For k = 3: The only valid pair satisfying $p < q$ is $p = 1$ and $q = 2$. Thus, $k = 3$ uniquely determines $p$ and $q$.

For k = 4: The possible pairs are $(1, 3)$ and $(2, 2)$. However, since $p < q$ is required, only the pair $(1, 3)$ is valid. Hence, $k = 4$ also uniquely determines $p$ and $q$.

For k = 5: The valid pairs satisfying $p < q$ are $(1, 4)$ and $(2, 3)$. Here, $k = 5$ does not uniquely determine $p$ and $q$, because there are two possible pairs.

The smallest value of k that does not determine $p$ and $q$ uniquely is $k = 5$.