I'm not sure it's advanced, but I got this task on a Number Theory course. Compute:
For 1. (easier one):
[MATH]\left(\frac{123}{917}\right) = \left(\frac{41 \cdot 3}{917}\right) = \left(\frac{41}{917}\right)\left(\frac{3}{917}\right) = (\ast)[/MATH]From Reciprocity Law [MATH]41 \equiv 1\ (\text{mod }4)[/MATH], so [MATH]\left(\frac{41}{917}\right) = \left(\frac{917}{41}\right) = \left(\frac{41 \cdot 22 + 15}{41}\right) = \left(\frac{15}{41}\right)[/MATH]RL again: [MATH]917 \equiv 1\ (\text{mod }4)[/MATH], so [MATH]\left(\frac{3}{917}\right) = \left(\frac{917}{3}\right) = \left(\frac{3 \cdot 305 + 2}{3}\right)[/MATH][MATH](\ast)=\left(\frac{15}{41}\right)\left(\frac{2}{3}\right)= (\ast)[/MATH]RL: [MATH]41 \equiv 1\ (\text{mod }4)[/MATH], so [MATH]\left(\frac{15}{41}\right) = \left(\frac{41}{15}\right)= \left(\frac{2 \cdot 15 + 11}{15}\right)= \left(\frac{11}{15}\right)[/MATH][MATH]\left(\frac{2}{3}\right) = -1[/MATH], because [MATH]3\equiv 3\ (\text{mod }8)[/MATH]RL: [MATH]11 \equiv 15\equiv 3\ (\text{mod }4)[/MATH], so [MATH]\left(\frac{11}{15}\right) = -\left(\frac{15}{11}\right)[/MATH][MATH](\ast)=-\left(\frac{15}{11}\right)(-1)= \left(\frac{3 \cdot 5}{11}\right) = \left(\frac{3}{11}\right)\left(\frac{5}{11}\right) = \left(\frac{11}{3}\right)\left(\frac{11}{5}\right) = \left(\frac{2}{3}\right)\left(\frac{1}{5}\right) = 1 \cdot 1 = 1[/MATH]
For 2. (I'm not sure this approach is right):
[MATH]\left(\frac{28}{p}\right) = \left(\frac{2 \cdot 2 \cdot 7}{p}\right) = \left(\frac{2}{p}\right)^2\left(\frac{7}{p}\right)[/MATH]So now [MATH]\left(\frac{2}{p}\right)^2[/MATH] is either 0 for even [MATH]p[/MATH], or [MATH](\pm1)^2 = 1[/MATH] for others, so [MATH]\left(\frac{28}{p}\right) = 0[/MATH] for even [MATH]p[/MATH] and [MATH]\left(\frac{7}{p}\right)[/MATH] for others. Now I look at all numbers: [MATH]p \equiv a\ (\text{mod }2\cdot 7)[/MATH]:
- [MATH]\left(\frac{123}{917}\right)[/MATH]
- [MATH]\left(\frac{28}{p}\right)[/MATH] for any [MATH]p \in \mathbb{Z}[/MATH]
For 1. (easier one):
[MATH]\left(\frac{123}{917}\right) = \left(\frac{41 \cdot 3}{917}\right) = \left(\frac{41}{917}\right)\left(\frac{3}{917}\right) = (\ast)[/MATH]From Reciprocity Law [MATH]41 \equiv 1\ (\text{mod }4)[/MATH], so [MATH]\left(\frac{41}{917}\right) = \left(\frac{917}{41}\right) = \left(\frac{41 \cdot 22 + 15}{41}\right) = \left(\frac{15}{41}\right)[/MATH]RL again: [MATH]917 \equiv 1\ (\text{mod }4)[/MATH], so [MATH]\left(\frac{3}{917}\right) = \left(\frac{917}{3}\right) = \left(\frac{3 \cdot 305 + 2}{3}\right)[/MATH][MATH](\ast)=\left(\frac{15}{41}\right)\left(\frac{2}{3}\right)= (\ast)[/MATH]RL: [MATH]41 \equiv 1\ (\text{mod }4)[/MATH], so [MATH]\left(\frac{15}{41}\right) = \left(\frac{41}{15}\right)= \left(\frac{2 \cdot 15 + 11}{15}\right)= \left(\frac{11}{15}\right)[/MATH][MATH]\left(\frac{2}{3}\right) = -1[/MATH], because [MATH]3\equiv 3\ (\text{mod }8)[/MATH]RL: [MATH]11 \equiv 15\equiv 3\ (\text{mod }4)[/MATH], so [MATH]\left(\frac{11}{15}\right) = -\left(\frac{15}{11}\right)[/MATH][MATH](\ast)=-\left(\frac{15}{11}\right)(-1)= \left(\frac{3 \cdot 5}{11}\right) = \left(\frac{3}{11}\right)\left(\frac{5}{11}\right) = \left(\frac{11}{3}\right)\left(\frac{11}{5}\right) = \left(\frac{2}{3}\right)\left(\frac{1}{5}\right) = 1 \cdot 1 = 1[/MATH]
For 2. (I'm not sure this approach is right):
[MATH]\left(\frac{28}{p}\right) = \left(\frac{2 \cdot 2 \cdot 7}{p}\right) = \left(\frac{2}{p}\right)^2\left(\frac{7}{p}\right)[/MATH]So now [MATH]\left(\frac{2}{p}\right)^2[/MATH] is either 0 for even [MATH]p[/MATH], or [MATH](\pm1)^2 = 1[/MATH] for others, so [MATH]\left(\frac{28}{p}\right) = 0[/MATH] for even [MATH]p[/MATH] and [MATH]\left(\frac{7}{p}\right)[/MATH] for others. Now I look at all numbers: [MATH]p \equiv a\ (\text{mod }2\cdot 7)[/MATH]:
- [MATH]a = 1 \implies \left(\frac{28}{p}\right) = 1 \cdot\left(\frac{7}{1}\right) = \left(\frac{1}{7}\right) = 1[/MATH]
- [MATH]a = 3 \implies \left(\frac{28}{p}\right) = 1 \cdot\left(\frac{7}{3}\right) = \left(\frac{1}{3}\right) = 1[/MATH]
- [MATH]a = 5 \implies \left(\frac{28}{p}\right) = 1 \cdot\left(\frac{7}{5}\right) = \left(\frac{2}{5}\right) = -1[/MATH], because [MATH]5\equiv 5\ (\text{mod }8)[/MATH]
- [MATH]a = 7 \implies \left(\frac{28}{p}\right) = 1 \cdot\left(\frac{7}{7}\right) = 0[/MATH]
- [MATH]a = 9 \implies \left(\frac{28}{p}\right) = 1 \cdot\left(\frac{7}{9}\right) = \left(\frac{9}{7}\right) = \left(\frac{2}{7}\right) = 1[/MATH], because [MATH]7\equiv 1\ (\text{mod }8)[/MATH]
- [MATH]a = 11 \implies \left(\frac{28}{p}\right) = 1 \cdot\left(\frac{7}{11}\right) = -\left(\frac{11}{7}\right) = -\left(\frac{4}{7}\right) = -\left(\frac{2}{7}\right)^2 = -1[/MATH]
- [MATH]a = 13 \implies \left(\frac{28}{p}\right) = 1 \cdot\left(\frac{7}{13}\right) = \left(\frac{13}{7}\right) = \left(\frac{6}{7}\right) = \left(\frac{2}{7}\right)\left(\frac{3}{7}\right) = \left(\frac{3}{7}\right) = -\left(\frac{7}{3}\right) = -\left(\frac{1}{3}\right) = -1[/MATH]