Let $C=\prod (q_i),\ q_i \in \mathbb P$ be a Carmichael number. It is further true of Carmichael numbers that they have at least three prime factors and are square free, i.e all $q_i$ are distinct. For any positive integer $a$ coprime to $C$, $a^{(C-1)}\equiv 1 \bmod C$. Moreover, by Euler's theorem, $a^{\phi (C)}\equiv 1 \bmod C$, where $\phi(n)$ represents the totient function of $n$. For Carmichael numbers, $\phi (C)=\prod (q_i-1)$. So for Carmichael numbers $$a^{(\prod (q_i)-1)}\equiv a^{(\prod (q_i-1))}\equiv 1 \bmod C$$ I was considering whether it would be possible to test whether a number is a semiprime by the evaluating relationships of the form $a^{(p-1)(q-1)}\equiv 1 \bmod pq,\ p,q \in \mathbb P$, and I realized that I did not know (and so far have been unable to prove to myself one way or the other) whether such a test would be compromised if one or both of $p,q$ were Carmichael numbers, much as Carmichael numbers in isolation compromise primality testing by Fermat's little theorem.
Question: In the simplest case, for an exponent $pC$, I know that $a^{\phi (pC)}\equiv 1 \bmod (pC)$, but is it also the case that $a^{(p-1)(C-1)}\equiv 1 \bmod (pC)$?
If so, one can imagine exponents having a large number of prime factors could be treated (for some purposes) as if they contained fewer factors by accumulating some of the prime factors into one or more Carmichael numbers.
Added by edit: I have performed some calculations using the Carmichael number $C=1105=5\cdot 13\cdot 17$ and the prime $p=7$, yielding the exponent $pC=7735$. For each of $a=2,3,11$ it seems that $a^{(p-1)(C-1)}\equiv a^{\phi(pC)}\equiv 1 \bmod 7735$. So Carmichael numbers as factors of an exponent can compromise a test for semiprimes based on $a^{(p-1)(q-1)}\equiv 1 \bmod pq$.
My choice of $1105$ was predicated on the fact that $1103$ is prime, and I was interested in whether such a test could be applied to twin prime candidates. So a result that $a^{(1103-1)(1105-1)}\equiv 1 \bmod (1103\cdot 1105)$ does not indicate that $1103,1105$ are twin primes, a fact that would be otherwise obvious in this particular instance because $1105$ is plainly divisible by $5$ and not prime. The example was chosen to test the method, not this particular case.