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I was doing a question yesterday on very elementary number theory, and I came across this pattern.

$2^1=2 \pmod {13}$

$2^2=4 \pmod {13}$

$2^3=8 \pmod {13}$

$2^4=3 \pmod {13}$

$2^5=6 \pmod {13}$

$2^6=12 \pmod {13}$

$2^7=11 \pmod {13}$

$2^8=9 \pmod {13}$

$2^9=5 \pmod {13}$

$2^{10}=10 \pmod {13}$

$2^{11}=7 \pmod {13}$

$2^{12}=1 \pmod {13}$

$2^{13}=2 \pmod {13}$

I noticed the remainder always is unique for every block multiples of powers 13. And it repeats!.

Also I tried this for base 3 and base 4 base 5. All showed similar patterns.

Why?

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    @GottfriedHelms All you need is that $a≡c$ and $b≡d \mod m$ implies $ab≡cd$. $c=a+me,d=b+mf,cd=ab+(be+af+mef)m$ [corrected for careless typo]2012-10-20

2 Answers 2

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The group of invertible classes modulo a prime $p$, i.e.:

$\Bbb F_p^\times := \{x \in \Bbb Z/ p \Bbb Z: \exists y: xy \equiv 1\}$

has cardinality $\phi(p) = p-1$, where $\phi$ is the Euler phi function. Thus, we see that the order of every element is a divisor of $p-1$ (by Little Fermat).

That $2$ has order $p-1$ in the cases you have tried is a coincidence. E.g. $p = 31$ will not work, for $2^6 \equiv 32 \equiv 1$, but $\Bbb F_{31}^\times$ has order $30$, so $2$ is not a generator.

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    I have problems logging into the chatroom. Error message : No referer was present - this may be due to a browser setting. I guess its ok thanks for offering me help farin! Back to proper school work now.2012-10-20
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Simpler answer: For all these modulo questions, it is enough to find some exponent that gives a modulo of 1. Consider

$2^n = 1 (mod 13)$ for some n. Then $2^{n + 1} = 2 (mod 13)$ which is same as $2^{0 + 1} = 2 (mod 13)$ and $2^{n + 2} = 4 (mod 13)$ which is $2^{1 + 1} = 4 (mod 13)$, in other words, they repeat. And also, notice that, if $2^n = k (mod 13)$ and $2^m = k (mod 13)$, then $2^{n - m} = 1 (mod 13)$ also. Since there are only 13 numbers below 13 that could be modulos, it is bound to repeat after some exponent n.

Salahuddin

http://maths-on-line.blogspot.in/