An RSA number $n=p\cdot q$, where $q=2\cdot d +1$, $d$ an odd integer, is given. Assuming $a \in \mathbb{Z}_n$ with $a^4=1$ and $a^2 \neq 1$. How can this information lead to finding $p$ and $q$?
I said that $\phi(n) =(p-1)\cdot 2d$, and the order of an element divides the order of the group implies that $4|(p-1)$ or $4|2d$, where $\phi$ is Euler's totient function. But because $q=2d+1$ this would mean $4|(4m+2)$ for some integer $m$, which cannot be. So, $4|(p-1) \Rightarrow p= 1$ mod $4$. I'm not sure where to go from here.
As you noted, there are no fourth roots of unity $\bmod q$, hence $a^2\equiv 1\pmod q$. Then $a^2\ne1 $ implies $a^2\not\equiv 1\pmod p$, i.e., $a^2\equiv -1\pmod p$. Thus $p\mid a^2+1$ and certainly $p=\gcd(n,a^2+1)$.