I am a bit confused with the question:
For what prime $p$, $\left(\frac{-21}{p}\right) = 1$?
I did something like that:
$$\left(\frac{-21}{p}\right) = \left(\frac{-1}{p}\right)\left(\frac{3}{p}\right)\left(\frac{7}{p}\right).$$
But I don't have any ideas what to do next. Thanks for any help.
Let $p$ be an odd prime not equal to $7$.
If $p\equiv 1\pmod{4}$, then $(7/p)=(p/7)$. But $(p/7)=1$ if $p\equiv 1,2,4\pmod{7}$. That gives, modulo $28$, $p\equiv 1, 9, 25$.
If $p\equiv 3\pmod{4}$, then $(7/p)=-(p/7)$. But $(p/7)=-1$ if $p\equiv 3,5,6\pmod{7}$. That gives, modulo $28$, $p\equiv 3, 19, 27$.
Now we can write down the $6$ congruence classes modulo $28$ for which $(7/p)=-1$.
It is not hard to show (and has been done more than once on MSE) that $(-3/p)=1$ if and only if $p\equiv 1\pmod{3}$.
Now we have a largish number of "Chinese Remainder Theorem" problems to solve modulo $84$. For we must combine the $6$ possibilities for $(7/p)=1$ with $p\equiv 1\pmod{3}$, and the $6$ possibilities for $(7/p)=-1$ with $p\equiv 2\pmod{3}$. All of them are pretty quick, but there are $12$.