Prove that number is square of natural We have number $a=246810...98100 $. Number $a$ consists of the following even numbers from $2 -100$ . Decide whether number is a square. 
I was given a hint to this task to sum all digits. 
Also I would be grateful if someone could explain whether this rule is true: If the sum of digits is divided by $k^2$ then this number is square of natural

Hint $\ $ Cast out nines to show the sum is not a square mod $\,9,\,$ hence not the square of an integer. To compute the sum very simply we can use Gauss's grade-school reflection method: pair up each term $\,n\,$ with its negation(inverse) $\:-n\equiv 108\!-\!n\pmod{108}$ so they sum to $\,108\equiv 0\pmod 9\,$ hence cancel out of the sum. In detail:
$\begin{eqnarray}{\rm mod}\ 9\!:\ \color{#c00}{10\equiv 1}\,&\Rightarrow&\quad 2\cdot \color{#c00}{10}^i + 4\cdot \color{#c00}{10}^j +\cdots + 100\cdot\color{#c00}{10}^k\\
&\equiv&\quad 2\cdot \color{#c00}{\ 1}^i\ + 4\cdot \color{#c00}{\ 1}^j\ +\cdots + 100\cdot\color{#c00}{\ 1}^k\\
&\equiv&\quad \quad2 + \quad 4 +\quad 6 +\quad 8 + \cdots + 50 + 52 +\!\smash[t]{\overbrace{ 54}^{\equiv\, 0}}\\
&      &+ \   \color{#0a0}{106 + 104 + 102}+ 100+\cdots + 58+56\ \ -\ \ (\color{#0a0}{106 + 104 - 102})\\
[{\rm by}\ \ 108\equiv 0]\ \  &\equiv&\qquad0 +\quad 0+\quad 0+\quad 0+\cdots +\,\ 0 +\,\ 0\ \ -\ \ \smash[b]{(\underbrace{\quad 7 +\quad 5 +\quad 3}_{\large \color{#c0d}{\equiv\ 15\ \equiv\ -3}})}  \\
&\equiv&\qquad 0-(\color{#c0d}{-3})\,\equiv\, 3\\
\end{eqnarray}
$
Therefore the sum has form $\,\color{#c0d}3\!+\!9n = 3(1\!+\!3n)\neq k^2,\,$ being divisible by $3$ but not by $\,3^2.$
Remark $\ $ This modular nonsquare-test works because squares remain squares mod $\,m,\,$ i.e. if $\,\bar a = (a\ {\rm mod}\ m)\,$ then $\,\overline{a^2} = \bar a^2,\,$ being a special case of $\,\overline{ac} = \bar a \bar c,\,$ which is a special case of the  Congruence product Rule. Contrapositively, if $\,a\,$ is nonsquare mod $\,m\,$ it is a nonsquare integer. 
