75th NMO Selection Tests

For $n=2$, we have $a_1+a_2=a_1{a}_2$, which rewrites as $(a_1-1)(a_2-1)=1$. We get $a_1=2,a_2=2$, so $(a_1^2-1)(a_2^2-1)=9$.

$$|a_1{a}_2\dots a_n|=|a_1+a_2+\cdots +a_n|\le |a_1|+|a_2|+\cdots +|a_n|.$$ Dividing by $|a_1{a}_2\dots a_n|>0$ we obtain $$1\le \frac{|a_1|}{|a_1{a}_2\cdots a_n|}+\frac{|a_2|}{|a_1{a}_2\cdots a_n|}+\cdots +\frac{|a_n|}{|a_1{a}_2\cdots a_n|}.$$ If all the numbers have absolute value greater than or equal to $2$, then the right-hand side of the above inequality is less than or equal to $\frac{n}{2^{n-1}}$, i.e., $2^{n-1}\le n$. For $n\ge 3$, we obtain (by induction or using Bernoulli's inequality) that $2^{n-1}>n$, which contradicts the previous inequality. Therefore, at least one of the numbers has absolute value $1$, which implies $(a_1^2-1)(a_2^2-1)\dots (a_n^2-1)=0$, a perfect square.

Remark. For every $n\ge 3$, there exist numbers satisfying the condition in the statement, for example $$\underset{n-2\text{ times}}{\underbrace{1+1+\cdots +1}}+2+n=\underset{n-2\text{ times}}{\underbrace{1\cdot 1\dots 1}}\cdot 2\cdot n.$$