Let $n$ be an arbitrary odd number; we want to show that $n$ has the form $4q+1$ or $4q+3$.
But what does an "odd number" mean? It means something of the form $2m+1$ (for a whole number $m$), so the assumption is really $$n=2m+1$$
Now, $m$ itself is either even or odd. If we can show that no matter whether $m$ even or od then $n$ has one of the required forms, we will be through. If $m$ is even then $m=2k$ for some $k$, so
Then $k$ works as the $q$ we were looking for, which completes this case of the proof.
On the other hand, if $m$ is odd, that means $m=2k+1$ for some $k$, and then
Then again $k$ works as $q$, and the proof is finished.
Alternative proof. We can also show the required implication in "contrapositive" form:
If $n$ does not have one of the forms $4q+1$ or $4q+3$, then $n$ is even.
We can always divide $n$ by 4 and get an integral quotient and remainder instead of continuing into decimans. This allows us to express any $n$ as
where $r$ is either zero or a positive number less than $4$, that is, one of the numbers 0, 1, 2, or 3. By assumption $r=1$ and $r=3$ cannot be the case, so we have either $n=4q+0$ or $n=4q+2$. In either case $n$ is the sum of two even numbers ($4q$ is always even, and 0 and 2 are also even), and that means that $n$ is even too.