Show that the square of any positive integer cannot be of the form $6m+2$ or $6m+5$ for any integer $m$.

(N/A) Let $a$ be an arbitrary positive integer. By Euclid's division algorithm,for positive integers $a$ and $6$,there exist non-negative integers $q$ and $r$ such that $a = 6q + r$,where $0 \leq r < 6$.
Squaring both sides,we get $a^2 = (6q + r)^2 = 36q^2 + 12qr + r^2 = 6(6q^2 + 2qr) + r^2$. Let this be equation $(i)$.
Case $I$: If $r = 0$,$a^2 = 6(6q^2) = 6m$,where $m = 6q^2$.
Case $II$: If $r = 1$,$a^2 = 6(6q^2 + 2q) + 1 = 6m + 1$,where $m = 6q^2 + 2q$.
Case $III$: If $r = 2$,$a^2 = 6(6q^2 + 4q) + 4 = 6m + 4$,where $m = 6q^2 + 4q$.
Case $IV$: If $r = 3$,$a^2 = 6(6q^2 + 6q) + 9 = 6(6q^2 + 6q + 1) + 3 = 6m + 3$,where $m = 6q^2 + 6q + 1$.
Case $V$: If $r = 4$,$a^2 = 6(6q^2 + 8q) + 16 = 6(6q^2 + 8q + 2) + 4 = 6m + 4$,where $m = 6q^2 + 8q + 2$.
Case $VI$: If $r = 5$,$a^2 = 6(6q^2 + 10q) + 25 = 6(6q^2 + 10q + 4) + 1 = 6m + 1$,where $m = 6q^2 + 10q + 4$.
Thus,the square of any positive integer can only be of the form $6m, 6m+1, 6m+3,$ or $6m+4$. Therefore,it cannot be of the form $6m+2$ or $6m+5$.