Math Problem Statement
if P is an odd prime then ∑_(n=0)^(p-1)▒(n/p) =0
Solution
To prove that , where is an odd prime and denotes the Legendre symbol, follow these steps:
Legendre Symbol:
The Legendre symbol is defined as:
\begin{cases} 1 & \text{if } n \text{ is a quadratic residue modulo } p, \\ -1 & \text{if } n \text{ is a quadratic non-residue modulo } p, \\ 0 & \text{if } n \equiv 0 \pmod{p}. \end{cases}$$ ### Sum of Legendre Symbols: We need to evaluate the sum: $$\sum_{n=0}^{p-1} \left(\frac{n}{p}\right).$$ This sum can be split as: $$\sum_{n=0}^{p-1} \left(\frac{n}{p}\right) = \left(\frac{0}{p}\right) + \sum_{n=1}^{p-1} \left(\frac{n}{p}\right).$$ Since $$\left(\frac{0}{p}\right) = 0$$, we have: $$\sum_{n=0}^{p-1} \left(\frac{n}{p}\right) = \sum_{n=1}^{p-1} \left(\frac{n}{p}\right).$$ ### Properties of the Legendre Symbol: 1. **Number of Residues and Non-Residues**: For any odd prime $$p$$, exactly half of the integers $$n$$ in $$\{1, 2, \dots, p-1\}$$ are quadratic residues, and the other half are non-residues. Therefore, there are $$(p-1)/2$$ values of $$n$$ such that $$\left(\frac{n}{p}\right) = 1$$, and $$(p-1)/2$$ values such that $$\left(\frac{n}{p}\right) = -1$$. 2. **Sum of Residues and Non-Residues**: Given the symmetry and the equal count of residues and non-residues, the sum $$\sum_{n=1}^{p-1} \left(\frac{n}{p}\right)$$ will consist of $$(p-1)/2$$ terms of 1 and $$(p-1)/2$$ terms of -1. Thus: $$\sum_{n=1}^{p-1} \left(\frac{n}{p}\right) = \frac{p-1}{2} \cdot 1 + \frac{p-1}{2} \cdot (-1) = \frac{p-1}{2} - \frac{p-1}{2} = 0.$$ ### Conclusion: Therefore, $$\sum_{n=0}^{p-1} \left(\frac{n}{p}\right) = 0$$. Would you like more details on any part of the proof? Here are some related questions: 1. How is the Legendre symbol defined for composite numbers? 2. What are quadratic residues and non-residues in more detail? 3. How does the symmetry of quadratic residues contribute to the sum being zero? 4. Can this property be generalized to non-prime numbers? 5. What are some applications of the Legendre symbol in number theory? **Tip**: Understanding the properties of quadratic residues and non-residues is crucial for studying more advanced topics in number theory, like quadratic reciprocity.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Number Theory
Legendre Symbol
Formulas
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Theorems
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Suitable Grade Level
Advanced Undergraduate