Pagina:Werkecarlf02gausrich.djvu/122

Demonstr. I. Accipiendo integros ${\textstyle \alpha ,\beta }$ ita, ut fiat ${\textstyle \alpha a+\beta b=\lambda }$, erit ${\textstyle \lambda i=\alpha b-\beta a+m(\beta +\alpha i)}$. Iam sit ${\textstyle A+Bi}$ numerus complexus propositus, ${\textstyle y}$ residuum minimum positivum ipsius ${\textstyle B}$ secundum modulum ${\textstyle \lambda }$, atque ${\textstyle x}$ residuum minimum positivum ipsius ${\textstyle A+(\alpha b-\beta a)\cdot {\frac {B-y}{\lambda }}}$ secundum modulum ${\textstyle {\frac {p}{\lambda }}}$, statuaturque

$$A+(\alpha b-\beta a)\cdot {\frac {B-y}{\lambda }}=x+{\frac {p}{\lambda }}\cdot k$$

Hinc erit

$${\begin{aligned}A+Bi-(x+yi)&={\frac {p}{\lambda }}\cdot k+(B-y)i-(\alpha b-\beta a){\frac {B-y}{\lambda }}\\&={\frac {p}{\lambda }}\cdot k+{\frac {B-y}{\lambda }}\cdot m(\beta +\alpha i)\\&=({\frac {a}{\lambda }}-{\frac {b}{\lambda }}\cdot i)km+{\frac {B-y}{\lambda }}(\beta +\alpha i)m\end{aligned}}$$

i.e. per ${\textstyle m}$ divisibilis, sive ${\textstyle A+Bi\equiv x+yi{\pmod {m}}\quad }$ Q. E. P.

II. Supponamus, secundum modulum ${\textstyle m}$ eidem numero complexo congruos esse duos numeros ${\textstyle x+yi}$, ${\textstyle x^{\prime }+y^{\prime }i}$, qui proin etiam inter se congrui erunt secundum modulum ${\textstyle m}$. A potiori itaque secundum modulum ${\textstyle \lambda }$ congrui erunt, adeoque ${\textstyle y\equiv y^{\prime }{\pmod {\lambda }}}$. Quodsi igitur uterque ${\textstyle y}$, ${\textstyle y^{\prime }}$ inter numeros ${\textstyle 0}$, ${\textstyle 1}$, ${\textstyle 2}$, ${\textstyle 3\ldots \lambda -1}$ contentus esse supponitur, necessario debet esse ${\textstyle y=y^{\prime }}$. Hoc pacto vero etiam fiet ${\textstyle x\equiv x^{\prime }{\pmod {m}}}$, i.e. ${\textstyle x-x^{\prime }}$ per ${\textstyle m}$, adeoque ${\textstyle {\frac {x-x^{\prime }}{\lambda }}}$ integer per ${\textstyle {\frac {a}{\lambda }}+{\frac {b}{\lambda }}\cdot i}$ divisibilis, sive

$${\frac {x-x^{\prime }}{\lambda }}\equiv 0{\pmod {{\frac {a}{\lambda }}+{\frac {b}{\lambda }}\cdot i}}$$

Hinc autem, quum ${\textstyle {\frac {a}{\lambda }}}$, ${\textstyle {\frac {b}{\lambda }}}$ sint numeri inter se primi, concluditur per partem secundam theorematis art. praec., ${\textstyle {\frac {x-x^{\prime }}{\lambda }}}$ etiam per normam numeri ${\textstyle {\frac {a}{\lambda }}+{\frac {b}{\lambda }}\cdot i}$, i.e. per numerum ${\textstyle {\frac {p}{\lambda \lambda }}}$ divisibilem fore, adeoque ${\textstyle x-x^{\prime }}$ per ${\textstyle {\frac {p}{\lambda }}}$. Quapropter si etiam uterque ${\textstyle x}$, ${\textstyle x^{\prime }}$ in complexu numerorum ${\textstyle 0}$, ${\textstyle 1}$, ${\textstyle 2}$, ${\textstyle 3\ldots {\frac {p}{\lambda }}-1}$ contentus esse supponitur, necessario erit ${\textstyle x=x^{\prime }}$, sive residua ${\textstyle x+yi}$, ${\textstyle x^{\prime }+y^{\prime }i}$ identica. Q. E. S.

Ceterum sponte patet, huc quoque referendum esse casum, ubi modulus est numerus realis, puta ${\textstyle b=0}$, et proin ${\textstyle \lambda =\pm a}$, nec non eum, ubi modulus est numerus pure imaginarius, puta ${\textstyle a=0}$, et proin ${\textstyle \lambda =\pm b}$. In utroque casu habetur ${\textstyle {\frac {p}{\lambda }}=\lambda }$.