Pagina:Werkecarlf02gausrich.djvu/133

Demonstr. Si ${\textstyle t}$ non esset divisor ipsius ${\textstyle u}$, sit ${\textstyle gt}$ multiplum ipsius ${\textstyle u}$ proxime maius quam ${\textstyle u}$, adeoque ${\textstyle gt-u}$ integer positivus minor quam ${\textstyle t}$. Ex ${\textstyle k^{t}\equiv 1}$, ${\textstyle k^{u}\equiv 1}$, sequitur ${\textstyle 0\equiv k^{gt}-k^{u}\equiv k^{u}(k^{gt-u}-1)}$, adeoque ${\textstyle k^{gt-u}\equiv 1}$, i.e. datur potestas ipsius ${\textstyle k}$ cum exponente minori quam ${\textstyle t}$ unitati congrua, contra hyp.

Tamquam corollarium hinc sequitur, ${\textstyle t}$ certo metiri numerum ${\textstyle p-1}$.

Numeros tales ${\textstyle k}$, pro quibus ${\textstyle t=p-1}$, etiam hic radices primitivas pro modulo ${\textstyle m}$ vocabimus: quales revera adesse iam ostendemus.

53.

Resolvatur numerus ${\textstyle p-1}$ in factores suos primos, ita ut habeatur $$p-1=a^{\alpha }b^{\beta }c^{\gamma }\ldots$$ designantibus ${\textstyle a}$, ${\textstyle b}$, ${\textstyle c}$ etc. numeros primos reales positivos inaequales. Sint ${\textstyle A}$, ${\textstyle B}$, ${\textstyle C}$ etc. integri (complexi) per ${\textstyle m}$ non divisibiles, atque resp. congruentiis $$x^{\frac {p-1}{a}}\equiv 1,x^{\frac {p-1}{b}}\equiv 1,x^{\frac {p-1}{c}}\equiv 1{\text{ etc.}}$$ secundum modulum ${\textstyle m}$ non satisfacientes, quales dari e theoremate art. 50 manifestum est. Denique sit ${\textstyle h}$ congruus secundum modulum ${\textstyle m}$ producto $$A^{\frac {p-1}{a^{\alpha }}}B^{\frac {p-1}{b^{\beta }}}C^{\frac {p-1}{c^{\gamma }}}\ldots$$ Tunc dico, ${\textstyle h}$ fore radicem primitivam.

Demonstr. Denotando per ${\textstyle t}$ exponentem infimae potestatis ${\textstyle h^{t}}$ unitati congruae, erit, si ${\textstyle h}$ non esset radix primitiva, ${\textstyle t}$ submultiplum ipsius ${\textstyle p-1}$, sive ${\textstyle {\frac {p-1}{t}}}$ integer unitate maior. Manifesto hic integer factores suos primos reales inter hos ${\textstyle a}$, ${\textstyle b}$, ${\textstyle c}$ etc. habebit: supponamus itaque, (quod licet), ${\textstyle {\frac {p-1}{t}}}$ esse divisibilem per ${\textstyle a}$, statuamusque ${\textstyle p-1=atu}$. Erit itaque, propter ${\textstyle h^{t}\equiv 1}$, etiam ${\textstyle h^{tu}\equiv 1}$ sive $$A^{{\frac {p-1}{a^{\alpha }}}\cdot {\frac {p-1}{a}}}B^{{\frac {p-1}{b^{\beta }}}\cdot {\frac {p-1}{a}}}C^{{\frac {p-1}{a^{\gamma }}}\cdot {\frac {p-1}{a}}}\ldots \equiv 1$$ At manifesto ${\textstyle {\frac {p-1}{ab^{\beta }}}}$ est integer, adeoque $$B^{{\frac {p-1}{b^{\beta }}}\cdot {\frac {p-1}{a}}}=(B^{p-1})^{\frac {p-1}{ab^{\beta }}}\equiv 1$$ perinde etiam