Show that the last two decimal digits of a perfect square must be one of the following pairs. Show that the last two decimal digits of a perfect square must be one of the following pairs: 
$00, e1, e4, 25, o6, e9$ ($e$ stand for even digit, $o$ for odd).
Solution is given, but I don't understand it at all.


So the last two digits of $n^2$ clearly depend on the last two digits of $n$. There are $100$ possibilities.
Now $(50+n)^2=2500+100n+n^2=100(25+n)+n^2$ and the multiple of $100$ shows that this part can't affect the last two digits, which are therefore the same as the last two digits of $n^2$. So numbers which differ by $50$ produce the same final pair of digits when squared. Since the cycle repeats after $50$ numbers there are now only $50$ possibilities to consider say the integers $0$ to $49$.
Then $(50-n)^2=2500-100n+n^2=100(25+n)+n^2$ so, for the same reason, $n$ and $50-n$ have the same final digits. In other words two numbers which sum to $50$ have the same final digits. We therefore pair $(1,49); (2,48) \dots (24,26)$ and therefore only have to consider the $26$ numbers $0$ to $25$.

Now consider $10a+5\pm b$ where $0\le a\le 2$ and $0\le b \le 5$
$$(10a+5\pm b)^2=100a^2+25+b^2+100a\pm 20ab\pm 10b$$
If we work modulo $20$ we have that the final digit and the parity of the penultimate digit are determined by that of $(b-5)^2$. Taking $b=0$ and $b=5$ as special cases giving $25$ and $00$ respectively. we need only test $|b-5|=1,2,3,4$ giving $e1, e4, e9, o6$
All the cases do in fact occur, as we find by squaring all the numbers $0 \to 25$
