One problem about number theory

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SUMMARY

The discussion centers on proving that if \(2^p - 1\) is prime, then \(p\) must also be prime. Participants suggest assuming \(p\) is not prime and expressing it as \(p = ab\), where \(a\) and \(b\) are nontrivial factors. The factorization \(2^{ab} - 1\) can be rewritten using the identity \(x^n - y^n\), which leads to further insights into the properties of prime numbers and their relationship with Mersenne primes.

PREREQUISITES
  • Understanding of prime numbers and their properties
  • Familiarity with Mersenne primes
  • Knowledge of algebraic identities, specifically \(x^n - y^n\)
  • Basic concepts of factorization in number theory
NEXT STEPS
  • Study the properties of Mersenne primes and their significance in number theory
  • Learn about algebraic identities, particularly the factorization of \(x^n - y^n\)
  • Explore the implications of composite numbers in prime factorization
  • Investigate the relationship between prime numbers and their exponents in expressions like \(2^p - 1\)
USEFUL FOR

Mathematics students, number theorists, and anyone interested in the properties of prime numbers and their applications in theoretical mathematics.

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Homework Statement



Let [tex]p\ge 2[/tex]. Prove if [tex]2^p-1[/tex] is prime, then [tex]p[/tex] must be prime.

Homework Equations





The Attempt at a Solution



I am a physics student. I need help from those studying mathematics. Thank you very much!
 
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Assume that p is NOT prime, then we can write p=ab, with a and b nontrivial.

Try to find a factorization of [tex]2^{ab}-1=(2^a)^b-1[/tex]...

(HINT: what does [tex]x^n-y^n[/tex] equal?)
 

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