Show that the square of an odd positive integer can be of the form $6q + 1$ or $6q + 3$ for some integer $q$.

(N/A) We know that any positive integer can be expressed in the form $6m, 6m + 1, 6m + 2, 6m + 3, 6m + 4,$ or $6m + 5$ for some integer $m$.
An odd positive integer must be of the form $6m + 1, 6m + 3,$ or $6m + 5$.
Now,we calculate the squares of these forms:
$1.$ $(6m + 1)^2 = 36m^2 + 12m + 1 = 6(6m^2 + 2m) + 1 = 6q + 1$,where $q = 6m^2 + 2m$ is an integer.
$2.$ $(6m + 3)^2 = 36m^2 + 36m + 9 = 36m^2 + 36m + 6 + 3 = 6(6m^2 + 6m + 1) + 3 = 6q + 3$,where $q = 6m^2 + 6m + 1$ is an integer.
$3.$ $(6m + 5)^2 = 36m^2 + 60m + 25 = 36m^2 + 60m + 24 + 1 = 6(6m^2 + 10m + 4) + 1 = 6q + 1$,where $q = 6m^2 + 10m + 4$ is an integer.
Thus,the square of any odd positive integer is always of the form $6q + 1$ or $6q + 3$.