why do last 4 digits of any square can be equal only when all of them equal 0? From the statement below, how can the author say that "every natural number can be written in the form $50000k\pm r$ 
with $0\le r\le 25000$" .Is there a rule or theory about that, or It is just a counterexample?From all the numbers,why 50 000 and 25000?Can someone enlighten me? It's from the journal Oblath's Problem 
Since every natural number can be written in the form $50000k\pm r$ 
with $0\le r\le 25000$, and 
$(50000k\pm r)^2\equiv r^2\pmod {100000}$, 
we compute $r^2$ for $r\le 25000$  and find that the last 4 digits of any square 
can be equal only when all of them equal 0,
 A: It is a general fact, known as "Euclidean division" (Wikipedia), that for any integers $a$ and $b$ (with $b\neq 0$), there are unique integers $k$ and $r$ such that 
$$a=bk+r\qquad\text{ and }\qquad 0\leq r<|b|$$
Usually, the number $r$ is referred to as the "remainder" of dividing $a$ by $b$. However, this theorem can be slightly modified (see here in the Wikipedia article) to say that there are unique integers $k$ and $r$ such that
$$a=bk+r\qquad\text{ and }\qquad -\left\lfloor {\frac {b}{2}}\right\rfloor \leq r<b-\left\lfloor {\frac {b}{2}}\right\rfloor$$
which, for the choice of $b=50000$, says that any integer $a$ can be written uniquely as $a=50000k+r$
with 
$$-25000\leq r<25000$$
Now, in the statement you cited, the author has just slightly removed the uniqueness, and instead made the weaker (but no less valid) statement that any $a$ can be written as $a=50000k+r$ with
$$-25000\leq r\leq 250000$$
A: When last 2 digits are equal, the last digit can be only 4 ($12^2=144$). Just search a number which become ○○4444.  3rd digit of $ab12^2$ become 4b+1, this can't be 4. So such a number does not exist. 
