Balkan 2012 shortlist

Solution. There are no two such consecutive terms. Assume that $a_n=x^2$, $a_{n+1}=y^2$ where $x,y$ are positive integers. Then $$(x+1{)}^2\le y^2=a_{n+1}=a_n+\tau (n)=x^2+\tau (n)\le x^2+2\sqrt{n}.$$ Therefore $x<\sqrt{n}$. The last inequality gives $a_n<n$, which is impossible since the sequence is strictly increasing and $a_1\ge 1$. We used the inequality $\tau (n)\le 2\sqrt{n}$ which follows immediately from the fact that the positive integer divisors of $n$ can be paired off (with the possible exception of $\sqrt{n}$) with one in each pair less than $\sqrt{n}$.