49th Mathematical Olympiad in Ukraine

a) Consider remainders of these squares divided by $3$. In the big number which we have generated each square can be represented as $k^2\cdot 10^n$, and the big number is the sum of such terms. If $k^2\equiv l(\mathrm{mod}3)$ then $k^2\cdot 10^n\equiv l(\mathrm{mod}3)$. It is well-known that $k^2\equiv 0(\mathrm{mod}3)$ is equivalent to $3|k$, and $k^2\equiv 1(\mathrm{mod}3)$ is equivalent to $3\nmid k$. There are $669$ numbers from $1$ to $2009$ that are divisible by $3$, and $1340$ there are not. As $1340\equiv 2(\mathrm{mod}3)$ the remainder of the big number modulo $3$ is $2$, which is impossible for a perfect square, which completes the proof.

b) Consider remainders of these numbers divided by $9$. In the big number which we have generated each number can be represented as $k\cdot 10^n$, and the big number is the sum of such terms. $k\equiv l(\mathrm{mod}9)$ if and only if $k\cdot 10^n\equiv l(\mathrm{mod}9)$. Thus we need to find the sum of the remainders which equals to the sum of all digits of the big number. If we break the number into the groups of nine $1$, $2$, $\dots$, $9$; $10$, $11$, $\dots$, $18$; $\dots$, $1999$, $2000$, $\dots$, $2007$, then the sum of the numbers in each group is divisible by $9$. The sum of the remaining numbers $2008$ and $2009$ is divisible by $3$ but is not divisible by $9$, which is impossible for a perfect square.