, 2012

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 $\sqrt{n}$.