Find all odd primes $p$ for which $15$ is a quadratic residue modulo $p$ I want to find all odd primes $p$ for which $15$ is a quadratic residue modulo $p$.
My thoughts so far: I want to find $p$ such that $ \left( \frac{15}{p} \right) = 1$. By multiplicativity of the Legendre symbol, this is equivalent to $ \left( \frac{5}{p} \right) \left( \frac{3}{p} \right) = 1 $. Using the Law of Quadratic Reciprocity, this is equivalent to finding $p$ such that $ - \left( \frac{p}{5} \right) \left( \frac{p}{3} \right) = 1$.
So there are two cases:


*

*$ \left( \frac{p}{5} \right) = -1, \left( \frac{p}{3} \right) = 1$.

*$ \left( \frac{p}{5} \right) = 1,  \left( \frac{p}{3} \right) = -1 $. 
For case (1.), the quadratic residues modulo $5$ are $1$ and $4$, so for $ \left( \frac{p}{5} \right ) = -1$, we must have that $p$ is $2$ or $3$ modulo $5$. We must also have that $p$ is $1$ modulo $3$ from the other condition. One of these pairs is incompatible, and we can solve to give $p$ is $13$ modulo $15$.
Similarly for case (2.)

Is this the correct approach? I'm unsure if each step in my working is an "if and only if". If $p$ is $13$ modulo $15$, is $15$ necessarily a quadratic residue modulo $p$?
Thanks!

You must also consider your primes mod 4, as that alters the behavior of reciprocity. So, in the order I found them, all values are
$$ 1,49; \; 17, 53; \; 11, 59; \; 7, 43 \pmod {60}.$$ In ordinary numerical order,
$$1, 7,11,17,43,49, 53,59 \pmod {60}.   $$
