Pagina:Werkecarlf02gausrich.djvu/61

THEOREMATIS FUNDAMENTALIS IN THEORIA RESIDUORUM QUADRATICORUM DEMONSTRATIO QUINTA

1.

In introductione iam declaravimus, demonstrationem quintam et tertiam ab eodem lemmate proficisci, quod commoditatis caussa, in signis disquisitioni praesenti adaptatis hoc loco repetere visum est.

Lemma. Sit ${\textstyle m}$ numerus primus (positivus impar), ${\textstyle M}$ integer per ${\textstyle m}$ non divisibilis; capiantur residua minima positiva numerorum

$$M,2M,3M,4M\ldots {\frac {1}{2}}(m-1)M$$

secundum modulum ${\textstyle m}$, quae partim erunt minora quam ${\textstyle {\frac {1}{2}}m}$, partim maiora: posteriorum multitudo sit ${\textstyle =n}$. Tunc erit ${\textstyle M}$ residuum quadraticum ipsius ${\textstyle m}$, vel nonresiduum, prout ${\textstyle n}$ par est, vel impar.

Demonstr. Sint e residuis illis ea, quae minora sunt quam ${\textstyle {\frac {1}{2}}m}$, haec ${\textstyle a}$, ${\textstyle b}$, ${\textstyle c}$, ${\textstyle d}$ etc., reliqua vero, maiora quam ${\textstyle {\frac {1}{2}}m}$, haec ${\textstyle a^{\prime }}$, ${\textstyle b^{\prime }}$, ${\textstyle c^{\prime }}$, ${\textstyle d^{\prime }}$ etc. Posteriorum complementa ad ${\textstyle m}$, puta ${\textstyle m-a^{\prime }}$, ${\textstyle m-b^{\prime }}$, ${\textstyle m-c^{\prime }}$, ${\textstyle m-d^{\prime }}$ etc. manifesto cuncta minora erunt quam ${\textstyle {\frac {1}{2}}m}$, atque tum inter se tum a residuis ${\textstyle a}$, ${\textstyle b}$, ${\textstyle c}$, ${\textstyle d}$ etc. diversa, quamobrem cum his simul sumta, ordine quidem mutato, identica erunt cum omnibus numeris ${\textstyle 1}$, ${\textstyle 2}$, ${\textstyle 3}$, ${\textstyle 4\ldots {\frac {1}{2}}(m-1)}$. Statuendo itaque productum

$$1.2.3.4\ldots {\frac {1}{2}}(m-1)=P$$

erit

$$P=abcd\ldots \times (m-a^{\prime })(m-b^{\prime })(m-c^{\prime })(m-d^{\prime })\ldots$$

adeoque

$$(-1)^{n}P=abcd\ldots \times (a^{\prime }-m)(b^{\prime }-m)(c^{\prime }-m)(d^{\prime }-m)\ldots$$

Porro fit, secundum modulum ${\textstyle m}$,

$$PM^{{\frac {1}{2}}(m-1)}\equiv abcd\ldots \times a^{\prime }b^{\prime }c^{\prime }d^{\prime }\ldots \equiv abcd\ldots \times (a^{\prime }-m)(b^{\prime }-m)(c^{\prime }-m)(d^{\prime }-m)\ldots$$

adeoque

$$PM^{{\frac {1}{2}}(m-1)}\equiv P(-1)^{n}$$

Hinc ${\textstyle M^{{\frac {1}{2}}(m-1)}\equiv \pm 1}$, accepto signo superiori vel inferiori, prout ${\textstyle n}$ par est vel impar, unde adiumento theorematis in Disquisitionibus Arithmeticis art. 106 demonstrati lemmatis veritas sponte demanat.