Given odd positive integer, $n=2m-1$, $n \equiv 1 \pmod 4 \implies m \equiv 1 \pmod 2$

Question

$n=2m-1\implies n+1=2m$.
So, $n\equiv 1\pmod 4, 2m=n+1\equiv 2\pmod 4\implies 2m=n+1\equiv 0\pmod 2$ $\implies n\equiv -1\pmod 2\implies n\equiv 1\pmod 2$.
But, how to find $m$ from this last line of equivalence relations is not clear.

Answer 1

$$n = 2m-1\equiv 1\pmod{4}\quad\Rightarrow\quad 2m \equiv 2\pmod{4}.$$ Now, note that in general, if $k\ne 0$ and $ak\equiv bk\mod{ck}$, this means that $ak$ is $bk$ plus a multiple of $ck$, say $rck$: $$ak = bk + rck.$$ Dividing through by $k$ gives $a = b+cr$, so that $a\equiv b\mod c$. Applying this to $2m\equiv 2\mod{(4 = 2\cdot 2)}$ gives $$m\equiv 1\pmod{2}.$$