If $p$ and $q$ are primes of the form $4k+3$, and $x^2 \equiv p \pmod q$ has no solutions, show $x^2 \equiv q \pmod p$ has two incongruent solutions

Question

Prove or disprove:

If $p$ and $q$ are primes of the form $4k+3$, and $x^2 \equiv p \pmod q$ has no solutions, show $x^2 \equiv q \pmod p$ has two incongruent solutions

I think it is a true statement $(p,q)=1$ and $p\equiv 3 \pmod 4$ and $q\equiv 3 \pmod 4$.

Since $x^2 \equiv p \pmod q$ has no solution then $(p/q)=-1$ and since $p\equiv 3 \pmod 4$ and $q\equiv 3 \pmod 4$ then $(q/p)=1$

Hence, $x^2$ ≡ q (mod p) has two incongruent solutions.

is this right?