Recall the following two facts,
Proposition 1: If $s$ is any integer then $s^2 \equiv (s+50)^2 \pmod{100}$.
Proposition 2: The following four modulo statements are true,
$\; 1^2 \equiv 9^2 \equiv 11^2 \equiv 19^2 \equiv 21^2 \equiv 29^2 \equiv 31^2 \equiv 39^2 \equiv 41^2 \equiv 49^2 \pmod{20}$
$\; 2^2 \equiv 8^2 \equiv 12^2 \equiv 18^2 \equiv 22^2 \equiv 28^2 \equiv 32^2 \equiv 38^2 \equiv 42^2 \equiv 48^2 \pmod{20}$
$\; 3^2 \equiv 7^2 \equiv 13^2 \equiv 17^2 \equiv 23^2 \equiv 27^2 \equiv 33^2 \equiv 37^2 \equiv 43^2 \equiv 47^2 \pmod{20}$
$\; 4^2 \equiv 6^2 \equiv 14^2 \equiv 16^2 \equiv 24^2 \equiv 26^2 \equiv 34^2 \equiv 36^2 \equiv 44^2 \equiv 46^2 \pmod{20}$
$\text{*****}$
It is immediate that if $n \equiv 0 \pmod5$ then $n^2$ can't end in two odd digits.
Also, if the last two digits $n^2$ are not both odd and $m \equiv n \pmod{20}$ then the last two digits of $m^2$ can't both be odd.
Putting all the above together we proceed with the calculations,
$\quad 1^2 \equiv 1 \pmod{20}$
$\quad 2^2 \equiv 4 \pmod{20}$
$\quad 3^2 \equiv 9 \pmod{20}$
$\quad 4^2 \equiv 16 \pmod{20}$
and we see that for every integer $n$ the last two digits of $n^2$ can't both be odd.
We can also derive the result given by Raffaele in his comment below the OP's question.
If the units digit of an integer $n$ is $0$ then the last two digits of $n^2$ is equal to $00$.
If the units digit of an integer $n$ is $5$ then the last two digits of $n^2$ is equal to $25$.
The remaining possibilities can be found by looking at the four sequences generated (extrapolated) from proposition 2:
$\quad 01, 21, 41, 61, 81$
$\quad 04, 24, 44, 64, 84$
$\quad 09, 29, 49, 69, 89$
$\quad 16, 36, 56, 76, 96$
So there are a total of $22$ different endings.
Exercise:
Show that each of these $22$ endings occurs as the last two digits of the square of some integer.