Zahlbereich/X^3-3X+1/Modulo p/p-te Potenz/Möglichkeiten/Aufgabe/Lösung

{}p=2\,.

Es ist

{}X^{2}=X^{2}-2\,.

{}p=5\,.

Es ist

{}{\begin{aligned}X^{5}&=X^{3}X^{2}\\&={\left(3X-1\right)}X^{2}\\&=3X^{3}-X^{2}\\&=3{\left(3X-1\right)}-X^{2}\\&=-X^{2}+9X-3\\&=-X^{2}-X+2.\end{aligned}}

{}p=7\,.

Es ist

{}{\begin{aligned}X^{7}&={\left(X^{3}\right)}^{2}X\\&={\left(3X-1\right)}^{2}X\\&={\left(9X^{2}-6X+1\right)}X\\&=2X^{3}+X^{2}+X\\&=2{\left(3X-1\right)}+X^{2}+X\\&=X^{2}-2.\end{aligned}}

{}p=11\,.

Es ist

{}{\begin{aligned}X^{11}&={\left(X^{3}\right)}^{3}X^{2}\\&={\left(3X-1\right)}^{3}X^{2}\\&={\left(27X^{3}-27X^{2}+9X-1\right)}X^{2}\\&={\left(5{\left(3X-1\right)}-5X^{2}+9X-1\right)}X^{2}\\&={\left(-5X^{2}+2X-6\right)}X^{2}\\&=-5X{\left(3X-1\right)}+2{\left(3X-1\right)}-6X^{2}\\&=-21X^{2}-2\\&=X^{2}-2.\end{aligned}}

{}p=13\,.

Es ist

{}{\begin{aligned}X^{13}&={\left(X^{3}\right)}^{4}X\\&={\left(3X-1\right)}^{4}X\\&={\left(81X^{4}-108X^{3}+54X^{2}-12X+1\right)}X\\&={\left(3X^{4}+9X^{3}+2X^{2}+X+1\right)}X\\&={\left(3{\left(3X-1\right)}X+9{\left(3X-1\right)}+2X^{2}+X+1\right)}X\\&={\left(11X^{2}-X+5\right)}X\\&=11X^{3}-X^{2}+5X\\&=-2{\left(3X-1\right)}-X^{2}+5X\\&=-X^{2}-X+2.\end{aligned}}

{}p=19\,.

Wegen

{}3^{3}-3\cdot 3+1=19=0\,

modulo ${}19$ besitzt ${}X^{3}-3X+1$ eine Nullstelle. Daraus folgt

{}X^{19}=X

.