Reductio ad absurdum.. So, let us assume that all those $a_n$ are rational, so we may write, for some coprime $p_n, q_n, \ \ a_n = \frac {p_n}{q_n}, n \geq 1.$ Hence, using the condition we get that $q_{n+1}^2 = q_n,$ so that from some point $a_n$ will be an integer. Now, let's assume that we've reached this point, then clearly $a_n$ cannot equal $1,$ whence $a_{n+1} < a_n.$ Then, from this point, the sequence is a strictly decreasing sequence of positive, which is a contradiction, so we're done.

or simply observe that by the condition each $ a_n + 1$ must be a perfect square, and $ (a_n + 1)*(a_{n + 1} + 1) = (a_n*a_{n + 1})^2$ will get $ a_n = 1$, which is a contradiction since 2 is not a perfect square. p.s. i don't and can't believe this is a putnam problem.

It's definitely not on the 1995 Putnam, at least. Check your sources again.