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.