Pagina:Werkecarlf02gausrich.djvu/94

Iam si potestas ${\textstyle (z^{4}+1)^{{\frac {1}{4}}(p-1)}}$ secundum theorema binomiale evolvitur, per lemma praec. fiet

$$\Sigma (z^{4}+1)^{{\frac {1}{4}}(p-1)}\equiv -2{\pmod {p}}$$

Sed residua minima omnium ${\textstyle z^{4}}$ exhibent omnes numeros ${\textstyle A}$, quovis quater occurrente; habebimus itaque inter residua minima ipsius ${\textstyle z^{4}+1}$

$${\begin{aligned}&4(00){\text{ ad }}A\\&4(01){\text{ ad }}B\\&4(02){\text{ ad }}C\\&4(03){\text{ ad }}D\end{aligned}}$$

pertinentia, quatuorque erunt ${\textstyle =0}$ (puta pro ${\textstyle z^{4}\equiv p-1}$). Hinc, considerando criteria complexuum ${\textstyle A}$, ${\textstyle B}$, ${\textstyle C}$, ${\textstyle D}$, deducimus

$$\Sigma (z^{4}+1)^{{\frac {1}{4}}(p-1)}\equiv 4(00)+4f(01)-4(02)-4f(03)$$

adeoque

$$-2\equiv 4(00)+4f(01)-4(02)-4f(03)$$

sive substitutis pro ${\textstyle (00)}$, ${\textstyle (01)}$ etc. valoribus in art. praec inventis,

$$-2\equiv -2a-2-2bf$$

Hinc itaque colligimus, semper fieri debere ${\textstyle a+bf\equiv 0}$, sive, multiplicando per ${\textstyle f}$,

$$b\equiv af$$

quae congruentia determinationi signi ipsius ${\textstyle b}$, si numerus ${\textstyle f}$ iam electus est, vel determinationi numeri ${\textstyle f}$, si signum ipsius ${\textstyle b}$ aliunde praescribitur, inservit.

20.

Postquam problema nostrum pro modulis formae ${\textstyle 8n+1}$ complete solvimus, progredimur ad casum alterum, ubi ${\textstyle p}$ est formae ${\textstyle 8n+5}$: quem eo brevius absolvere licebit, quod omnia ratiocinia parum a praecedentibus differunt.

Quum pro tali modulo ${\textstyle -1}$ ad classem ${\textstyle C}$ pertineat, complementa numerorum complexuum ${\textstyle A}$, ${\textstyle B}$, ${\textstyle C}$, ${\textstyle D}$ ad summam ${\textstyle p}$, in classibus ${\textstyle C}$, ${\textstyle D}$, ${\textstyle A}$, ${\textstyle B}$ resp. contenta erunt. Hinc facile colligitur