Prove that the square of every integer is of the form $3m$ or $3m+1$,where $m \in \mathbb{Z}$.

(N/A) Let $a$ be any integer. According to Euclid's division lemma,for any integer $a$ and divisor $b=3$,there exist unique integers $q$ and $r$ such that $a = 3q + r$,where $0 \le r < 3$.
Thus,the possible values for $r$ are $0, 1, 2$.
Case $1$: If $r = 0$,then $a = 3q$. Squaring both sides,$a^2 = (3q)^2 = 9q^2 = 3(3q^2)$. Let $m = 3q^2$,then $a^2 = 3m$.
Case $2$: If $r = 1$,then $a = 3q + 1$. Squaring both sides,$a^2 = (3q + 1)^2 = 9q^2 + 6q + 1 = 3(3q^2 + 2q) + 1$. Let $m = 3q^2 + 2q$,then $a^2 = 3m + 1$.
Case $3$: If $r = 2$,then $a = 3q + 2$. Squaring both sides,$a^2 = (3q + 2)^2 = 9q^2 + 12q + 4 = 9q^2 + 12q + 3 + 1 = 3(3q^2 + 4q + 1) + 1$. Let $m = 3q^2 + 4q + 1$,then $a^2 = 3m + 1$.
In all cases,the square of an integer is of the form $3m$ or $3m+1$.