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QUESTION
Given that the numbers 719, 853 and 971 are all prime, calculate
$\left(\frac{719}{853}\right)$ and $\left(\frac{853}{971}\right)$.
ANSWER
$\left(\frac{719}{853}\right)=\left(\frac{853}{719}\right)$ by the
quadratic reciprocity law, as $853\equiv1$ mod 4. Hence
$\left(\frac{719}{853}\right)=\left(\frac{853}{719}\right)=
\left(\frac{134}{719}\right)=\left(\frac{2}{719}\right)
\left(\frac{67}{719}\right)$ (reducing mod 179 and factorising).
Now $\left(\frac{2}{719}\right)=1$ as $719\equiv7\equiv-1$ mod 8
(th.7.3), while
$\left(\frac{67}{719}\right)=-\left(\frac{719}{67}\right)$ by
quadratic reciprocity, as 719 and 67 are both $\equiv3$ mod 4. On
reducing mod 67 we see that
$\left(\frac{67}{719}\right)=-\left(\frac{719}{67}\right)=
-\left(\frac{49}{67}\right)=-\left(\frac{7^2}{67}\right)=-1$ (as
$7^2$ is clearly a square!) Thus
$\left(\frac{719}{853}\right)=1.-1=-1$ and 719 is not a square mod
853.
[An alternative approach is to begin by replacing 719 by its least
absolute residue $-134$ mod 853, and using
$\left(\frac{719}{853}\right)=\left({-134}{853}\right)=
\left(\frac{-1}{853}\right)\left(\frac{134}{853}\right)$ etc. You
might like to try this to see which method is quicker.]
$\left(\frac{853}{971}\right)=\left(\frac{971}{853}\right)$ by the
quadratic reciprocity law, as $853\equiv1$ mod 4. Thus, on
reducing mod 853, and factoring, we get
$\left(\frac{853}{971}\right)=\left(\frac{971}{853}\right)=
\left(\frac{118}{853}\right)=\left(\frac{2}{853}\right)
\left(\frac{59}{853}\right)$. Now $\left(\frac{2}{853}\right)=-1$
as $853\equiv5\equiv-3$ mod 8 (th.7.3). As 59 is prime, we may
again assume quadratic reciprocity, and as $853\equiv1$ mod 4, we
get $\left(\frac{59}{853}\right)=\left(\frac{853}{59}\right)=
\left(\frac{27}{59}\right)=\left(\frac{3^2}{59}\right)
\left(\frac{3}{59}\right)=\left(\frac{3}{59}\right)$ ( by reducing
mod 59, factoring and using $\left(\frac{a^2}{p}\right)=1$.) Now 3
is prime, and as 59 and 3 are both congruent to 3 mod 4, quadratic
reciprocity gives
$\left(\frac{3}{59}\right)=-\left(\frac{59}{3}\right)=-
\left(\frac{2}{3}\right)=-(-1)=1$ (using th.7.3 again).
Thus $\left(\frac{59}{853}\right)=1$ and so
$\left(\frac{853}{971}\right)=(-1).1=-1$, and 853 is not a square
mod 971.
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