Show that the square of any positive integer cannot be of the form $5q+2$ or $5q+3$ for any integer $q$.

(A) Let $a$ be an arbitrary positive integer.
By Euclid's division algorithm,for integers $a$ and $5$,there exist non-negative integers $m$ and $r$ such that $a = 5m + r$,where $0 \leq r < 5$.
Squaring both sides,we get $a^2 = (5m + r)^2 = 25m^2 + 10mr + r^2 = 5(5m^2 + 2mr) + r^2$.
Let $q' = 5m^2 + 2mr$,then $a^2 = 5q' + r^2$.
We test all possible values of $r \in \{0, 1, 2, 3, 4\}$:
Case $I$: If $r=0$,$a^2 = 5(5m^2) = 5q$,which is of the form $5q$.
Case $II$: If $r=1$,$a^2 = 5(5m^2 + 2m) + 1 = 5q + 1$,which is of the form $5q+1$.
Case $III$: If $r=2$,$a^2 = 5(5m^2 + 4m) + 4 = 5q + 4$,which is of the form $5q+4$.
Case $IV$: If $r=3$,$a^2 = 5(5m^2 + 6m) + 9 = 5(5m^2 + 6m + 1) + 4 = 5q + 4$,which is of the form $5q+4$.
Case $V$: If $r=4$,$a^2 = 5(5m^2 + 8m) + 16 = 5(5m^2 + 8m + 3) + 1 = 5q + 1$,which is of the form $5q+1$.
In all cases,$a^2$ is of the form $5q, 5q+1,$ or $5q+4$. Thus,it cannot be of the form $5q+2$ or $5q+3$.