Austrian Mathematical Olympiad

For $n \ge 1$ we have
$$a_{n+1} - a_n = (a_n - a_{n-1}) + (a_{n-1} - a_{n-2}) + \dots + (a_1 - a_0) = a_n - a_0,$$
which yields $a_{n+1} = 2a_n - a_0$. The sequence can therefore be written in the form
$$a_0 = 2012,\ a_1 = 2012 + d,\ a_2 = 2012 + 2d,\ \dots,\ a_n = 2012 + 2^{n-1}d,\ \dots$$
For $n \ge 3$ we therefore have $a_n = 4 \cdot (503 + 2^{n-3}d)$. Since $503 + 2^{n-3}d \equiv 3 \pmod 4$ holds for $n \ge 5$, we see that $a_n$ can never be a perfect square for $n \ge 5$. The only numbers in the sequence that can possibly be perfect squares are therefore
$$a_0 = 2012,\ a_1 = 2012 + d,\ a_2 = 2012 + 2d,\ a_3 = 4(503 + d) \text{ or } a_4 = 4(503 + 2d)$$
with $1 \le d \le 43$. Since $2012 = a_0 < a_4 \le 2012 + 8 \cdot 43 = 2356$, the only perfect squares that can occur in the sequences must lie between 2012 and 2356, i.e.
$$\begin{align*} 45^2 &= 2025 = 2012 + 13, \\ 46^2 &= 2116 = 2012 + 104 = 2012 + 2 \cdot 52 = 2012 + 4 \cdot 26 = 2012 + 8 \cdot 13, \\ 47^2 &= 2209 = 2012 + 197 \text{ or} \\ 48^2 &= 2304 = 2012 + 292 = 2012 + 2 \cdot 146 = 2012 + 4 \cdot 73. \end{align*}$$
Since we must have $d \le 43$, the only sequences of the required type containing perfect squares are those with $d=13$ (which contains 2025 and 2116) and with $d=26$ (which also contains 2116). $\square$