Pagina:Werkecarlf02gausrich.djvu/125

$${\begin{array}{r|l}x&y\\\hline -1&3,4,5\\0&0,1,2,3,4,5\\+1&1,2,3,4,5,6\\+2&1,2,3,4,5,6\\+3&2,3,4,5,6\\+4&2,3,4\\\end{array}}$$

Simili modo pro residuis absolute minimis, ${\textstyle \xi }$ et ${\textstyle \eta }$ alicui numerorum ${\textstyle -14}$, ${\textstyle -13}$, ${\textstyle -12\ldots +14}$ aequales esse debent; hinc ${\textstyle 29x}$ nequit esse extra limites ${\textstyle -7.14}$ et ${\textstyle +7.14}$, adeoque ${\textstyle x}$ alicui numerorum ${\textstyle -3}$, ${\textstyle -2}$, ${\textstyle -1}$, ${\textstyle 0}$, ${\textstyle 1}$, ${\textstyle 2}$, ${\textstyle 3}$ aequalis esse debet. Pro ${\textstyle x=-3}$ erit ${\textstyle 2y=\xi -5x=\xi +15}$ alicui numerorum ${\textstyle 1}$, ${\textstyle 2}$, ${\textstyle 3\ldots 29}$ aequalis, ${\textstyle 5y=\eta +2x=\eta -6}$ autem alicui horum ${\textstyle -20}$, ${\textstyle -19}$, ${\textstyle -18\ldots +8}$: hinc prodit pro ${\textstyle y}$ valor unicus ${\textstyle +1}$. Tractando eodem modo valores reliquos ipsius ${\textstyle x}$, habemus schema omnium residuorum absolute minimorum:

$${\begin{array}{r|l}x&y\\\hline -3&+1\\-2&-2,-1,0,+1,+2\\-1&-3,-2,-1,0,+1,+2\\0&-2,-1,0,+1,+2\\+1&-2,-1,0,+1,+2,+3\\+2&-2,-1,0,+1,+2,\\+3&-1\\\end{array}}$$

45.

In applicatione methodi secundae duos casus distinguere conveniet.

In casu priori, ubi ${\textstyle a}$ et ${\textstyle b}$ divisorem communem non habent, fiat ${\textstyle \alpha a+\beta b=1}$, sitque ${\textstyle k}$ residuum minimum positivum ipsius ${\textstyle \beta a-\alpha b}$ secundum modulum ${\textstyle p}$. Hinc aequationes identicae

$$a(\beta a-\alpha b)=\beta p-b(\alpha a+\beta b),\quad b(\beta a-\alpha b)=-\alpha p+a(\alpha a+\beta b)$$

docent, esse ${\textstyle ak\equiv -b}$, ${\textstyle bk\equiv a{\pmod {p}}}$. Statuendo itaque ut supra ${\textstyle ax+by=\xi }$,