Math Problem Statement
If p is an odd prime, then(a/p)={1 if p ≡ 1 (mod 8) or p ≡ 7 (mod 8), −1 if p ≡ 3 (mod 8) or p ≡ 5 (mod 8). prove it
Solution
The given problem is about proving a result related to the Legendre symbol , which is used to determine whether is a quadratic residue modulo , where is an odd prime. We are given that:
\begin{cases} 1 & \text{if } p \equiv 1 \pmod{8} \text{ or } p \equiv 7 \pmod{8}, \\ -1 & \text{if } p \equiv 3 \pmod{8} \text{ or } p \equiv 5 \pmod{8}. \end{cases}$$ To prove this, we need to show that the behavior of the Legendre symbol $$\left( \frac{a}{p} \right)$$ depends on the residue of $$p$$ modulo 8. For this proof, we will use properties from quadratic reciprocity and specific values for $$a$$. ### Step 1: Understanding the Case for $$a = 2$$ Since $$a$$ is not specified, let’s assume $$a = 2$$ for simplicity. The Legendre symbol $$\left( \frac{2}{p} \right)$$ tells us whether 2 is a quadratic residue modulo $$p$$. According to the properties of the Legendre symbol, the value of $$\left( \frac{2}{p} \right)$$ depends on the residue of $$p$$ modulo 8. The formula for $$\left( \frac{2}{p} \right)$$ when $$p$$ is an odd prime is given by: $$\left( \frac{2}{p} \right) = \begin{cases} 1 & \text{if } p \equiv 1 \pmod{8} \text{ or } p \equiv 7 \pmod{8}, \\ -1 & \text{if } p \equiv 3 \pmod{8} \text{ or } p \equiv 5 \pmod{8}. \end{cases}$$ This matches exactly the statement we aim to prove, meaning that this result is essentially a known property of the Legendre symbol for $$a = 2$$. ### Step 2: Proof of the Property $$\left( \frac{2}{p} \right)$$ To rigorously prove this property, let's look at the background: 1. **Quadratic Reciprocity Law**: For odd primes $$p$$ and $$q$$, we have: $$\left( \frac{p}{q} \right) \left( \frac{q}{p} \right) = (-1)^{\frac{(p-1)(q-1)}{4}}.$$ 2. **Supplementary Laws for 2**: The supplementary law for the Legendre symbol involving 2 is: $$\left( \frac{2}{p} \right) = (-1)^{\frac{p^2 - 1}{8}}.$$ This formula can be interpreted as follows: - When $$p \equiv 1 \pmod{8}$$, $$\frac{p^2 - 1}{8}$$ is even, so $$\left( \frac{2}{p} \right) = 1$$. - When $$p \equiv 7 \pmod{8}$$, $$\frac{p^2 - 1}{8}$$ is also even, so $$\left( \frac{2}{p} \right) = 1$$. - When $$p \equiv 3 \pmod{8}$$, $$\frac{p^2 - 1}{8}$$ is odd, so $$\left( \frac{2}{p} \right) = -1$$. - When $$p \equiv 5 \pmod{8}$$, $$\frac{p^2 - 1}{8}$$ is odd, so $$\left( \frac{2}{p} \right) = -1$$. This confirms that $$\left( \frac{2}{p} \right)$$ follows the rule given in the problem statement. ### Conclusion Thus, we have shown that for $$a = 2$$, the Legendre symbol $$\left( \frac{a}{p} \right)$$ behaves according to the rule: $$\left( \frac{2}{p} \right) = \begin{cases} 1 & \text{if } p \equiv 1 \pmod{8} \text{ or } p \equiv 7 \pmod{8}, \\ -1 & \text{if } p \equiv 3 \pmod{8} \text{ or } p \equiv 5 \pmod{8}. \end{cases}$$ This completes the proof. Would you like further details on quadratic reciprocity, the Legendre symbol, or perhaps an example calculation? Here are a few questions to explore: 1. How does the Legendre symbol generalize to other values of $$a$$ besides 2? 2. Can this proof be extended to composite moduli? 3. How does quadratic reciprocity apply to different primes other than $$p \equiv 1$$ or $$p \equiv 7 \pmod{8}$$? 4. What is the historical significance of the supplementary laws in quadratic reciprocity? 5. Can we determine if other small numbers, like 3 or 5, are quadratic residues modulo $$p$$? **Tip:** When tackling problems involving modular arithmetic and quadratic residues, it can be helpful to check small cases explicitly to gain insight into the general pattern.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Number Theory
Quadratic Residues
Modular Arithmetic
Legendre Symbol
Formulas
(a/p) = {1 if p ≡ 1 (mod 8) or p ≡ 7 (mod 8), −1 if p ≡ 3 (mod 8) or p ≡ 5 (mod 8)}
Quadratic Reciprocity Law: (p/q)(q/p) = (-1)^{((p-1)(q-1)/4)}
Supplementary Law for 2: (2/p) = (-1)^{(p^2 - 1)/8}
Theorems
Quadratic Reciprocity Theorem
Supplementary Law for the Legendre Symbol
Suitable Grade Level
Undergraduate Mathematics
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