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QUESTION

Use Gauss' Lemma to decide for each of the following pairs $(a,p)$
whether or not $a$ is a square mod $p$.

(i) (5,23)\hspace{1.5cm} (ii) (10,17)\hspace{1.5cm} (iii) (10,13).



ANSWER

\begin{description}

\item[(i)]
Here $S$ consists of the first $\frac{(23-1)}{2}=11$ multiples of
5, viz. $S=\{5,10,15,20,25,30,35,40,45,50,55\}$. Reducing these to
their least positive residues mod 23 gives the set
$s'=\{5,10,\underline{15},\underline{20},2,7, \underline{12},
\underline{17},\\ \underline{22},4,9\}$. We have underlined those
exceeding $\frac{23}{2}$, and we note that there are 5 of them.
Thus $n=5, (-1)^5=-1$, so $\left(\frac{5}{23}\right)=-1$ and 5 is
non-square mod 23.

\item[(ii)]
Here $\frac{(P-1)}{2}=\frac{16}{2}=8$, so we want the first 8
multiples of 10. Thus $S=\{10,20,30,40,50,60,7 0,80\}$. Reducing
to least positive residues mod 17 gives
$S'=\{\underline{10},3,\underline{13},6,
\underline{16},9,2,\underline{12}\}$ and the ones exceeding
$\frac{17}{2}$ have been underlined. Again there are 5 of them, so
$n=5, (-1)^5=-1$ and $\left(\frac{10}{17}\right)=-1$. Thus 10 is a
non-square mod 17.

\item[(iii)]
Here $\frac{(p-1)}{2}=\frac{12}{2}=6$, so
$S=\{10,20,30,40,50,60\}$. Reducing mod 13,
$S'=\{\underline{10},\underline{7},4,1,\underline{11},
\underline{8}\}$, where the entries bigger then $\frac{13}{2}$ are
underlined. Thus $n=4, (-1)^4=1$, and
$\left(\frac{10}{13}\right)=1$. Thus 10 is a square mod 13. (In
fact it is $6^2$.)

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