Fibonacci-Folgen und Lucas-Folgen: Lösungen

Lösungen

• Zeige anhand von ${\displaystyle a={\frac {P+{\sqrt {P^{2}-4Q}}}{2}}}$ und ${\displaystyle b={\frac {P-{\sqrt {P^{2}-4Q}}}{2}}}$, das ${\displaystyle a+b=P\ }$ und ${\displaystyle a\cdot b=Q}$ gilt.
A. ${\displaystyle a+b=P\ }$

${\displaystyle a+b={\frac {P+{\sqrt {P^{2}-4Q}}}{2}}+{\frac {P-{\sqrt {P^{2}-4Q}}}{2}}}$

${\displaystyle a+b={\frac {P+{\sqrt {P^{2}-4Q}}+P-{\sqrt {P^{2}-4Q}}}{2}}}$

${\displaystyle a+b={\frac {2P}{2}}}$

${\displaystyle a+b=P\ }$

B. ${\displaystyle a\cdot b=Q}$

${\displaystyle a\cdot b={\frac {P+{\sqrt {P^{2}-4Q}}}{2}}\cdot {\frac {P-{\sqrt {P^{2}-4Q}}}{2}}}$

${\displaystyle a\cdot b={\frac {(P+{\sqrt {P^{2}-4Q}})\cdot (P-{\sqrt {P^{2}-4Q}})}{4}}}$

${\displaystyle a\cdot b={\frac {P^{2}-({\sqrt {P^{2}-4Q}})^{2}}{4}}}$

${\displaystyle a\cdot b={\frac {P^{2}-(P^{2}-4Q)}{4}}}$

${\displaystyle a\cdot b={\frac {P^{2}-P^{2}+4Q}{4}}}$

${\displaystyle a\cdot b={\frac {4Q}{4}}}$

${\displaystyle a\cdot b=Q\ }$

• Zeige die Gleichheit von ${\displaystyle V_{p}(a+1,a)\equiv V_{1}(a+1,a)\mod p}$ und ${\displaystyle a^{p}\equiv a\mod p}$.