MHB P-1 consecutive numbers coprime to n

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SUMMARY

The discussion centers on the properties of consecutive integers that are coprime to a given integer \( n > 1 \), specifically focusing on its smallest prime factor \( p \). It is established that there can be at most \( p - 1 \) consecutive positive integers coprime to \( n \). Furthermore, it is demonstrated that this maximum cannot be reduced, as \( p - 1 \) consecutive integers can indeed be found that are coprime to \( n \). The discussion also touches on the greatest common divisor \( \text{gcd}(p - 1, n) \) and the modular arithmetic property \( 2^n \not\equiv 1 \mod n \).

PREREQUISITES
  • Understanding of coprime integers and their properties
  • Familiarity with prime factorization and smallest prime factors
  • Knowledge of Euler's totient function, \( \varphi(n) \)
  • Basic concepts of modular arithmetic
NEXT STEPS
  • Study the properties of Euler's totient function, \( \varphi(n) \)
  • Explore the implications of the Chinese Remainder Theorem on coprime integers
  • Investigate the relationship between prime factors and consecutive integers
  • Learn about modular exponentiation and its applications in number theory
USEFUL FOR

Mathematicians, number theorists, and students studying properties of integers and modular arithmetic will benefit from this discussion.

shmounal
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a) Let an integer $n > 1$ be given, and let $p$ be its smallest prime factor. Show
that there can be at most $p − 1$ consecutive positive integers coprime to $n$.
b) Show further that the number $p − 1$ in (a) cannot be decreased, by exhibiting
$p − 1$ consecutive positive integers coprime $n$.
c) What is gcd$(p − 1, n)$?
d)Show that $2^n \not\equiv 1 (mod n)$.

I think the first part has something to with the fact that two positive integers are coprime iff they have no prime factors in common. As if there were more than $p-1$ consecutive numbers then they would have a coprime in common. Not sure how to word this convincingly!

Not sure for b). Guessing I'd say c) is $p-1$, though not sure. d).. no clue!

If you can point me in the right direction I'd appreciate it, thanks.
 
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shmounal said:
a) Let an integer $n > 1$ be given, and let $p$ be its smallest prime factor. Show
that there can be at most $p − 1$ consecutive positive integers coprime to $n$.
b) Show further that the number $p − 1$ in (a) cannot be decreased, by exhibiting
$p − 1$ consecutive positive integers coprime $n$.
c) What is gcd$(p − 1, n)$?
d)Show that $2^n \not\equiv 1 (mod n)$.

I think the first part has something to with the fact that two positive integers are coprime iff they have no prime factors in common. As if there were more than $p-1$ consecutive numbers then they would have a coprime in common. Not sure how to word this convincingly!

Not sure for b). Guessing I'd say c) is $p-1$, though not sure. d).. no clue!

If you can point me in the right direction I'd appreciate it, thanks.

Considering that the number of coprimes of n is the Euler's function...

$\displaystyle \varphi(n)= n\ \prod_{p|n} (1-\frac{1}{p})$ (1)

... if $p_{0}$ is the smallest prime dividing n then is...

$\displaystyle \varphi(n)= \frac{n}{p_{0}}\ (p_{0}-1)\ \prod_{p|\frac{n}{p_{0}}} (1-\frac{1}{p})$ (2)

... and in any case the term $p_{0}-1$ is a factor of $\varphi(n)$...

Kind regards

$\chi$ $\sigma$
 

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