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11-00-UsingPrimitiveRootsAndIndexToSolveCongruences.tex
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11-00-UsingPrimitiveRootsAndIndexToSolveCongruences.tex
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\documentclass[12pt]{article}
\usepackage{pmmeta}
\pmcanonicalname{UsingPrimitiveRootsAndIndexToSolveCongruences}
\pmcreated{2013-03-22 16:20:58}
\pmmodified{2013-03-22 16:20:58}
\pmowner{alozano}{2414}
\pmmodifier{alozano}{2414}
\pmtitle{using primitive roots and index to solve congruences}
\pmrecord{4}{38483}
\pmprivacy{1}
\pmauthor{alozano}{2414}
\pmtype{Example}
\pmcomment{trigger rebuild}
\pmclassification{msc}{11-00}
\pmrelated{PrimitiveRoot}
\pmrelated{Congruence}
\pmrelated{PolynomialCongruence}
\endmetadata
% this is the default PlanetMath preamble. as your knowledge
% of TeX increases, you will probably want to edit this, but
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% almost certainly you want these
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% Some sets
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\newcommand{\Ints}{\mathbb{Z}}
\newcommand{\Reals}{\mathbb{R}}
\newcommand{\Complex}{\mathbb{C}}
\newcommand{\Rats}{\mathbb{Q}}
\newcommand{\Gal}{\operatorname{Gal}}
\newcommand{\Cl}{\operatorname{Cl}}
\newcommand{\ind}{\operatorname{ind}}
\begin{document}
The aim of the following example is to illustrate how to use primitive roots and the index of an integer to solve seemingly complicated congruences.
For this example, let $p=13$ and let us attempt to solve $$7x^{10}\equiv 5 \mod 13.$$ Since every prime has a primitive root, we can easily find one. In particular, the number $2$ is a primitive root for $p=13$. Indeed, the powers of $2$ are the following modulo $13$:
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
% after \\: \hline or \cline{col1-col2} \cline{col3-col4} ...
$x$ & $x^2$ & $x^3$ & $x^4$ & $x^5$ & $x^6$ & $x^7$ & $x^8$ & $x^9$
& $x^{10}$ & $x^{11}$ & $x^{12}$ \\
\hline
$2$ & $4$ & $8$ & $3$ & $6$ & $12$ & $11$ & $9$ & $5$ & $10$ & $7$ & $1$ \\
\hline
\end{tabular}
\end{center}
The table above allows us to build a table of indices with base $2$ (for the definition of index and its properties which will be used below, see \PMlinkid{this entry}{PropertiesOfTheIndexOfAnIntegerWithRespectToAPrimitiveRoot}):
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
% after \\: \hline or \cline{col1-col2} \cline{col3-col4} ...
$x$ & $1$ & $2$ & $3$ & $4$ & $5$ & $6$ & $7$ & $8$ & $9$
& $10$ & $11$ & $12$ \\
\hline
$\ind x$ & $12$ & $1$ & $4$ & $2$ & $9$ & $5$ & $11$ & $3$ & $8$ & $10$ & $7$ & $6$ \\
\hline
\end{tabular}
\end{center}
Now we can use the properties of index to solve the equation $7x^{10}\equiv 5 \mod 13$. By taking indices on both sides we obtain
$$\ind (7x^{10})\equiv \ind 7 +10\ind x \equiv \ind 5 \mod 12$$
and so $\ind x \equiv \ind 5 - \ind 7 \equiv 9-11\equiv 10 \mod 12$. The equivalence $10\ind x \equiv 10 \mod 12$ implies $5\ind x\equiv 5 \mod 6$ and hence $\ind x \equiv 1 \mod 6$. Lifting this solution to modulo $12$ we obtain $\ind x \equiv 1$ or $7$ modulo $12$, and by the table of indices, $x\equiv 2$ or $11$ modulo $13$ are the unique solutions of the congruence.
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\end{document}