Pagina:Werkecarlf02gausrich.djvu/121

Demonstr. I. Sint ${\textstyle \alpha }$, ${\textstyle \beta }$ integri tales qui faciant ${\textstyle \alpha a+\beta b=1}$, unde erit

$$i=\alpha b-\beta a+m(\beta +\alpha i)$$

Proposito itaque numero integro complexo ${\textstyle A+Bi}$, habebimus

$$A+Bi=A+(\alpha b-\beta a)B+m(\beta B+\alpha Bi)$$

Quare denotando per ${\textstyle h}$ residuum minimum positivum numeri ${\textstyle A+(\alpha b-\beta a)B}$ secundum modulum ${\textstyle p}$, statuendoque

$$A+(\alpha b-\beta a)B=h+kp=h+m(ak-bki)$$

erit

$$A+Bi=h+m(\beta B+ak+(\alpha B-bk)i)$$

sive

$$A+Bi\equiv h{\pmod {m}}.\quad {\text{ Q. E. P. }}$$

II. Quoties eidem numero complexo duo numeri reales ${\textstyle h}$, ${\textstyle h^{\prime }}$ secundum modulum ${\textstyle m}$ congrui sunt, etiam inter se congrui erunt. Statuamus itaque ${\textstyle h-h^{\prime }=m(c+di)}$, unde fit

$$(h-h^{\prime })(a-bi)=p(c+di)$$

adeoque

$$(h-h^{\prime })a=pc,\quad (h-h^{\prime })b=-pd$$

nec non, propter ${\textstyle a\alpha +b\beta =1}$,

$$h-h^{\prime }=p(c\alpha -d\beta ),{\text{ i.e. }}h\equiv h^{\prime }{\pmod {p}}$$

Quapropter ${\textstyle h}$ et ${\textstyle h^{\prime }}$, siquidem sunt inaequales, ambo simul in complexu numerorum ${\textstyle 0}$, ${\textstyle 1}$, ${\textstyle 2}$, ${\textstyle 3\ldots p-1}$ contenti esse nequeunt. Q. E. S.

41.

Theorema. Secundum modulum complexum ${\textstyle m=a+bi}$, cuius norma ${\textstyle aa+bb=p}$, et pro quo ${\textstyle a}$, ${\textstyle b}$ non sunt inter se primi, sed divisorem communem maximum ${\textstyle \lambda }$ habent (quem positive acceptum supponimus), quilibet numerus complexus congruus est residuo ${\textstyle x+yi}$ tali, ut ${\textstyle x}$ sit aliquis numerorum ${\textstyle 0,1,2,3\ldots {\frac {p}{\lambda }}-1}$, atque ${\textstyle y}$ aliquis horum ${\textstyle 0}$, ${\textstyle 1}$, ${\textstyle 2}$, ${\textstyle 3\ldots \lambda -1}$, et quidem unico tantum inter omnia ${\textstyle p}$ residua, quae tali forma gaudent.