Consequence of a change on the Guillou-Quisquater Protocol In the Guillou-Quisquater Protocol, the prover convinces the verifier that he knows an $e$-th root of an element $y \in \mathbb{Z}^*_n$ ($p, q, e$ are primes, $n = pq$ and $e$ is coprime with $\varphi(n)$). The protocol between prover $P$ and verifier $V$ (see here for instance) is:


$P$ picks $r$ at random in $\mathbb{Z}^*_n$ and sends $a = r^e \mod n$ to $V$.
$V$ picks $c$ at random in $\mathbb{Z}_e$ and sends it to $P$.
$P$ computes $z = a x^c$ and sends it to $V$.
$V$ accepts iff $z^e = a y^c$.


What happens if $e$ divides $\varphi(n)$ and the protocol still succeeds (i.e. $V$ has received an $a$, sent back $c$, and get $z$ s.t. $z^e = a y^c$)? I don't think it still proves that $P$ knows an $e$-th root of $y$... It is even possible that $y$ doesn't have any $e$-th root (since $e$ and $\varphi(n)$ aren't coprime). But does it proves something (new) about $y$?
Yes it proves that $y \ne 1 \pmod{n}$ is not an $e$-th root. That is, no $x$ exists such that $x^e = y \pmod{n}$. In particular, if $e | \phi(n)$ then for any $x$ it holds that
$\left( x^{ \frac{\phi(n)} {e}} \right)^e = 1 \ne y$.

This probably means one last final conclusion designed as a homework.

Regarding GQ protocol: prover will be unable to pick an $r$ such that
$a = r^e \ne 1$.