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icystrike
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Find all odd primes [tex]p[/tex], if any, so that [tex]p[/tex] divides [tex]\sum_{n=1}^{103} n^{p-1}[/tex]
Odd Primes Divisible by Sum of n^p-1 from n=1 to 103
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SUMMARY
The discussion centers on identifying odd primes \( p \) such that \( p \) divides the sum \( \sum_{n=1}^{103} n^{p-1} \). According to Fermat's Little Theorem, \( n^{p-1} \equiv 1 \mod p \), leading to the conclusion that \( \sum_{n=1}^{103} n^{p-1} \equiv 103 \mod p \). Consequently, since 103 is prime, it is established that 103 is the only odd prime satisfying this condition. The discussion also highlights that for primes \( p \leq 103 \), \( p \) can divide \( n \), resulting in \( n^{p-1} \equiv 0 \mod p \).
PREREQUISITES- Fermat's Little Theorem
- Modular arithmetic
- Understanding of prime numbers
- Basic summation techniques
- Study advanced applications of Fermat's Little Theorem
- Explore properties of prime numbers in modular arithmetic
- Investigate the implications of \( n^{p-1} \equiv 0 \mod p \)
- Learn about the distribution of prime numbers and their divisibility
Mathematicians, number theorists, and students interested in prime number properties and modular arithmetic applications.
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