Prove the fractions aren't integers 
Prove that if $p$ and $q$ are distinct primes then $\dfrac{pq-1}{(p-1)(q-1)}$ is never an integer. Is it similarly true that if $p,q,r$ are distinct primes then $\dfrac{pqr-1}{(p-1)(q-1)(r-1)}$ is also never an integer?

I think using a modular arithmetic argument here would help. In other words, we must have for the first fraction $pq-1 \equiv 0 \pmod{(p-1)(q-1)}$. Then I am unsure how to proceed next since we can't really use the Chinese Remainder Theorem. 

Yves Daoust's answer covers the two-prime case as well as the three-prime case except when $p \in \{2,3\}$.  Here's a simple way to dispense with the remaining cases efficiently:

Suppose $\frac{pqr-1}{(p-1)(q-1)(r-1)}$ is an integer.  Then neither $q$ nor $r$ is congruent to $1$ mod $p$.

This is easy to see because the numerator is not divisible by $p$, so the denominator cannot have any factors of $p$.
The above observation immediately eliminates the case $p=2$.  For $p=3$, look at the numerator and denominator mod $3$.  The numerator is $2\pmod 3$, and so is the denominator (because $q$ and $r$ must be $2$ mod $3$ by above observation).  Thus the quotient, if it is an integer, must be $\equiv 1\pmod3$, and since it's $>1$ it must be $\ge 4$.  But the computation in Yves's answer shows that it can't possibly be that large.
[Note that $q$ and $r$ are forced to be $2 \pmod 3$ because they can't be $0 \pmod 3$: this is where we use the fact that $q$ and $r$ are prime, so the argument is consistent with the existence of non-prime solutions $(2,4,8)$ and $(3,5,15)$.]
