Show that the square of any positive integer is either of the form $4q$ or $4q+1$ for some integer $q$.

(N/A) Let $a$ be an arbitrary positive integer. By Euclid's division algorithm,for integers $a$ and $4$,there exist non-negative integers $m$ and $r$ such that $a = 4m + r$,where $0 \leq r < 4$.
Squaring both sides,we get $a^2 = (4m + r)^2 = 16m^2 + 8mr + r^2 = 4(4m^2 + 2mr) + r^2$.
Case $I$: If $r = 0$,$a^2 = 4(4m^2) = 4q$,where $q = 4m^2$.
Case $II$: If $r = 1$,$a^2 = 4(4m^2 + 2m) + 1 = 4q + 1$,where $q = 4m^2 + 2m$.
Case $III$: If $r = 2$,$a^2 = 16m^2 + 16m + 4 = 4(4m^2 + 4m + 1) = 4q$,where $q = 4m^2 + 4m + 1$.
Case $IV$: If $r = 3$,$a^2 = 16m^2 + 24m + 9 = 16m^2 + 24m + 8 + 1 = 4(4m^2 + 6m + 2) + 1 = 4q + 1$,where $q = 4m^2 + 6m + 2$.
Thus,the square of any positive integer is of the form $4q$ or $4q + 1$.