Question

Quadratic Reciprocity ) Compute () (We know many different ways to complete this computation: beute foece, Euler's Criterion Gauss Lemma, Quadratic Recipeocity. Be sre to nderstand how to apply all of these methods) ae (1.2). ee (1.2.34). a (1.2.34,5,6). aE1.2.3,5,67 a (2.3.7,8,15, 17.24.26 (2,3.524.73,221,225,627), p= 1237 -3 p29 Compucte (we no mang different wos to complete this Computafion bute Foce Buler's Crterion Gauss' lenma Cuadratic Reciprocity. Be sure ts undavstond has t apply al these muthods) -a E L21 a 6 2,,549] p aet 2.3.78.s,A, 14,14) ps aG ibs 24, 41

Answer 1

We have to compute . Basic rules we will follow are as follows:

a

1. $\left(\dfrac{a}{p}\right)=\left(\dfrac{b}{p}\right)$ if

а — 6(d p) тос

2.$\left(\dfrac{p}{q}\right) = \left(\dfrac{q}{p}\right)$ if .​

p 1(mod 4) or q 1(mod 4)

Otherwise, if
p 3(mod 8) and q 3(mod 8)
, we have
P P
.

3.

p-1)/2 (mod p) a P

4.
2 = (-1)(p°-1)/8 (mod p) t1(mod 8) 1(if p
and
= 1(if p+3(mod 8))

5.$\left(\dfrac{x^2}{p}\right) = 1 (if~p\nmid x) ,=0 ~otherwise$

6. $\left(\dfrac{ab}{p}\right) = \left(\dfrac{a}{p}\right)\left(\dfrac{b}{p}\right)$

Using the above things, computations can be done as follows:

$\left(\dfrac{1}{3}\right) = 1^{(3-1)/2}(mod~3)=1$

$\left(\dfrac{2}{3}\right) = -1$ since

3 3(mod 8)

$\left(\dfrac{1}{5}\right) = 1^{(5-1)/2}(mod~3)=1$

$\left(\dfrac{2}{5}\right) = -1$ since

5-3(mod 8)

$\left(\dfrac{3}{5}\right) = 3^{(5-1)/2}(mod~ 5)=3^2(mod~5)=9(mod~5) = -1$

$\left(\dfrac{4}{5}\right)=\left(\dfrac{2^2}{5}\right)=1$ since $5\nmid2$

$\left(\dfrac{1}{7}\right) = 1$

$\left(\dfrac{2}{7}\right) = 1 (\because 7\equiv-1(mod~8))$

$\left(\dfrac{3}{7}\right) = -\left(\dfrac{7}{3}\right) (\because 3 \equiv3(mod~4) ~and ~7\equiv3(mod~4))$$=-\left(\dfrac{1}{3}\right) =-(1)=-1$

$\left(\dfrac{5}{7}\right) = \left(\dfrac{7}{5}\right) (\because 5\equiv1(mod~4))\\~~~~~~~~~~~~=\left(\dfrac{2}{5}\right)(\because 7\equiv2(mod~5))\\~~~~~~~~~~~~=-1(\because5\equiv-3(mod~8))$

$\left(\dfrac{6}{7}\right) = \left(\dfrac{2}{7}\right)\left(\dfrac{3}{7}\right) = (1)(-1)=-1$

$\left(\dfrac{1}{11}\right) = 1^{(11-1)2}=1$

$\left(\dfrac{2}{11}\right) = -1 ~(\because11\equiv3(mod~8))$

$\left(\dfrac{3}{11}\right) = - \left(\dfrac{11}{3}\right)(\because 11\equiv3(mod~4)~and~3\equiv3(mod~4))\\~~~~~~~~~~~~~~=-\left(\dfrac{2}{3}\right)(\because11\equiv2(mod~3))\\~~~~~~~~~~~~~~=-(-1) (\because 3\equiv3(mod~8))\\~~~~~~~~~~~~~~=1$

$\left(\dfrac{5}{11}\right) = \left(\dfrac{11}{5}\right)(\because 5\equiv1(mod~4))\\~~~~~~~~~~~~~~=\left(\dfrac{1}{5}\right)(\because11\equiv1(mod~5))\\~~~~~~~~~~~~~~=1$

$\left(\dfrac{6}{11}\right) = \left(\dfrac{2}{11}\right)\left(\dfrac{3}{11}\right)\\~~~~~~~~~~~~~~=(-1)(1)\\~~~~~~~~~~~~~~=-1$

7 (. 11 3(mod 4) and 7 3(mod 4) 7 11 (114(mod 11)) 22 7 -(1 72)

Four of the given questions are answered. The remaining can be computed similarly.

I will demonstrate the last one since the numbers seem very large.

$\left(\dfrac{627}{1237}\right) = \left(\dfrac{1237}{627}\right) (\because 1237\equiv1(mod~4))\\~~~~~~~~~~~~~~~~~=\left(\dfrac{610}{627}\right)(\because 1237\equiv610(mod~627))\\~~~~~~~~~~~~~~~~~= \left(\dfrac{2}{627}\right)\left(\dfrac{5}{627}\right)\left(\dfrac{61}{627}\right)\\~~~~~~~~~~~~~~~~~=(-1)\left(\dfrac{627}{5}\right)\left(\dfrac{627}{61}\right)(\because 627\equiv3(mod~8), 5\equiv1(mod~4),61\equiv1(mod~4))\\~~~~~~~~~~~~~~~~~~=(-1)\left(\dfrac{2}{5}\right)\left(\dfrac{17}{61}\right)(\because 627\equiv2(mod~5),627\equiv17(mod~61))\\~~~~~~~~~~~~~~~~~~=(-1)(-1)\left(\dfrac{61}{17}\right)(\because5\equiv3(mod~8),17\equiv1(mod~4))\\~~~~~~~~~~~~~~~~~~=\left(\dfrac{10}{17}\right)(\because 61\equiv10(mod~17))\\~~~~~~~~~~~~~~~~~~=\left(\dfrac{2}{17}\right)\left(\dfrac{5}{17}\right)=(1)\left(\dfrac{17}{5}\right)(\because17\equiv1(mod~8) ,17\equiv1(mod~4))\\£~~~~~~~~~~~~~~~~~~=\left(\dfrac{2}{5}\right)(\because17\equiv2(mod~5))\\~~~~~~~~~~~~~~~~~~=-1(\because 5\equiv-3(mod~8)$

Just by following the same technique, remaining can be calculated.