P-1 consecutive numbers coprime to n

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In summary: $\varphi(n)=n\ \prod_{p|n} (1-\frac{1}{p})=n\ \prod_{p|\frac{n}{p_{0}}} (1-\frac{1}{p})$ (1) $\displaystyle \chi(\varphi(n))=n\sum_{p|\frac{n}{p_{0}}} (1-\frac{1}{p})$ (3)$\displaystyle \sigma(\varphi(n))=\sum_{p|\frac{n}{p_{0}}} (1-\frac{1}{p})$ (4)$
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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

$\chi$ $\sigma$
 

Related to P-1 consecutive numbers coprime to n

1. What are P-1 consecutive numbers coprime to n?

P-1 consecutive numbers coprime to n are a sequence of P-1 integers that are all coprime to a given number n, meaning they have no common factors other than 1.

2. Why are P-1 consecutive numbers coprime to n important in mathematics?

These numbers are important because they have many applications in number theory, cryptography, and other branches of mathematics. They also have connections to other important concepts such as Euler's totient function and the Chinese remainder theorem.

3. How do you find P-1 consecutive numbers coprime to n?

To find P-1 consecutive numbers coprime to n, you can use the formula a + kn, where a is any integer coprime to n and k is a positive integer less than P-1. This will give you a sequence of P-1 numbers that are all coprime to n.

4. Can P-1 consecutive numbers coprime to n be negative?

Yes, P-1 consecutive numbers coprime to n can be negative. As long as the numbers are coprime to n, they can be either positive or negative.

5. What is the relationship between P-1 consecutive numbers coprime to n and the prime factors of n?

There is no direct relationship between P-1 consecutive numbers coprime to n and the prime factors of n. However, the number of P-1 consecutive numbers coprime to n can give insights into the prime factorization of n, as well as the prime factors of n can help to find P-1 consecutive numbers coprime to n.

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