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Problem 2: Power of 2 modulo 9

by henderson, Feb 11, 2016, 5:52 PM

$\bf\color{red}Problem \ 2$ If $m$ is an odd positive integer, show that
$2^{2m}-2^m+1\equiv 3 \text{(mod 9)} .$ $\bf\color{red}My \ solution$ We need to consider three cases:
$\textbf{Case 1: }$ $m=6n+1:$
$2^{2{(6n+1)}}-2^{6n+1}+1\equiv 4-2+1\equiv 3 (\text{mod 9}).$ $\textbf{Case 2: }$ $m=6n+3:$
$2^{2{(6n+3)}}-2^{6n+3}+1\equiv 1-8+1\equiv 3 (\text{mod 9}).$ $\textbf{Case 3: }$ $m=6n+5:$
$2^{2{(6n+5)}}-2^{6n+5}+1\equiv 7-5+1\equiv 3 (\text{mod 9}).$ Note: I showed $m$ as $6n+t,$ where $t$ is an odd integer, because $2^6\equiv 1 (\text{mod 9)}.$

This post has been edited 8 times. Last edited by henderson, Sep 12, 2016, 2:44 PM

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