. From $a_{1}>a_{2} \geqslant$ $d=\operatorname{gcd}\left(a_{1}, b_{k}\right)$ it follows that $a_{1} \neq b_{k}$. Now the have a pair $a_{1}, b_{k} \operatorname{such}$ that $\operatorname{gcd}\left(a_{1}, b_{k}\right)=d$ but $a_{1}+b_{k}>a_{1}+a_{2}$; that contradicts the choice of $a_{1}$ and $a_{2}$.

Final answer: No such sequence exists