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Modular arithmetic

  1. Aug 16, 2014 #1
    1. The problem statement, all variables and given/known data
    Hi,
    I am currently working on modular arithmetic and recently I have been investigating on the effects of
    exponentiation on the system.It will be helpful if someone can share some ideas on the following problem.

    According to the division algorithm:

    a= nQ+r,
    where a is the dividend,n is the divisor,Q is the quotient and r is the remainder respectively.

    Let a,n and r be constants.If a is raised to the power of k,where k is an arbitrary positive integer,

    a^k = (nQ+r)^k

    By using the binomial theorem,(nQ+r)^k = nQ' + r^k

    if n<= r^k< n^2 , the quotient will be increased by 1.

    So if r^k =/= n,the division process continues,the initial dividend a^k multiplies by a to the power of m,after that the remainder changes again.In equation form:

    (a^k)(a^m) = [n(Q'+1) + (r^k)-n)] [nQ+r^m]
    = nQ" + [(r^k) - n](r^m)-n] , the final term is the new remainder.

    The process stops if the following is satisfied:

    {[(r^e - n)r^b - n]r^c - n}r^d - n (......similarly,could be more terms) = 0
    the power of a at which the algorithm stops = e+b+c+d

    Rearranging the terms gives:
    r^(e+b+c+d) = n + nr^d + nr^c + nr^b + nr^e

    Or alternatively,in some occasion,the remainder of a^(e+b+c+d) is the same as the of a,so:

    {[(r^e - n)r^b - n]r^c - n}r^d - n = r
    r^(e+b+c+d) = r + n + nr^d + nr^c + nr^b + nr^e

    My question is:
    Is there a way to find the value of e+b+c+d for a given value of a,so that we are able to find the
    value of the power of a at which the remainder = 0 or recurs upon dividing by a constant n?
    In addition,for a given a,is it true that there must be an interger k such that n|a^k (i.e. r = 0) ?
    That is:
    r^(e+b+c+d) = n + nr^d + nr^c + nr^b + nr^e

    where k = e+b+c+d

    Example:
    I know it is quite messy,I hope it will make it easier to understand by giving an example:
    Let a = 11,n = 7
    so if Q = 1,r = 4 ( 11 = 7(1) + 4 )

    and a^2 = 121,keeping n to be constant yields:
    121 = 7(17) + 2, r = 2

    For a particular value of a,is there any method to find the power to a at which r recurs(i.e. the remainder = 4 again) and/or r = 0 ?



    If there is any mistake please point out,and thanks for reading the wordy paragraphs.



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Aug 16, 2014
  2. jcsd
  3. Aug 16, 2014 #2
    The question is confusing. I'll interpret it as follows:

    "Suppose we have that [itex]a\equiv b\mod n[/itex]. Can we find an [itex]m[/itex] such that [itex]a^m\equiv 0\mod n[/itex] or such that the remainder remains the same for [itex]m'>m[/itex]?"

    The answer is, generally, no. Consider [itex]2\mod 6[/itex]. Then,

    [tex]2^2\equiv 4\mod 6 \\ 2^3\equiv 2\mod 6 \\ 2^4\equiv 4\mod 6 \\ \vdots \\ 2^{2k}\equiv 4\mod 6 \\ 2^{2k+1}\equiv 2\mod 6.[/tex]
     
  4. Aug 16, 2014 #3

    haruspex

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    As I read the question, it's
    "Suppose we have that [itex]a\equiv b\mod n[/itex]. Can we find an [itex]m[/itex] such that [itex]a^m\equiv 0\mod n[/itex] or such that the remainder is b? And how to find the smallest such m?"
    As you say, there will not in general be an m that makes the remainder 0. davon806, consider factorisations of a and n.
    For the second question, if a and am are congruent to b mod n, what equation does give you?
     
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