Prove the fractions aren't integers

Suppose, for the sake of contradiction, such distinct $p$ and $q$ exist. First of all observe that the statement implies that $p-1|pq-1$. So,

$$p-1|pq-1-q(p-1) \implies p-1|q-1$$

Similarly we get,

$$q-1|p-1$$

These observations imply that $p-1 = q-1$. This implies that $p = q$. Contradiction. They aren't distinct.

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.

$$\frac{pq-1}{(p-1)(q-1)}-1=\frac1{p-1}+\frac1{q-1}$$cannot be an integer, as the only possible values would be $2$ ($p=q=2$) or $1$ ($p=q=3$).

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Similarly,

$$\frac{pqr-1}{(p-1)(q-1)(r-1)}-1\\ =\frac1{p-1}+\frac1{q-1}+\frac1{r-1}+\frac1{(p-1)(q-1)}+\frac1{(q-1)(r-1)}+\frac1{(r-1)(p-1)}$$ cannot be an integer for $p,q,r>4$ as the sum wouldn't exceed $\dfrac{15}{16}$.

The remaining possibilities for $p<q<r$ are $2,q,r$ and $3,q,r$. The maximum value of the sum is achieved for $2,3,5$ and equals $\dfrac{21}8$, hence the only possible integer values are $2$ and $1$.

To be continued.