A number theory problem in RSA
If we know $e$, then it's easy.

If don't have to know the factors, but it does make it easier - as fgrieu pointed out, either $(p-1)/2$ or $(q-1)/2$ will appear as one of the prime factors, that will allow us to immediately recover $p, q$

Once we have that, then we know that we want:

$$y^{p_0} \equiv 1 \pmod p$$
$$y^{p_0} \equiv 1 \pmod q$$

If we have $y \equiv 4 \pmod p$ (or any other quadratic residue) and $p_0 = (p-1)/2$ (which is prime), then we satisfy the first equation.

If we have $y \equiv 1 \pmod q$, then we satisfy the second equation.

It's easy to find a $y$ that meets both modulii (as $p, q$ are relatively prime) using the Chinese Remainder Theorem, and so that's our answer.