Use Euclid's division lemma to show that the square of any positive integer is either of the form $3m$ or $3m+1$ for some integer $m$.

(N/A) Let $a$ be any positive integer and $b=3$.
By Euclid's division lemma,$a = 3q + r$,where $q \geq 0$ and $r \in \{0, 1, 2\}$.
Case $1$: If $r=0$,then $a = 3q$. Squaring both sides,$a^2 = (3q)^2 = 9q^2 = 3(3q^2) = 3m$,where $m = 3q^2$.
Case $2$: If $r=1$,then $a = 3q+1$. Squaring both sides,$a^2 = (3q+1)^2 = 9q^2 + 6q + 1 = 3(3q^2 + 2q) + 1 = 3m + 1$,where $m = 3q^2 + 2q$.
Case $3$: If $r=2$,then $a = 3q+2$. Squaring both sides,$a^2 = (3q+2)^2 = 9q^2 + 12q + 4 = 9q^2 + 12q + 3 + 1 = 3(3q^2 + 4q + 1) + 1 = 3m + 1$,where $m = 3q^2 + 4q + 1$.
Thus,the square of any positive integer is always of the form $3m$ or $3m+1$.