Carmichael Number: Proving $(d+p)(p-1)$ Divisor of $q-1$

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SUMMARY

The discussion centers on proving that for a Carmichael number \( n = pqr \), if \( q-1 \) divides \( (d+p)(p-1) \) under the conditions \( q-1|pr-1 \) and \( r-1|pq-1 \), then \( q-1 \) is indeed a divisor of \( (d+p)(p-1) \). The participants clarify that \( pq-1 = d(r-1) \) where \( d \) is defined as \( [2;p-1] \), and they emphasize the importance of understanding these relationships to grasp the proof fully. The discussion highlights the interconnectedness of these divisibility conditions in the context of Carmichael numbers.

PREREQUISITES
  • Understanding of Carmichael numbers and their properties
  • Familiarity with divisibility rules in number theory
  • Knowledge of modular arithmetic and its applications
  • Basic concepts of prime factorization and its implications
NEXT STEPS
  • Study the properties of Carmichael numbers in detail
  • Explore the implications of divisibility in number theory
  • Learn about modular arithmetic and its role in proofs
  • Investigate related theorems in number theory, such as Fermat's Little Theorem
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This discussion is beneficial for mathematicians, number theorists, and students studying advanced topics in number theory, particularly those interested in the properties and proofs related to Carmichael numbers.

peteryellow
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$n=pqr$ is a Carmichael number. If $q-1|pr-1$ and $r-1|pq-1$ then show that q-1 is a divisor in
$(d+p)(p-1).$
 
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ok here it is the clear version of it.

n=pqr is carmicahelnumber. if we have that r-1|pq-1 then I have shown that
pq-1=d(r-1) where d is [2;p-1]. Moreover I have shown that q-1|d(r-1)-p+1.
Now I want to show that if q-1|pr-1 is also fulfilled then q-1 is divisor in (d+p)(p-1). Do you understand it?
 
Hi!
I suggest you to read more before publish your problems since this is a very well known question (Wikipedia).
 

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