Pagina:Werkecarlf02gausrich.djvu/90

eum qui reddit ${\textstyle \beta \delta \equiv 1}$ (qui manifesto erit e complexu ${\textstyle D}$), et pro ${\textstyle \delta ^{\prime }}$ residuum minimum positivum producti ${\textstyle \alpha \delta }$ (quod itidem erit e complexu ${\textstyle D}$); perinde patet regressus a solutione data congruentiae ${\textstyle 1+\delta +\delta ^{\prime }\equiv 0}$ ad solutionem congruentiae ${\textstyle 1+\alpha +\beta \equiv 0}$, si ${\textstyle \beta }$ accipitur ita, ut fiat ${\textstyle \beta \delta \equiv 1}$, simulque statuitur ${\textstyle \alpha \equiv \beta \delta ^{\prime }}$. Hinc concludimus, utramque congruentiam aequali solutionum multitudine gaudere, sive esse ${\textstyle (01)=(33)}$.

Simili modo e congruentia ${\textstyle 1+\alpha +\gamma \equiv 0}$ deducimus ${\textstyle \gamma ^{\prime }+\gamma ^{\prime \prime }+1\equiv 0}$, si ${\textstyle \gamma ^{\prime }}$ accipitur e complexu ${\textstyle C}$ ita ut fiat ${\textstyle \gamma \gamma ^{\prime }\equiv 1}$, atque ${\textstyle \gamma ^{\prime \prime }}$ ex eodem complexu congruus producto ${\textstyle \alpha \gamma ^{\prime }}$. Unde facile colligimus, has duas congruentias aequalem solutionum multitudinem admittere, sive esse ${\textstyle (02)=(22)}$.

Perinde e congruentia ${\textstyle 1+\alpha +\delta \equiv 0}$ deducimus ${\textstyle \beta +\beta ^{\prime }+1\equiv 0}$, accipiendo ${\textstyle \beta }$, ${\textstyle \beta ^{\prime }}$ ita ut fiat ${\textstyle \beta \delta \equiv 1,\beta \alpha \equiv \beta ^{\prime }}$, eritque adeo ${\textstyle (03)=(11)}$.

Denique e congruentia ${\textstyle 1+\beta +\gamma \equiv 0}$ simili modo tum congruentiam ${\textstyle \delta +1+\beta ^{\prime }\equiv 0}$, tum hanc ${\textstyle \gamma ^{\prime }+\delta ^{\prime }+1\equiv 0}$ derivamus, atque hinc concludimus ${\textstyle (12)=(13)=(23)}$.

Nacti sumus itaque, inter sedecim incognitas nostras, undecim aequationes, ita ut illae ad quinque reducantur, schemaque ${\textstyle S}$ ita exhiberi possit:

$${\begin{array}{llll}h,&i,&k,&l\\i,&l,&m,&m\\k,&m,&k,&m\\l,&m,&m,&i\end{array}}$$

Facile vero tres novae aequationes conditionales adiiciuntur. Quum enim quemvis numerum complexus ${\textstyle A}$, excepto ultimo ${\textstyle p-1}$, sequi debeat numerus ex aliquo complexuum ${\textstyle A}$, ${\textstyle B}$, ${\textstyle C}$ vel ${\textstyle D}$, habebimus

$$(00)+(01)+(02)+(03)=2n-1$$

et perinde

$${\begin{aligned}&(10)+(11)+(12)+(13)=2n\\&(20)+(21)+(22)+(23)=2n\\&(30)+(31)+(32)+(33)=2n\end{aligned}}$$

In signis modo introductis tres primae aequationes suppeditant:

$${\begin{aligned}h+i+k+l&=2n-1\\i+l+2m&=2n\\k+m&=n\end{aligned}}$$