Write whether the square of any positive integer can be of the form $3m + 2$,where $m$ is a natural number. Justify your answer.

(N/A) No,the square of any positive integer cannot be of the form $3m + 2$.
By Euclid's division lemma,any positive integer $b$ can be expressed as $b = 3q + r$,where $0 \leq r < 3$. Thus,any positive integer is of the form $3k, 3k + 1$,or $3k + 2$.
Case $1$: If $b = 3k$,then $b^2 = (3k)^2 = 9k^2 = 3(3k^2) = 3m$,where $m = 3k^2$.
Case $2$: If $b = 3k + 1$,then $b^2 = (3k + 1)^2 = 9k^2 + 6k + 1 = 3(3k^2 + 2k) + 1 = 3m + 1$,where $m = 3k^2 + 2k$.
Case $3$: If $b = 3k + 2$,then $b^2 = (3k + 2)^2 = 9k^2 + 12k + 4 = 9k^2 + 12k + 3 + 1 = 3(3k^2 + 4k + 1) + 1 = 3m + 1$,where $m = 3k^2 + 4k + 1$.
In all cases,the square of a positive integer is either of the form $3m$ or $3m + 1$. Therefore,it cannot be of the form $3m + 2$.