$A$ positive integer is of the form $3q + 1$,where $q$ is a natural number. Can you write its square in any form other than $3m + 1$,i.e.,$3m$ or $3m + 2$ for some integer $m$? Justify your answer.

(B) No. According to Euclid's Division Lemma,any positive integer can be expressed in the form $3q, 3q + 1,$ or $3q + 2$ for some integer $q$.
Let us examine the squares of these forms:
$1.$ If the integer is $3q$,then its square is $(3q)^2 = 9q^2 = 3(3q^2) = 3m$,where $m = 3q^2$.
$2.$ If the integer is $3q + 1$,then its square is $(3q + 1)^2 = 9q^2 + 6q + 1 = 3(3q^2 + 2q) + 1 = 3m + 1$,where $m = 3q^2 + 2q$.
$3.$ If the integer is $3q + 2$,then its square is $(3q + 2)^2 = 9q^2 + 12q + 4 = 9q^2 + 12q + 3 + 1 = 3(3q^2 + 4q + 1) + 1 = 3m + 1$,where $m = 3q^2 + 4q + 1$.
Thus,the square of any positive integer is always of the form $3m$ or $3m + 1$. It can never be of the form $3m + 2$.