consider the number -1, i can say it is a special number because the powers of (-1) is always 1 or -1.Generally $(-1)^n = 1$ if n(mod 2) = 0 and $(-1)^n = -1$ if n(mod 2) = 1. my question is that is there any number x such that,
- $x^n = 1$ if n(mod 3) = 0
- $x^n = 1$ if n(mod 3) = 1
- $x^n = -2$ if n(mod 3) = 2
The relation between this problem and modular arithmetic is that we can define a definite formula for n(mod k). from the approach I obtained the formula for n(mod 2) which is,
n(mod 2) = $\frac{1\space+\space(-1)^{n-1}}{2}$
the above formula comes from the sequence 0, 1, 0, 1, 0,.... this sequence starts from 0th term.In this sequence every even term is zero and every odd term is one. hence the nth term of the sequence is n(mod 2).
Proof of the Nth Term of the Sequence
Let $S\small_ n = 0, 1, 0, 1, 0...$
$S_n = \sum_{i=0}^{n-1} (-1)^i$
1.$S_n + S_{n-1}$ = 1
2.$S_n - S_{n-1} = (-1)^{n-1}$
Adding 1. and 2. we get,
$S_ n = \dfrac{1+(-1)^{n-1}}{2}$
n(mod 2) = $S_n = \dfrac{1+(-1)^{n-1}}{2}$
To find the formula for n(mod 3) we need to find the $n$-th term of the sequence $0, 1, 2, 0, 1, 2...$
But for that we need to find x such that,
- $x^n = 1$ if n(mod 3) = 0
- $x^n = 1$ if n(mod 3) = 1
- $x^n = -2$ if n(mod 3) = 2
Then the formula for the sequence $A_n = 0, 1, 2, 0, 1, 2...$ will be defined as, $$A_n = \sum_{i=0}^{n-1} x^i = \frac{x^n - 1}{x-1}$$
it think it would be a better information to extend my research in modular arithmetic. I am high school graduate and i didn't have any other higer mathematical knowledege than high school.