Pagina:Werkecarlf02gausrich.djvu/46

II. in casu eo, ubi ${\textstyle n=8}$, vel altior potestas binarii cum exponente impari, fieri

$${\begin{array}{l}l=1,{\text{ si fuerit }}h{\text{ formae }}8\mu +1,\\l=-1,{\text{ si fuerit }}h{\text{ formae }}8\mu +5,\\l=+i,{\text{ si fuerit vel }}h{\text{ formae }}8\mu +3{\text{, atque }}k{\text{ formae }}4\mu +1,\\{\phantom {l=+i}}\quad {\text{ vel }}h{\text{ formae }}8\mu +7,{\text{ atque }}k{\text{ formae }}4\mu +3,\\l=-i,{\text{ si fuerit vel }}h{\text{ formae }}8\mu +3,{\text{ atque }}k{\text{ formae }}4\mu +3,\\{\phantom {l=-i}}\quad {\text{ vel }}h{\text{ formae }}8\mu +7,{\text{ atque }}k{\text{ formae }}4\mu +1.\end{array}}$$

Per praecc. determinatio summae ${\textstyle W}$ pro iis casibus, ubi ${\textstyle n}$ est numerus primus vel numeri primi potestas, complete perfecta est: superest itaque, ut eos quoque casus absolvamus, ubi ${\textstyle n}$ e pluribus numeris primis compositus est, huc viam nobis sternet theorema sequens.

25.

Theorema. Sit n productum e duobus integris positivis inter se primis ${\textstyle a}$, ${\textstyle b}$, statuaturque

$${\begin{aligned}&P=1+r^{aa}+r^{4aa}+r^{9aa}+{\text{etc.}}+r^{(b-1)^{2}aa}\\&Q=1+r^{bb}+r^{4bb}+r^{9bb}+{\text{etc.}}+r^{(a-1)^{2}bb}\end{aligned}}$$

Tum dico fore ${\textstyle W=PQ}$.

Demonstr. Designet ${\textstyle \alpha }$ indefinite numeros ${\textstyle 0}$, ${\textstyle 1}$, ${\textstyle 2}$, ${\textstyle 3\ldots a-1}$, ${\textstyle \beta }$ indefinite numeros ${\textstyle 0}$, ${\textstyle 1}$, ${\textstyle 2}$, ${\textstyle 3\ldots b-1}$, ${\textstyle \nu }$ indefinite numeros ${\textstyle 0}$, ${\textstyle 1}$, ${\textstyle 2}$, ${\textstyle 3\ldots n-1}$. Tunc patet esse

$$P=\Sigma r^{aa\beta \beta },\quad Q=\Sigma r^{bb\alpha \alpha },\quad W=\Sigma r^{\nu \nu }$$

Hinc erit ${\textstyle PQ=\Sigma r^{aa\beta \beta +bb\alpha \alpha }}$, substituendo pro ${\textstyle \alpha }$ et ${\textstyle \beta }$ omnes valores, omnibus modis inter se combinatos; hinc porro propter ${\textstyle 2ab\alpha \beta =2\alpha \beta n}$, erit ${\textstyle PQ=\Sigma r^{(a\beta +b\alpha )^{2}}}$. Sed nullo negotio perspicitur, singulos valores ipsius ${\textstyle a\beta +b\alpha }$ inter se diversos esse, atque alicui valori ipsius ${\textstyle \nu }$ aequales. Hinc erit ${\textstyle PQ=\Sigma r^{\nu \nu }=W}$.
Ceterum notandum est, ${\textstyle r^{aa}}$ esse radicem propriam aequationis ${\textstyle x^{b}-1=0}$, atque ${\textstyle r^{bb}}$ radicem propriam aequationis ${\textstyle x^{a}-1=0}$.