<s>[INST] 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. 
 [/INST] I will assume that
$p \le q$,
not just
$p < q$.
Also,
you don't need the assumption
about primality.
I will show that 
the only solutions are
$p=q=2$
and
$p=q=3$.
If
$p=2$
then
$\dfrac{pq-1}{(p-1)(q-1)}
=\dfrac{2q-1}{q-1}
=\dfrac{2q-2+1}{q-1}
=2+\dfrac{1}{q-1}
$
which is not an integer
unless $q = 2$.
If
$p=3$
then
$\begin{array}\\
\dfrac{pq-1}{(p-1)(q-1)}
&=\dfrac{3q-1}{2(q-1)}\\
&=\dfrac{3q-3+2}{2q-2}\\
&=\dfrac32+\dfrac{1}{q-1}\\
\end{array}
$
which is not an integer
if $q \ge 4$.
If $q=3$,
then
$\dfrac{pq-1}{(p-1)(q-1)}
=\dfrac{3\cdot 3-1}{2\cdot 2}
=2
$
which is an integer.
If $p > 3$
then
$\begin{array}\\
\dfrac{pq-1}{(p-1)(q-1)}
&=\dfrac{pq-1}{pq-p-q+1}\\
&=\dfrac{pq-p-q+1+p+q-2}{pq-p-q+1}\\
&=1+\dfrac{p+q-2}{pq-p-q+1}\\
\end{array}
$
so we are done if
$p+q-2
<pq-p-q+1
$
or
$0
<pq-2p-2q+3
=(p-2)(q-2)-1
$
and this is true.
</s>