finding reduced residue system

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Arif Ahmed
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finding reduced residue system

Unread post by Arif Ahmed » Thu May 24, 2012 10:15 pm

Let $a_1$,$a_2$,..........$a_r$ be a reduced residue system modulo $m$ and let $n\geq1$.Prove that $a_1^n$,$a_2^n$,......$a_r^n$ is a reduced residue system modulo $m$ if and only if $(n, \phi(m))=1$

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SANZEED
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Re: finding reduced residue system

Unread post by SANZEED » Sat May 26, 2012 6:14 pm

Firstly, Let \[S_{0}=\left \{ a_{1},....,a_{r} \right \}\] and \.Now $S$ is a reduced system. So there are no integers in $S_{0}$ such that \[(a_{i},m)> 1\].Hence \[(a_{i}^{n},m)= 1\]. for all \[i< r\].Now we will prove a lemma .
Lemma: Let $R$ be a reduced system of residues mod $m$ and let $a$ be any given integer such that \[(a,m)=1\],hen there exists a given integer in $R$ such that a unique integer,corresponding to $a$,say $b$ such that \[a\equiv b(mod m)\].

Proof:Let $r$ be the least residue of $a(mod m)$ so that we have \[a\equiv r(mod m), 0\leq r< m\] and $(r,m)=(a,m)=1$.But the set of least residues of of $R$ is just the set of all the integers less than $m$ and coprime to $m$.hat means that $r$ is an element of the set of the residues of the elements of$R$ mod $m$.it implies that there exist a uniqe integer $b$ in $R$ such that \[a\equiv b(mod m)\].

From the lemma above we can directly say that $S$ is a reduced system.

There is surely a problem.It is true that \[(n,\phi (m))=1\] is necessary, but where is the bug of my solution?will anyone debug it please?
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shehab ahmed
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Re: finding reduced residue system

Unread post by shehab ahmed » Sat May 26, 2012 8:26 pm

what is the meaning of reduced residue system?i don't know

Arif Ahmed
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Re: finding reduced residue system

Unread post by Arif Ahmed » Sun May 27, 2012 11:32 am

Let $m$ be positive.A reduced residue system modulo $m$ is a set of integers such that every number relatively prime to $m$ is congruent modulo $m$ to a unique element of the set.

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Phlembac Adib Hasan
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Re: finding reduced residue system

Unread post by Phlembac Adib Hasan » Sun May 27, 2012 4:37 pm

SANZEED wrote:Firstly, Let $S_{0}=\left \{ a_{1},....,a_{r} \right \}$ and $S=\left \{ a_{1}^{n},....,a_{r}^{n} \right \}$.Now $S$ is a reduced system. So there are no integers in $S_{0}$ such that $(a_{i},m)> 1$.Hence $(a_{i}^{n},m)= 1$. for all $i< r$.Now we will prove a lemma .
Lemma: Let $R$ be a reduced system of residues mod $m$ and let $a$ be any given integer such that $(a,m)=1$,hen there exists a given integer in $R$ such that a unique integer,corresponding to $a$,say $b$ such that $a\equiv b(mod m)$.

Proof:Let $r$ be the least residue of $a(mod m)$ so that we have $a\equiv r(mod m), 0\leq r< m$ and $(r,m)=(a,m)=1$.But the set of least residues of of $R$ is just the set of all the integers less than $m$ and coprime to $m$.hat means that $r$ is an element of the set of the residues of the elements of$R$ mod $m$.it implies that there exist a uniqe integer $b$ in $R$ such that $a\equiv b(mod m)$.

From the lemma above we can directly say that $S$ is a reduced system.

There is surely a problem.It is true that $(n,\phi (m))=1$ is necessary, but where is the bug of my solution?will anyone debug it please?
What have you proved? :lol:
BTW, there is no need to state or prove this lemma in such a way.(Actually it's not a lemma,direct definition of reduced residue system)
And my question, how are you using this 'lemma'?If I assume the first one is a typo,(if I assume it is $S_0$)there is still problem.Your 'lemma' has no validity then.
Just see, what you have to prove
\[S\; \; \text {is a reduced residue system}\; \; \Leftrightarrow \; \; (n,\phi(m))=1\](My debugging is not so cool 8-) )
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SANZEED
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Re: finding reduced residue system

Unread post by SANZEED » Mon May 28, 2012 1:01 am

Sorry, but actually I wanted to show that $S_{0}$ is a reduced system,not $S$. Now what will you say? :?:
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Re: finding reduced residue system

Unread post by *Mahi* » Mon May 28, 2012 1:16 am

SANZEED wrote:Sorry, but actually I wanted to show that $S_{0}$ is a reduced system,not $S$. Now what will you say? :?:
That is already given.
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Phlembac Adib Hasan
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Re: finding reduced residue system

Unread post by Phlembac Adib Hasan » Mon May 28, 2012 7:58 am

SANZEED wrote:Sorry, but actually I wanted to show that $S_{0}$ is a reduced system,not $S$. Now what will you say? :?:
Same question as Mahi vaia.Please read what I wrote before:
If I assume the first one is a typo,(if I assume it is $S_0$)there is still problem.Your 'lemma' has no validity then.
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shehab ahmed
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Re: finding reduced residue system

Unread post by shehab ahmed » Mon May 28, 2012 10:56 pm

আমার সলিউশনটা কেউ দেখ।আমি শিওর এখানে কোথাও না কোথাও ভুল আছে
first let, $(n,\varphi (m))=1$ and $p$ be any prime factor of $m$ and $p^a||m$
we are given that all $a_i$'s are incongruent and co-prime to $m$.
Now,for sake of argument, let $a_i^n$ and $a_j^n$ are congruemt modulo $m$ that is modulo $p^a$.let, $g$ is a primitive root of $p^a$.
let, $a_i=g^m$ and $a_j=g^k$
it implies that $mn\equiv kn\pmod {\varphi p^a}$. Now, $(n,\varphi (p^a))=1$ so $m\equiv k \pmod \varphi (p^a)$. As $p$ can be any prime factor of $m$,continuing this argument for every prime factor $p$ of $m$,we can deduce that $a_i\equiv a_j\pmod m$ which is absurd.
For the converse, I have also used contradiction. Let, $(n,\varphi (m))>1$ so there is a prime $p$ such that $p^a||m$ and $(n,\varphi (p^a))>1$ consider that prime. By our assumption, $a_i^n$ is congruent to a unique $a_j\pmod m$ that is $\pmod p^a$.so $mn\equiv k\pmod\varphi (p^a)$. Let $d$ be a common prime divisor of $p^a$ and $n$. Then, $d|k$,but we have the freedom to choose $k$ such that $d$ is co prime to $k$ and the result follows from here
Last edited by Masum on Tue May 29, 2012 12:19 pm, edited 1 time in total.
Reason: LaTeXed

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Masum
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Re: finding reduced residue system

Unread post by Masum » Tue May 29, 2012 12:22 pm

There are nice tutorials about using LaTeX. You should read them, it is not hard at all. Just using two dollars efficiently. You certainly don't want that only you in the world would understand your post.
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