A deck of $n>1$ cards is given. A positive integer is written on each card. The deck has the property that the arithmetic mean of the numbers on each pair of cards is also the geometric mean of the numbers on some collection of one or more cards.

For which $n$ does it follow that the numbers on the cards are all equal?

Step 1 (Setup): Let the card values be $a_1\ge a_2\ge \cdots \ge a_n>0$. WLOG $\gcd (a_1,\dots ,a_n)=1$. Suppose not all equal.

Step 2 (Prime divisor argument): Since not all equal and $\gcd =1$, let $p$ be a prime dividing $a_1$. Let $k$ be the smallest index with $p\nmid a_k$.

Step 3 (AM-GM condition): The arithmetic mean of $a_1$ and $a_k$ is $\frac{a_1+a_k}{2}$. Since $p|a_1$ and $p\nmid a_k$, we have $p\nmid \frac{a_1+a_k}{2}$ (assuming $p$ is odd; handle $p=2$ separately). This AM equals the GM of some subcollection.

Step 4 (GM constraint): The GM of the subcollection $S$ is $({\prod }_{i\in S}{a}_i{)}^{1/|S|}$. Since $p\nmid \text{AM}$, and the GM equals the AM, we need $p\nmid \text{GM}$. This means $p\nmid a_i$ for all $i\in S$, so all cards in $S$ have index $\ge k$.

Step 5 (Size contradiction): Since all cards in $S$ have values $\le a_k$, their GM $\le a_k$. But $\text{AM}=\frac{a_1+a_k}{2}>a_k$ (since $a_1>a_k$, as $p|a_1$ and $p\nmid a_k$ with $\gcd =1$). Contradiction.

Conclusion: All card values must be equal, for all $n>1$. $■$