A sentence I was reading reads "Let $p$ be a prime factor of $n$. Take another prime $l$ with $l\equiv1\pmod{2p}$ so that $-1$ is a $p$-th power residue mod $l$."
Was it saying that for any prime $l$ satisfying $l\equiv1\pmod{2p}$, $-1$ is a $p$-th power residue mod $l$? Or was it saying to take a prime $l$ such that both (1) $l\equiv1\pmod{2p}$ and (2) $-1$ is a $p$-th power residue mod $l$ hold?
If $l$ is odd and $r$ is a primitive root modulo $l$, then $r^{l-1}\equiv 1 \pmod{l} $. So if $x = r^{(l-1)/2},$ we have $x^2 \equiv 1 \pmod{l}.$ The only solutions are $x\equiv \pm 1\ pmod{l}$, but since $r$ is a primitive root, we must have $x\equiv -1 \pmod{l}.$
Now note that $l\equiv 1 \pmod{2p}$ means that $l-1 = 2pd$ for some $d,$ which can be written $(l-1)/2 =pd.$ Then you have
$$-1 \equiv x \equiv r^{(l-1)/2} \equiv r^{pd} \equiv (r^d)^p \pmod{l}, $$
showing that your first condition implies the second.