Show that the square of any positive integer cannot be of the form $3 m+2$, where $m$ is a natural number..

By Euclid's lemma, $b=a q+r, 0 \leq r

Here, b is a positive integer and a = 3.

$\therefore b=3 q+r$, for $0 \leq r<3$

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

Now,

$(3 q)^{2}=9 q^{2}=3 m$, where $m=3 q^{2}$

$(3 q+1)^{2}=9 q^{2}+6 q+1=3\left(3 q^{2}+2 q\right)+1=3 m+1$, where $m=3 q^{2}+2 q$

$(3 q+2)^{2}=9 q^{2}+12 q+4=3\left(3 q^{2}+4 q+1\right)+1=3 m+1$, where $m=3 q^{2}+4 q+1$

Therefore, the square of a positive integer cannot be of the form 3m+2, where m is a natural number.