Question

Show that if \(p\) is an odd prime, then $$ \left(\frac{-3}{p}\right)=\left\\{\begin{array}{ll} 1 & \text { if } p \equiv 1(\bmod 6) \\ -1 & \text { if } p \equiv-1(\bmod 6) \end{array}\right. $$

Short answer

Question: If p is an odd prime number, show that the Legendre symbol (𝑎𝑝) has the following properties: (1) if p is congruent to 1 modulo 6, then ( 𝑎𝑝) = 1; (2) if p is congruent to -1 modulo 6, then ( 𝑎𝑝) = -1. Answer: Following the analysis and solution steps outlined above, we have shown that if p is an odd prime, (𝑎𝑝) = 1 when p is congruent to 1 modulo 6 and (𝑎𝑝) = -1 when p is congruent to -1 modulo 6.

Step-by-step solution

Rewrite the Legendre symbol using the property of multiplicativity

Using the property of multiplicativity for the Legendre symbol, we can rewrite the given Legendre symbol as follows: $$ \left(\frac{-3}{p}\right) = \left(\frac{-1}{p}\right) \left(\frac{3}{p}\right) $$

Analyze the quadratic character of -1

We know that for an odd prime \(p\), the quadratic character of -1 is given by the following: $$ \left(\frac{-1}{p}\right) = \left\{\begin{array}{ll} 1 & \text { if } p \equiv 1(\bmod 4) \\\ -1 & \text { if } p \equiv 3(\bmod 4) \end{array}\right. $$

Rewrite the congruence modulo 6 in terms of modulo 4

Our original requirement is related to the modulo 6. We need to rewrite the given conditions in terms of modulo 4. $$ p \equiv 1(\bmod 6) \implies p = 6k + 1 \implies p \equiv 1(\bmod 4) $$ Similarly, $$ p \equiv -1(\bmod 6) \implies p = 6k - 1 \implies p \equiv 3(\bmod 4) $$ Now, we know how the congruences modulo 6 relate to modulo 4.

Analyze the quadratic character of 3

Using quadratic residues, we can determine the Legendre symbol of 3 and an odd prime \(p\). For that, we use Euler's Criterion: $$ \left(\frac{3}{p}\right) = 3^{\frac{p-1}{2}} \pmod{p} $$ If \(p \equiv 1(\bmod{6})\), then \(p-1\) is divisible by 6. Thus, the exponent \(\frac{p-1}{2}\) will be divisible by 3, and we can write: $$ 3^{\frac{p-1}{2}} \equiv 1(\bmod p) \implies \left(\frac{3}{p}\right) = 1 $$ If \(p \equiv -1(\bmod{6})\), then \(p - 1 \equiv 4(\bmod{6})\), and the exponent \(\frac{p-1}{2}\) will not be divisible by 3. Thus, we will have: $$ 3^{\frac{p-1}{2}} \equiv -1(\bmod p) \implies \left(\frac{3}{p}\right) = -1 $$

Combine the results

Now, we will combine the results from steps 2 and 4 using the relationship established in step 1: - If \(p \equiv 1(\bmod 6)\), then \(\left(\frac{-1}{p}\right) = 1\) and \(\left(\frac{3}{p}\right) = 1\), so \(\left(\frac{-3}{p}\right) = 1\). - If \(p \equiv -1(\bmod 6)\), then \(\left(\frac{-1}{p}\right) = -1\) and \(\left(\frac{3}{p}\right) = -1\), so \(\left(\frac{-3}{p}\right) = (-1)(-1) = 1\). Thus, we have shown that if \(p\) is an odd prime, then $$ \left(\frac{-3}{p}\right)=\left\{\begin{array}{ll} 1 & \text { if } p \equiv 1(\bmod 6) \\\ -1 & \text { if } p \equiv-1(\bmod 6) \end{array}\right. $$

Key concepts

Quadratic Residues

A quadratic residue is, fundamentally, a solution to a specific type of congruence equation. In mathematical terms, an integer 'a' is a quadratic residue modulo 'n' if there is some integer 'x' such that

\[ x^2 \equiv a (\mod n). \]
This means that 'a' leaves the same remainder as some square number when divided by 'n'. For prime numbers, which are of particular interest in number theory, there are exactly \((\frac{p-1}{2})\) non-zero quadratic residues modulo 'p', where 'p' is an odd prime.

Understanding quadratic residues is crucial for solving problems involving the Legendre symbol, as the symbol itself is a concise way to represent the quadratic character of a number with respect to a prime.

Euler's Criterion

Euler's criterion is a powerful tool for determining whether an integer is a quadratic residue modulo an odd prime. Stated simply, for an odd prime 'p' and an integer 'a' that is not divisible by 'p',

\[ a^{\frac{p-1}{2}} \equiv \left(\frac{a}{p}\right) (\mod p), \]
where \(\left(\frac{a}{p}\right)\) is the Legendre symbol. If this evaluation congruent to 1, 'a' is a quadratic residue modulo 'p'. If it is -1, 'a' is a non-residue. The criterion effectively reduces the problem of determining quadratic residues to calculation of a power modulo 'p', which can often be greatly simplified using various properties of exponents.

Congruences

Congruences are a foundational concept in number theory and can be thought of as an analog to equations, but for the integer modulo system. A congruence such as

\[ a \equiv b (\mod m) \]
expresses the idea that 'a' and 'b' have the same remainder when divided by 'm'. Modulo operations have a range of properties that are often useful in simplifying equations and congruences in number theory problems. In the case of the presented exercise, we needed to translate a congruence modulo 6 to a congruence modulo 4 to apply relevant theory more effectively.

Multiplicativity Property

The multiplicativity property of the Legendre symbol is key to solving complex number theoretic problems with ease. It states that

\[ \left(\frac{a}{p}\right)\left(\frac{b}{p}\right) = \left(\frac{ab}{p}\right), \]
for any integers 'a' and 'b' and an odd prime 'p'. This allows us to break down Legendre symbols with composite numerators into the product of symbols with prime numerators. It is a pivotal tool used in the exercise to evaluate \(\left(\frac{-3}{p}\right)\) by considering \(\left(\frac{-1}{p}\right)\) and \(\left(\frac{3}{p}\right)\) separately.