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QUESTION

Find all the solutions of each of the following congruences,
expressing your answers in terms of congruence classes, $[x]$:

\begin{description}

\item[(i)]
$7x\equiv29$ mod (51),

\item[(ii)]
$6(x^5+x^3+3001)\equiv23$ mod (48),

\item[(iii)]
$x^2-x+2\equiv0$ mod (4).

\end{description}



ANSWER

\begin{description}

\item[(i)]
Since HCF(7,51)=1 there is exactly one congruence class $[x]$ of
solutions. We can find this either by using the Euclidean
algorithm to find $u$ and $v$ such that $7u+51v=1$ then
$[x]=[29u]$ or we can find the solution as follows. The equation
is the same mod (51) as $-44x\equiv-22$ mod (51) which has the
same solutions as $2x\equiv1$ mod (510. The solution to this is
$[x]=[26]$, which is correct since
$26\times7=182=153+29=3\times51+29$, as required.

\item[(ii)]
Since HCF(6,48)=6 does not divide 23 there are no solutions. Put
another way, if $6(x^5+x^3+3001)=23+48n$ then taking remainder mod
(6) on both sides gives the contradiction that $23\equiv0$ mod
(6).

\item[(iii)]
It suffices to try substituting the values $[x]=[0],[1],[2],[3]$
since the congruence class mod (4) of the left hand side depends
only on $[x]$. Here is a table:

\begin{tabular}{c|c|c|c}
$[x]$&$[x^2]$&$[x]$&$[x^2-x+2]$\\
$\left[0\right]$&$\left[0\right]$&$\left[0\right]$&$\left[
2\right]$\\
$\left[1\right]$&$\left[1\right]$&$\left[1\right]$&$\left[2\right]$\\
$\left[2\right]$&$\left[ 0\right]$&$\left[
2\right]$&$\left[0\right]$\\
$\left[3\right]$&$\left[1\right]$&$\left[3\right]$&$\left[0\right]$
\end{tabular}

so the solutions are all integers $x\equiv2,3$ mod (4).

\end{description}




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