Pagina:Werkecarlf02gausrich.djvu/129

ubi ${\textstyle h}$, ${\textstyle h^{\prime }}$ erunt integri, et quidem, si designatione in Disquiss. Ar. art. 27 introducta uti placet

$${\begin{aligned}&h=\pm \left[k^{\prime },k^{\prime \prime },k^{\prime \prime \prime }\ldots k^{(n-1)}\right]=\pm \left[k^{(n-1)},k^{(n-2)}\ldots k^{\prime \prime },k^{\prime }\right]\\&h^{\prime }=\mp \left[k,k^{\prime },k^{\prime \prime },k^{\prime \prime \prime }\ldots k^{(n-1)}\right]=\mp \left[k^{(n-1)},k^{n-2)}\ldots k^{\prime \prime },k^{\prime },k\right]\end{aligned}}$$

valentibus signis superioribus vel inferioribus, prout ${\textstyle n}$ par est vel impar. Hoc theorema ita enunciamus:

Divisor communis maximus duorum numerorum complexorum ${\textstyle m}$, ${\textstyle m^{\prime }}$ redigi potest ad formam ${\textstyle hm+h^{\prime }m^{\prime }}$, ita ut ${\textstyle h}$, ${\textstyle h^{\prime }}$ sint integri.

Manifesto enim hoc non solum de eo divisore communi maximo valet, ad quem algorithmus art. praec. deduxit, sed etiam de tribus illi associatis, pro quibus loco coëfficientium ${\textstyle h}$, ${\textstyle h^{\prime }}$ accipere oportebit vel hos ${\textstyle hi}$, ${\textstyle h^{\prime }i}$ vel ${\textstyle -h}$, ${\textstyle -h^{\prime }}$, vel ${\textstyle -hi}$, ${\textstyle -h^{\prime }i}$.

Quoties itaque numeri ${\textstyle m}$, ${\textstyle m^{\prime }}$ inter se primi sunt, satisfieri poterit aequationi

$$1=hm+h^{\prime }m^{\prime }$$

Propositi sint e.g. numeri ${\textstyle 31+6i=m}$, ${\textstyle 11-20i=m^{\prime }}$. Hic invenimus

$${\begin{array}{clccl}k&={\phantom {+1-\;}}i,&\quad &m^{\prime \prime }&=+11-5i\\k^{\prime }&=+1-i,&\quad &m^{\prime \prime \prime }&=+{\phantom {0}}5-4i\\k^{\prime \prime }&=+2,&\quad &m^{\prime \prime \prime \prime }&=+{\phantom {0}}1+3i\\k^{\prime \prime \prime }&=-1-2i,&\quad &m^{\prime \prime \prime \prime }&={\phantom {+00}}+i\\k^{\prime \prime \prime \prime }&=+3-i&\quad &&\end{array}}$$

atque hinc

$${\begin{aligned}&{\left[k^{\prime },k^{\prime \prime },k^{\prime \prime \prime }\right]=-6-5i}\\&{\left[k,k^{\prime },k^{\prime \prime },k^{\prime \prime \prime }\right]=+4-10i}\end{aligned}}$$

et proin

$$m^{\prime \prime \prime \prime \prime }=i=(6+5i)m+(4-10i)m^{\prime }$$

nec non

$$1=(5-6i)m+(-10-4i)m^{\prime }$$

quod calculo instituto confirmatur.

48.

Per praecedentia omnia, quae ad theoriam congruentiarum primi gradus in arithmetica numerorum complexorum requiruntur, praeparata sunt: sed quum illa