Show that the square of an odd positive integer is of the form $8m+1$ for some whole number $m$.

(N/A) Any positive odd integer can be represented in the form $2q+1$,where $q$ is a whole number.
Squaring this integer,we get:
$(2q+1)^2 = 4q^2 + 4q + 1$
$(2q+1)^2 = 4q(q+1) + 1$ $...(1)$
Since $q$ and $q+1$ are consecutive integers,their product $q(q+1)$ is always even. Therefore,we can write $q(q+1) = 2m$ for some whole number $m$.
Substituting this into equation $(1)$:
$(2q+1)^2 = 4(2m) + 1 = 8m + 1$.
Thus,the square of any odd positive integer is of the form $8m+1$ for some whole number $m$.