, 2014

Impossible. We use proof by contradiction. Suppose such an infinite sequence and positive integer $N$ exist. Let $y_k = \sum_{i=1}^{k} a_i 10^{i-1}$, then for all $k > N$, there exists a positive integer $x_k$ such that $y_k = x_k^2$.

1. For all $n$, let $\gamma_n$ be the largest positive integer satisfying $5^{\gamma_n}|x_n$. We will prove: for all $n > N$, $2\gamma_n \ge n$.
Suppose there exists $n < N$ such that $2\gamma_n < n$, then
$$y_{n+1} = 10^n a_{n+1} + y_n = 5^{2\gamma_n} \left( 2^n 5^{n-2\gamma_n} a_{n+1} + \frac{y_n}{5^{2\gamma_n}} \right).$$
Since $y_n/5^{2\gamma_n}$ cannot be divisible by 5, we must have $\gamma_{n+1} = \gamma_n < n < n+1$, hence $\gamma_n = \gamma_{n+1} = \dots = \gamma$.
On the other hand, for all $k \ge n$,
$$(x_{k+1} - x_k)(x_{k+1} + x_k) = y_{k+1} - y_k = a_{k+1} 10^k.$$
Since $(x_{k+1} - x_k) + (x_{k+1} + x_k) = 2x_{k+1}$, by the definition of $\gamma = \gamma_{k+1}$, one of these two must not be divisible by $5^{\gamma+1}$. Furthermore, since $5^k|(x_{k+1} - x_k)(x_{k+1} + x_k)$, one of them must be divisible by $5^{k-\gamma}$, hence
$$5^{k-\gamma} \le \max\{x_{k+1} - x_k, x_{k+1} + x_k\} < 2x_{k+1} = 2\sqrt{y_{k+1}} < 2 \times 10^{(k+1)/2},$$
This implies $5^{2k} < 4 \times 5^{2\gamma} \times 10^{k+1}$, that is, $(5/2)^k < 40 \times 5^{2\gamma}$. This cannot hold for sufficiently large $k$, giving a contradiction.

2. Now consider all $k > \max\{N/2, 2\}$. From 1. we know $2\gamma_{2k+1} \ge 2k+1$ and $2\gamma_{2k+2} \ge 2k+2$, so $\gamma_{2k+1} \ge k+1$ and $\gamma_{2k+2} \ge k+1$. Also, from $y_{2k+2} = a_{2k+2}10^{2k+1} + y_{2k+1}$ we know $5^{2k+2}|y_{2k+2} - y_{2k+1} = a_{2k+2}10^{2k+1}$, so $5|a_{2k+2} \Rightarrow a_{2k+2} = 5$. Therefore
$$(x_{2k+2} - x_{2k+1})(x_{2k+2} + x_{2k+1}) = y_{2k+2} - y_{2k+1} = 5 \times 10^{2k+1} = 2^{2k+1}5^{2k+2}.$$
Let $A_k = x_{2k+2}/5^{k+1}$, $B_k = x_{2k+1}/5^{k+1}$; note that by 1., these must be positive integers, and
$$(A_k - B_k)(A_k + B_k) = 2^{2k+1} \qquad (3)$$
Note that $a_1 \ne 0$, so $A_k$ and $B_k$ are both odd (otherwise $y_{2k+2}$ or $y_{2k+1}$ would be a multiple of 10, which is not allowed). This means that one of $A_k - B_k$ and $A_k + B_k$ is not divisible by 4. Hence by (1), $A_k - B_k = 2$ and $A_k + B_k = 2^{2k}$, so $A_k = 2^{2k-1} + 1$, giving
$$x_{2k+2} = 5^{k+1}A_k = 10^{k+1}2^{k-2} + 5^{k+1} > 10^{k+1}$$
(since $k \ge 2$) but this means $y_{2k+2} > 10^{2k+2}$, a contradiction!