Let $r$ be the difference of arithmetic progression. $r|p^{24}-p^{23}\implies r|p-1$ since $r|p\implies r|q$ and $r|p^{24}-q^{24}$ where both cannot hold at the same time. $r|p-1\implies r|p^{23}-1\implies r|p^{24}-p\implies p$ is in aritmetic progression. We can apply same process for $q$. @Below, if we say $r=\frac{a}{b}$ since $r$ is rational, we can apply same with $a$ I think.

bin_sherlo wrote: Let $r$ be the difference of arithmetic progression. $r|p^{24}-p^{23}\implies r|p-1$ since $r|p\implies r|q$ and $r|p^{24}-q^{24}$ where both cannot hold at the same time. $r|p-1\implies r|p^{23}-1\implies r|p^{24}-p\implies p$ is in aritmetic progression. We can apply same process for $q$. In fact, $r$ is not necessary integer.

Notice that the integer terms within a arithmetic progression form themselves an arithmetic progression. Let the ratio of this progression be $r$, then we must have $$p^{23}\equiv p^{24}\equiv q^{23}\equiv q^{24}\pmod{r}$$Clearly $p$ and $q$ do not divide $r$ so we get $p\equiv q\equiv 1$ and we are done as $$p\equiv p^{23} \equiv p^{24}\equiv q \equiv q^{23} \equiv q^{24}\equiv 1 \pmod{r}$$