Pagina:Werkecarlf02gausrich.djvu/93

$${\begin{aligned}&8h=4n-3a-5\\&8i=4n+a-2b-1\\&8k=4n+a-1\\&8l=4n+a+2b-1\\&8m=4n-a+1\end{aligned}}$$

Si loco ipsius ${\textstyle n}$ modulum ${\textstyle p}$ introducere malumus, schema ${\textstyle {S}}$, singulis terminis ad evitandas fractiones per 16 multiplicatis, ita se habet:

$${\begin{array}{l|l|l|l}p-6a-11&p+2a-4b-3&p+2a-3&p+2a+4b-3\\p+2a-4b-3&p+2a+4b-3&p-2a+1&p-2a+1\\p+2a-3&p-2a+1&p+2a-3&p-2a+1\\p+2a+4b-3&p-2a+1&p-2a+1&p+2a-4b-3\end{array}}']$$

19.

Superest, ut signum ipsi ${\textstyle b}$ tribuendum assignare doceamus. Iam supra, art. 10, monuimus, distinctionem inter complexus ${\textstyle B}$ et ${\textstyle D}$, per se non essentialem, ab electione numeri ${\textstyle f}$ pendere, pro quo alterutra radix congruentiae ${\textstyle xx\equiv -1}$ accipi debet, illasque inter se permutari, si loco alterius radicis altera adoptetur. Iam quum inspectio schematis modo allati doceat, similem permutationem cum mutatione signi ipsius ${\textstyle b}$ cohaerere, praevidere licet, nexum inter signum ipsius ${\textstyle b}$ atque numerum ${\textstyle f}$ exstare debere. Quem ut cognoscamus, ante omnia observamus, si, denotante ${\textstyle \mu }$ integrum non negativum, pro ${\textstyle z}$ accipiantur omnes numeri ${\textstyle 1}$, ${\textstyle 2}$, ${\textstyle 3\ldots p-1}$, fieri secundum modulum ${\textstyle p}$, vel ${\textstyle \Sigma z^{\mu }\equiv 0}$, vel ${\textstyle \Sigma z^{\mu }\equiv -1}$, prout ${\textstyle \mu }$ vel non-divisibilis sit per ${\textstyle p-1}$, vel divisibilis. Pars posterior theorematis inde patet, quod pro valore ipsius ${\textstyle \mu }$ per ${\textstyle p-1}$ divisibili, habetur ${\textstyle z^{\mu }\equiv 1}$: partem priorem vero ita demonstramus. Denotante ${\textstyle g}$ radicem primitivam, omnes ${\textstyle z}$ convenient cum residuis minimis omnium ${\textstyle g^{y}}$, accipiendo pro ${\textstyle y}$ omnes numeros ${\textstyle 0}$, ${\textstyle 1}$, ${\textstyle 2}$, ${\textstyle 3\ldots p-2}$, eritque adeo ${\textstyle \Sigma z^{\mu }\equiv \Sigma g^{\mu y}}$. Sed fit

$$\Sigma g^{\mu y}={\frac {g^{\mu (p-1)-1}}{g^{\mu }-1}},{\text{ adeoque }}(g^{\mu }-1)\Sigma z^{\mu }\equiv g^{\mu (p-1)}-1\equiv 0$$

Hinc vero sequitur, quoniam pro valore ipsius ${\textstyle \mu }$ per ${\textstyle p-1}$ non-divisibili ${\textstyle g^{\mu }}$ ipsi ${\textstyle 1}$ congruus sive ${\textstyle g^{\mu }-1}$ per ${\textstyle p}$ divisibilis esse nequit, ${\textstyle \Sigma z^{\mu }\equiv 0}$. Q. E. D.