Pagina:Werkecarlf02gausrich.djvu/65

THEOREMATIS FUNDAMENTALIS IN THEORIA RESIDUORUM QUADRATICORUM DEMONSTRATIO SEXTA

1.

Theorema. Designante ${\textstyle p}$ numerum primum (positivum imparem), ${\textstyle n}$ integrum positivum per ${\textstyle p}$ non divisibilem, ${\textstyle x}$ quantitatem indeterminatam, functio

$$1+x^{n}+x^{2n}+x^{3n}+{\text{etc.}}+x^{np-n}$$

divisibilis erit per

$$1+x+xx+x^{3}+{\text{etc.}}+x^{p-1}$$

Demonstr. Accipiatur integer positivus ${\textstyle g}$ ita ut fiat ${\textstyle gn\equiv 1{\pmod {p}}}$, statuaturque ${\textstyle gn=1+hp}$. Tunc erit

$${\begin{aligned}{\frac {1+x^{n}+x^{2n}+x^{3n}+{\text{etc.}}+x^{np-n}}{1+x+xx+x^{3}+{\text{etc.}}+x^{p-1}}}&={\frac {(1-x^{np})(1-x)}{(1-x^{n})(1-x^{p})}}\\&={\frac {(1-x^{np})(1-x^{gn}-x+x^{hp+1})}{(1-x^{n})(1-x^{p})}}\\&={\frac {1-x^{np}}{1-x^{p}}}\cdot {\frac {1-x^{gn}}{1-x^{n}}}-{\frac {x(1-x^{np})}{1-x^{n}}}\cdot {\frac {1-x^{hp}}{1-x^{p}}}\end{aligned}}$$

adeoque manifesto functio integra. Q. E. D.

Quaelibet itaque functio integra ipsius ${\textstyle x}$ per ${\textstyle {\frac {1-x^{np}}{1-x^{n}}}}$ divisibilis, etiam divisibilis erit per ${\textstyle {\frac {1-x^{p}}{1-x}}}$.

2.

Designet ${\textstyle \alpha }$ radicem primitivam positivam pro modulo ${\textstyle p}$, i.e. sit ${\textstyle \alpha }$ integer positivus talis, ut residua minima positiva potestatum ${\textstyle 1}$, ${\textstyle \alpha }$, ${\textstyle \alpha \alpha }$, ${\textstyle \alpha ^{3}\ldots \alpha ^{p-2}}$ secundum modulum ${\textstyle p}$ sine respectu ordinis cum numeris ${\textstyle 1}$, ${\textstyle 2}$, ${\textstyle 3}$, ${\textstyle 4\ldots p-1}$ identica fiant. Designando porro per ${\textstyle fx}$ functionem

$$x+x^{\alpha }+x^{\alpha \alpha }+x^{\alpha ^{3}}+{\text{etc.}}+x^{\alpha ^{p-2}}+1$$

patet, ${\textstyle fx-1-x-xx-x^{3}-{\text{etc.}}-x^{p-1}}$ divisibilem fore per ${\textstyle 1-x^{p}}$, adeoque a potiori per ${\textstyle {\frac {1-x^{p}}{1-x}}=1+x+xx+x^{3}+{\text{etc.}}+x^{p-1}}$, per quam itaque functionem ipsa quoque ${\textstyle fx}$ divisibilis erit. Hinc vero sequitur, quum ${\textstyle x}$ exprimat quantitatem indeterminatam, esse quoque ${\textstyle f(x^{n})}$ divisibilem per ${\textstyle {\frac {1-x^{np}}{1-x^{n}}}}$, et proin (art. praec.) etiam per ${\textstyle {\frac {1-x^{p}}{1-x}}}$, quoties quidem ${\textstyle n}$ sit integer per ${\textstyle p}$ non divisibilis. Contra, quoties ${\textstyle n}$ est integer per ${\textstyle p}$ divisibilis, singulae partes functionis ${\textstyle f(x^{n})}$ uni-