Prove that the square of any odd positive integer is of the form $8m + 1$,where $m$ is a non-negative integer.

Let $a$ be an odd positive integer. By Euclid's division lemma,any positive integer $a$ can be expressed as $a = 4q + r$,where $r \in \{0, 1, 2, 3\}$.
Since $a$ is odd,$r$ must be $1$ or $3$.
Case $1$: If $a = 4q + 1$,then $a^2 = (4q + 1)^2 = 16q^2 + 8q + 1 = 8(2q^2 + q) + 1$. Let $m = 2q^2 + q$,then $a^2 = 8m + 1$.
Case $2$: If $a = 4q + 3$,then $a^2 = (4q + 3)^2 = 16q^2 + 24q + 9 = 16q^2 + 24q + 8 + 1 = 8(2q^2 + 3q + 1) + 1$. Let $m = 2q^2 + 3q + 1$,then $a^2 = 8m + 1$.
In both cases,the square of an odd positive integer is of the form $8m + 1$.