<s>[INST] Given odd positive integer, $n=2m-1$, $n \equiv 1 \pmod 4 \implies m \equiv 1 \pmod 2$ $n=2m-1\implies n+1=2m$. 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.       
 [/INST] $$n = 2m-1\equiv 1\pmod{4}\quad\Rightarrow\quad
  2m \equiv 2\pmod{4}.$$
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
$$m\equiv 1\pmod{2}.$$
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