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QUESTION
Using you answers to questions 1 and 2, find all solution of the
following equations:-
(i) $x^5\equiv4$ mod 27\hspace{1.5cm} (ii) $x^3\equiv9$ mod
187\hspace{1.5cm} (iii) $x^4\equiv 5$ mod 18.
ANSWER
\begin{description}
\item[(i)]
By q.2, a primitive root mod 27 can be chosen, We choose 2 here.
(5 would do just as well.) Now 4 can be written as $2^2$ mod 27
(or $5^4$ if you are using 5 as your primitive root). We may then
write $x\equiv2^k$ mod 27 ($5^k$ in the other case). The equation
then reads $2^{5k}\equiv2^2$ mod 27, i.e. $2^{5k-2}\equiv1$ mod
27. As the order of 2 mod 27 is $\phi(27)=18$, we obtain
$5k-2\equiv0$ mod 18, or $5k\equiv2$ mod 18. (If you used 5 as
your primitive root, you should have $5k\equiv4$ mod 18 here).
As gcd(5,18)=1, there is a unique root mod 18 to this
congruence,which, by using $5k\equiv2\equiv20$ mod 18, we can see
is 4. Thus $k\equiv4$ mod 18, and there is a unique root to
$x^5\equiv4$ mod 27, namely $2^4\equiv 16$ mod 27 (Using 5 as a
primitive root, we get $k\equiv8$ mod 18 and hence arrive at the
same conclusion concerning $x$.)
\item[(ii)]
$x^3\equiv9$ mod 187. Now 187=11.17, and as there is no primitive
root mod 11.17, we'll begin by solving separately the two
congruences $x^3\equiv9$ mod 11 and $x^3\equiv 9$ mod 17. From
question2, 2 is a primitive root mod 11, and by calculating powers
of 2 we find that $9\equiv2^6$ mod 11. We are thus solving
$x^3\equiv2^6$ mod 11, so setting $x=2^k$ we get $2^{3k}\equiv2^6$
mod 11, or $2^{3k-6}\equiv1$ mod 11. It follows that the order of
2 mod 11 (i.e.10) must divide $3k-6$, and so we get $3k\equiv6$
mod 10. Since gcd(3,10)=1, this congruence has a unique solution,
which we see, on dividing by 3, is $k\equiv2$ mod 10. Thus the
only solution of $x^3\equiv9$ mod 11 is $x\equiv 2^2\equiv4$ mod
11.
From question 1(iii), 5 is a primitive root mod 17, so this time
we write 9 as a power of 5 mod 17. By trial and error (i.e. by
calculating powers of 5 mod 17), we find that $9\equiv5^{10}$ mod
17. (Using $9\equiv-8$ mod 17, and the equations $5^8\equiv-1$ mod
16, and $5^2\equiv8$ mod 17 from question 1 achieves this
quickly!) Thus setting $x=5^k$ our equation now reads
$5^{3k}\equiv5^{10}$ mod 17, or $5^{3k-10}\equiv1$ mod 17. We may
now deduce $3k-10\equiv0$ mod $\phi(17)$, and as $\phi(17)=16$,
this reads $3k\equiv 10$ mod 16.
Since gcd(3,16)=1, this congruence has a unique solution which we
may obtain, e.g., by multiplying through by 5 to get
$-k\equiv50\equiv2$ mod 16, so that $k\equiv-2\equiv14$ mod 16.
Thus (using the calculations in question 1),
$x\equiv5^{14}\equiv5^8.5^4.5^2\equiv-1.13.8\equiv-1.-1.8\equiv32\equiv15$
mod 17. Thus the unique solution of $x^3\equiv9$ mod 17 is
$x\equiv 15$ mod 17.
If $c$ is a simultaneous solution of $x\equiv4$ mod 11 and
$x\equiv 15$ mod 17, then $c^3\equiv9$ mod 11 and $c^3\equiv 9$
mod 17, so that $c^3\equiv9$ mod 187. Moreover, any root of
$x^3\equiv9$ mod 187 satisfies both $x^3\equiv9$ mod 11 and
$x^3\equiv 9$ mod 17, and so $x\equiv 4$ mod 11 and $x\equiv 15$
mod 17. By the Chinese Remainder Theorem the two congruences
$x\equiv4$ mod 11 and $x\equiv15$ mod 177 have a unique
simultaneous solution mod 187, and so the equation $x^3\equiv9$
mod 187 has a unique solution. If we note that $4\equiv15$ mod 11,
we see that 15 satisfies both congruences, so it is the
simultaneous solution we seek. Hence the unique solution of
$x^3\equiv9$ mod 187 is $x\equiv15$ mod 187.
\item[(iii)]
By question 2, 5 is a primitive element mod 18, and $\phi(18)=6$.
Setting $x\equiv5^k$ mod 18, we need to solve $5^{4k}\equiv5$
mod18, i.e. $5^{4k-1}\equiv1$ mod 18. This gives $4k\equiv1$ mod
6, but as gcd(4,6)=2, which does not divide 1, this congruence has
no solutions. Thus $x^4\equiv5$ mod 18 has no solutions.
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