Conjecture on the floor-function sum for a product of powers of primes modulo 4

Let $p$ and $q$ be distinct primes, let $\alpha,\beta\geq 1$, and define

$$S_n:=\frac{1}{n}\sum_{k=1}^{n-1}\operatorname{Rem}(k^2\div n).$$

Also set $h^*(-p)=h(-p)$ for $p\geq 7$ and $h^*(-3)=1/3$. The mixed-congruence class-number conjecture. If $n=p^\alpha q^\beta$ with $p\equiv1\pmod{4}$ and $q\equiv3\pmod{4}$, then

$$S_n=\frac{n-p^{\lfloor\alpha/2\rfloor}q^{\lfloor\beta/2\rfloor}}{2}-\frac{q^{\lfloor(\beta+1)/2\rfloor}-1}{q-1}\left(\left(\frac{p^{\lfloor(\alpha+2)/2\rfloor}-1}{p-1}-\frac{p^{\lfloor\alpha/2\rfloor}-1}{p-1}\left(\frac pq\right)\right)h^*(-q)-\frac{p^{\lfloor(\alpha+1)/2\rfloor}-1}{p-1}h^*(-pq)\right).$$

This is one of the identities suggested by extensive computations for products having distinct prime factors, extending the previously proved prime-power cases; its resolution is not given in the supplied text.