Suppose $p,q$ are distinct primes with least quadratic non-residues $n\_p$ and $n\_q$ respectively. Can one bound the least $n$ for which $\left(\frac{n}{p}\right)=\left(\frac{n}{q}\right)=-1$ in terms of $n\_p$ and $n\_q$?


I had originally thought this would be a consequence of quadratic reciprocity and the Chinese Remainder Theorem, and so would be bounded by $n\_pn\_q$, but I can't seem to work it out. 



Least simultaneous quadratic non-residue
Unless I'm mistaken, this is quite easy and only involves the multiplicative property of the quadratic residue.


If $n\_p=n\_q$ then we are done, because this very number is a simultaneous quadratic nonresidue. Otherwise we can without loss of generality suppose $n\_p<n\_q$. Then $(n\_p|p)=-1$ while $(n\_p|q)=1$. Also we know $(n\_q|q)=-1$.
If we also have $(n\_q|p)=-1$ then $n\_q$ is a simultaneous quadratic nonresidue.
Otherwise $(n\_q|p)=1$. Then by multiplicativity we have $(n\_pn\_q|p)=(n\_pn\_q|q)=-1$, so $n\_pn\_q$ is a simultaneous quadratic nonresidue.


