A positive integer is of the form 3q +1,

No, by Euclid's Lemma, $b=a q+r, 0 \leq r

Here, $b$ is any positive integer $a=3, b=3 q+r$ for $0 \leq r<3$

So, this must be in the form $3 q, 3 q+1$ or $3 q+2$.

$\begin{array}{lll}\text { Now, } & (3 q)^{2}=9 q^{2}=3 m & {\left[\text { here, } m=3 q^{2}\right]}\end{array}$

and $\quad(3 q+1)^{2}=9 q^{2}+6 q+1$'

$=3\left(3 q^{2}+2 q\right)+1=3 m+1 \quad$ [where, $\left.m=3 q^{2}+2 q\right]$

Also,

$(3 q+2)^{2}=9 q^{2}+12 q+4$

$=9 q^{2}+12 q+3+1$

$=3\left(3 q^{2}+4 q+1\right)+1$

$=3 m+1$ [here, $\left.m=3 q^{2}+4 q+1\right]$

Hence, square of a positive integer is of the form $3 \mathrm{~g}+1$ is always in the form $3 \mathrm{~m}+1$ for some integer $\mathrm{m}$.