Legendre Symbol Rules for (3/p) and (6/p)

# Example Questions Regarding Legendre Symbols

Let's first recall the following properties of Legendre symbols:

• A) The Legendre symbol $(a/p)$ is defined only if p is an odd prime and a is not a multiple of p ($p \nmid a$).
• B) If $a \equiv b \pmod p$, then $(a/p) = (b/p)$.
• C) $(a^2/p) = 1$ since $x^2 \equiv a^2 \pmod p$ has solutions, namely the least residue of a (mod p).
• D) $(ab/p) = (a/p)(b/p)$.
• E) $(-1/p) = 1$ if $p \equiv 1 \pmod 4$.
• F) $(-1/p) = -1$ if $p \equiv 3 \pmod 4$.
• G) $(p/q) = -(q/p)$ if $p \equiv q \equiv 3 \pmod 4$. Otherwise $(p/q) = (q/p)$.
• H) $(2/p) = 1$ if $p \equiv 1 \pmod 8$ or $p \equiv 7 \pmod 8$.
• I) $(2/p) = -1$ if $p \equiv 3 \pmod 8$ or $p \equiv 5 \pmod 8$.

We are now going to derive some rules for the Legendre symbols (3/p) and (6/p).

## Legendre Symbol (3/p)

Determine a rule for the Legendre symbol $(3/p)$ where p ≠ 3 with the rule being in modulo 12.

We first note that p must be an odd prime by the definition of a Legendre symbol. Hence p is NOT congruent to 2, 4, 6, 8, or 10 (mod 12). Additionally, since p ≠ 3, it thus follows that p is NOT congruent to 3, 6, or 9 (mod 12). Hence our prime p will be congruent to 1, 5, 7, or 11 (mod 12).

Case 1: Suppose that $p \equiv 1 \pmod {12}$. It thus follows that $p \equiv 1 \pmod {4}$, and $p \equiv 1 \pmod {3}$. Hence by property (G), $(3/p) = (p/3)$. Additionally, by property (A) and property (C), we get that $(3/p) = (p/3) = (1/3) = 1$. We will now look at the other three cases.

Case 2: Suppose that $p \equiv 5 \pmod {12}$. It thus follows that $p \equiv 1 \pmod {4}$, and $p \equiv 2 \pmod {3}$. Hence $(3/p) = (p/3) = (2/3) = -1$ by properties (H), (A), and (I).

Case 3: Suppose that $p \equiv 7 \pmod {12}$. It thus follows that $p \equiv 3 \pmod {4}$, and $p \equiv 1 \pmod {3}$. Hence $(3/p) = -(p/3) = -(1/3) = -1$ by properties (H), (A), and (C).

Case 4: Suppose that $p \equiv 11 \pmod {12}$. It thus follows that $p \equiv 3 \pmod {4}$ and $p \equiv 2 \pmod {3}$. Hence $(3/p) = -(p/3) = -(2/3) = 1$ by properties (H), (A), (I).

We can now summarize these cases in the following manner:

(1)
\begin{align} (3/p) = \begin{Bmatrix} 1 & \mathbf{if} \quad p \equiv 1 \pmod {12} \quad \mathbf{or} \quad p \equiv 11 \pmod {12}\\ -1 & \mathbf{if} \quad p \equiv 5 \pmod {12} \quad \mathbf{or} \quad p \equiv 7 \pmod {12}\\ \end{Bmatrix} \end{align}

## Legendre Symbol (6/p)

Determine a rule for the Legendre symbol (6/p) where p ≠ 3 with the rule being in modulo 24.

We should first note that $(6/p) = (2/p)(3/p)$ by property(D). Once again, since p is an odd prime, p cannot be congruent to 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, or 22. Additionally, since p ≠ 3, p cannot be congruent to 3, 6, 9, 12, 15, 18, or 21. Hence p can be congruent to 1, 5, 7, 11, 13, 17, 19, or 23 (mod 24).

Case 1: Suppose that $p \equiv 1 \pmod {24}$. Thus $p \equiv 1 \pmod {8}$ and $p \equiv 1 \pmod {12}$. Hence $(6/p) = (1)(1) = 1$.

Case 2: Suppose that $p \equiv 5 \pmod {24}$. Thus $p \equiv 5 \pmod {8}$ and $p \equiv 5 \pmod {12}$. Hence $(6/p) = (-1)(-1) = 1$.

Case 3: Suppose that $p \equiv 7 \pmod {24}$. Thus $p \equiv 7 \pmod {8}$ and $p \equiv 7 \pmod {12}$. Hence $(6/p) = (1)(-1) = -1$.

Case 4: Suppose that $p \equiv 11 \pmod {24}$. Thus $p \equiv 3 \pmod {8}$ and $p \equiv 11 \pmod {12}$. Hence $(6/p) = (-1)(1) = -1$.

Case 5: Suppose that $p \equiv 13 \pmod {24}$. Thus $p \equiv 5 \pmod {8}$ and $p \equiv 1 \pmod {12}$. Hence $(6/p) = (-1)(1) = -1$.

Case 6: Suppose that $p \equiv 17 \pmod {24}$. Thus $p \equiv 7 \pmod {8}$ and $p \equiv 5 \pmod {12}$. Hence $(6/p) = (1)(-1) = -1$.

Case 7: Suppose that $p \equiv 19 \pmod {24}$. Thus $p \equiv 3 \pmod {8}$ and $p \equiv 7 \pmod {12}$. Hence $(6/p) = (-1)(-1) = 1$.

Case 8: Suppose that $p \equiv 23 \pmod {24}$. Thus $p \equiv 7 \pmod {8}$ and $p \equiv 11 \pmod {12}$. Hence $(6/p) = (1)(1) = 1$.

We can now summarize the following cases with this general rule:

(2)
\begin{align} (6/p) = \begin{Bmatrix} 1 & \mathbf{if} \quad p \equiv 1, 5, 19, \: \mathbf{or} \: 23 \pmod {24} \\ -1 & \mathbf{if} \quad p \equiv 7, 11, 13, \: \mathbf{or} \: 17 \pmod {24} \end{Bmatrix} \end{align}