cubic residues mod pq, pq distinct primes I have shown that the number of cubic residues mod $pq$ (including zero) is $(\frac{p-1}{3}+1)(\frac{q-1}{3}+1)$ where $p \equiv q \equiv 1$ (mod 3), using the Chinese Remainder Theorem. But how can I show that there are $(\frac{p-1}{3})(\frac{q-1}{3})$ non-zero cubic residues mod $pq$? It seems that I have to subtract (# non-zero cubic residues (mod $p$) + # non-zero cubic residues (mod $q$) + $1$) from the number of cubic residues mod $pq$ (including zero), which I cannot seem to show.
 A: Well, you need to go into the proof,to go into the proof, the Chinese remainder theorem :
$$\mathbb{Z}_{pq}\rightarrow \mathbb{Z}_p\times\mathbb{Z}_q $$
$$a\mapsto (a_p,a_q):=(a\text{ mod } p,a\text{ mod } q) $$
This map being an isomorphism we get that $a$ is a cubic residue if and only if both $a_p$ and $a_q$ are. 
Now among cubic residues mod a prime $r$ number you have the non-zero and zero so you get $\frac{r-1}{3}+1$ cubic residues (provided that $3$ divides $r-1$). 
So on the whole you get $\frac{p-1}{3}+1$ choices for $a_p$ and $\frac{q-1}{3}+1$ choices for $a_q$ leading to the formula you gave us (I think this is what you have done so far).
Now for your question, among cubic residues there is only one which is zero : $0$. So that the number of non-zero cubic residues is $(\frac{p-1}{3}+1)(\frac{q-1}{3}+1)-1$. 
What the number $(\frac{p-1}{3})(\frac{q-1}{3})$ represents is another thing it is the number of cubic residues $a$ for which neither $a_p=0$ nor $a_q=0$ so this is the number of invertible (in $\mathbb{Z}_{pq}$) cubic residues.
