Solution:
Let’s say that $n$ be any arbitrary positive odd integer.
Upon dividing $n$ by 4, let $m$ be the quotient and $r$ be the remainder.
Therefore, by Euclid’s division lemma, we get
$n=4 m+r$, where $0 \leq r<4$
As $0 \leq r<4$ and $r$ is an integer, $r$ can take values $0,1,2,3$.
$\Rightarrow n=4 m$ or $n=4 m+1$ or $\mathrm{n}=4 m+2$ or $n=4 m+3$
But $n \neq 4 m$ or $n \neq 4 m+2(\because 4 m, 4 m+2$ are multiples of 2, so an even integer whereas $n$ is an odd integer)
$\Rightarrow n=4 m+1 \text { or } n=4 m+3$
As a result, any positive odd integer is of the form $(4 m+1)$ or $(4 m+3)$, where $m$ is some integer.