For what odd prime p is -3 a quadratic residues? Non-residue?

Question

For what odd prime p is -3 a quadratic residues? Non-residue?

• Having a bit of trouble with this question, we are currently covering a section on quadratic reciprocity and didn't really see anything in my notes that helped me solve this. Any help is greatly appreciated.

Answer 1

We have $-3$ is a QR of the odd prime $p\ne 3$ in two situations:

(i) $-1$ is a QR of $p$ and $3$ is a QR of $p$ or

(ii) $-1$ is a NR of $p$ and $3$ is a NR of $p$.

Case (i): $-1$ is a QR of $p$ if $p\equiv 1\pmod{4}$. Since $p\equiv 1\pmod{4}$, by Quadratic Reciprocity we have that the Legendre symbol $(3/p)$ is equal to $(p/3)$. Note that $(p/3)=1$ if $p\equiv 1\pmod{3}$, and $(p/3)=-1$ if $p\equiv -1\pmod{3}$.

So Case (i) occurs if $p\equiv 1\pmod{4}$ and $p\equiv 1\pmod{3}$, or equivalently if $p\equiv 1\pmod{12}$.

Case (ii): We must have $p\equiv 3\pmod{4}$. In this case, Reciprocity says that since $p$ and $3$ are both of the form $4k+3$, we have $(3/p)=-(p/3)$. Thus $(3/p)=-1$ if $p\equiv 1\pmod{3}$.

So Case (ii) happens if $p\equiv 3\pmod{4}$ and $p\equiv 1\pmod{3}$, or equivalently if $p\equiv 7\pmod{12}$.

We can put the two cases together, and say that $-3$ is a quadratic residue of the odd prime $p$ if $p\equiv 1\pmod{6}$.

Thus $-3$ is a quadratic non-residue of the odd prime $p$ if $p\equiv 5\pmod{6}$.

Answer 2

Hint: use the Legendre symbol. Then you have that $(\frac{-3}{p})=(\frac{-1}{p})(\frac{3}{p})$, and use what you know about $(\frac{-1}{p})$ and quadratic reciprocity.

Answer 3

By the multiplicity property of the Legendre symbol, $(\frac{-3}{p})=(\frac{-1}{p})*(\frac{3}{p})$.

Now $(\frac{-1}{p})=1$ if $p=1\mod 4$ and $(\frac{-1}{p})=-1$ if $p= 3 \mod 4$

Further, $(\frac{3}{p})=1$ if $p=1 $ or $ 11 \mod12$ and $(\frac{3}{p})=-1$ if $p=5$ or $ 7\mod 12$.