Show that the square of any positive integer cannot be of the form $$(5q+2)$$ or $$(5q+3)$$ for any integer $$q$$.
Let a be any positive integer
By Euclid's division lemma
$$a = bm + r$$
$$a = 5m + r$$ (when $$b = 5$$)
So, $$r$$ can be $$0,1,2,3,4$$
Case 1:
$$\therefore a = 5m$$ (when $$r = 0$$)
$$a^2 = 25m^2$$
$$a^2 = 5 (5m^2)=5q$$
where $$q = 5m^2$$
Case 2:
when $$r=1$$
$$a = 5m+1$$
$$a^2 = (5m + 1)^2 = 25m^2 + 10m + 1$$
$$a^2 = 5(5m^2 + 2m)+1$$
$$= 5q+1$$ where $$q = 5m^2 + 2m$$
Case 3:
$$a = 5m+2$$
$$a^2 = 25m^2 + 20m + 4$$
$$a^2 = 5(5m^2 + 4m)+4$$
$$5q + 4$$
where $$q = 5m^2 + 4m$$
Case 4:
$$a = 5m + 3$$
$$a^2 = 25m^2 + 30m + 9$$
$$=25m^2 + 30m + 5+4$$
$$= 5(5m^2 + 6m +1)+4$$
$$= 5q+4$$
where $$q = 5m^2+ 6m+1$$
Case 5:
$$a = 5m+4$$
$$a^2 = 25m^2 + 40m + 16 = 25m^2 + 40m + 15 + 1$$
$$= 5(5m^2 + 8m + 3)+1$$
$$= 5q+1$$ where $$q = 5m^2 + 8m + 3$$
From these cases, we see that square of any positive no can't be of the form $$5q +2, \, 5q+3$$.