Show that the cube of any positive integer is of the form $4m, 4m+1$ or $4m+3$ for some integer $m$.

(N/A) Let $a$ be any positive integer. By Euclid's division lemma,for $a$ and $b=4$,there exist unique integers $q$ and $r$ such that $a = 4q + r$,where $0 \leq r < 4$.
Taking the cube of both sides:
$a^3 = (4q + r)^3 = 64q^3 + 3(4q)^2r + 3(4q)r^2 + r^3$
$a^3 = 64q^3 + 48q^2r + 12qr^2 + r^3$
$a^3 = 4(16q^3 + 12q^2r + 3qr^2) + r^3$ ... $(i)$
Case $I$: If $r = 0$,then $a^3 = 4(16q^3) = 4m$,where $m = 16q^3$.
Case $II$: If $r = 1$,then $a^3 = 4(16q^3 + 12q^2 + 3q) + 1^3 = 4m + 1$,where $m = 16q^3 + 12q^2 + 3q$.
Case $III$: If $r = 2$,then $a^3 = 4(16q^3 + 12q^2(2) + 3q(2^2)) + 2^3 = 4(16q^3 + 24q^2 + 12q) + 8 = 4(16q^3 + 24q^2 + 12q + 2) = 4m$,where $m = 16q^3 + 24q^2 + 12q + 2$.
Case $IV$: If $r = 3$,then $a^3 = 4(16q^3 + 12q^2(3) + 3q(3^2)) + 3^3 = 4(16q^3 + 36q^2 + 27q) + 27 = 4(16q^3 + 36q^2 + 27q + 6) + 3 = 4m + 3$,where $m = 16q^3 + 36q^2 + 27q + 6$.
Thus,the cube of any positive integer is of the form $4m, 4m+1$ or $4m+3$.