Pagina:Werkecarlf02gausrich.djvu/16

VI. Hinc statim sequitur ${\textstyle (k,p)=}$ $${\begin{aligned}&\left[{\frac {2k}{p}}\right]+\left[{\frac {4k}{p}}\right]+\left[{\frac {6k}{p}}\right]\ldots +\left[{\frac {(p-1)k}{p}}\right]\\-2&\left[{\frac {k}{p}}\right]-2\left[{\frac {2k}{p}}\right]-2\left[{\frac {3k}{p}}\right]\ldots -2\left[{\frac {{\frac {1}{2}}(p-1)k}{p}}\right].\end{aligned}}$$

VII. Ex VI. et I. nullo negotio derivatur $$(k,p)+(-k,p)={\frac {1}{2}}(p-1)$$ Unde sequitur, ${\textstyle -k}$ vel eandem vel oppositam relationem ad ${\textstyle p}$ habere (quatenus huius residuum aut non-residuum quadraticum est) ut ${\textstyle +k}$, prout ${\textstyle p}$ vel formae ${\textstyle 4n+1}$ fuerit, vel formae ${\textstyle 4n+3}$. In casu priori manifesto ${\textstyle -1}$ residuum, in posteriori non-residuum ipsius ${\textstyle p}$ erit.
VIII. Formulam in VI. traditam sequenti modo transformabimus. Per III. fit $$\left[{\frac {(p-1)k}{p}}\right]\equiv k-1-\left[{\frac {k}{p}}\right],\left[{\frac {(p-3)k}{p}}\right]=k-1-\left[{\frac {3k}{p}}\right],\left[{\frac {(p-5)k}{p}}\right]=k-1-\left[{\frac {5k}{p}}\right]\ldots$$ Applicando hasce substitutiones ad ${\textstyle {\frac {p\mp 1}{4}}}$ membra ultima seriei superioris in illa expressione, habebimus

primo, quoties ${\textstyle p}$ est formae ${\textstyle 4n+1}$ $${\begin{aligned}(k,p)=&{\frac {1}{4}}(k-1)(p-1)\\&-2\left\{\left[{\frac {k}{p}}\right]+\left[{\frac {3k}{p}}\right]+\left[{\frac {5k}{p}}\right]\ldots +\left[{\frac {{\frac {1}{2}}(p-3)k}{p}}\right]\right\}\\&-\left\{\left[{\frac {k}{p}}\right]+\left[{\frac {2k}{p}}\right]+\left[{\frac {3k}{p}}\right]\ldots +\left[{\frac {{\frac {1}{2}}(p-1)k}{p}}\right]\right\}\end{aligned}}$$

secundo, quoties ${\textstyle p}$ est formae ${\textstyle 4n+3}$ $${\begin{aligned}(k,p)=&{\frac {1}{4}}(k-1)(p+1)\\&-2\left\{\left[{\frac {k}{p}}\right]+\left[{\frac {3k}{p}}\right]+\left[{\frac {5k}{p}}\right]\ldots +\left[{\frac {{\frac {1}{2}}(p-1)k}{p}}\right]\right\}\\&-\left\{\left[{\frac {k}{p}}\right]+\left[{\frac {2k}{p}}\right]+\left[{\frac {3k}{p}}\right]\ldots +\left[{\frac {{\frac {1}{2}}(p-1)k}{p}}\right]\right\}\end{aligned}}$$

IX. Pro casu speciali ${\textstyle k=+2}$ e formulis modo traditis sequitur ${\textstyle (2,p)={\frac {1}{4}}(p\mp 1)}$, sumendo signum superius vel inferius, prout ${\textstyle p}$ est formae ${\textstyle 4n+1}$ vel ${\textstyle 4n+3}$. Erit itaque ${\textstyle (2,p)}$ par, adeoque ${\textstyle 2{R}p}$, quoties ${\textstyle p}$ est formae ${\textstyle 8n+1}$ vel ${\textstyle 8n+7}$; contra erit ${\textstyle (2,p)}$ impar atque ${\textstyle 2Np}$, quoties ${\textstyle p}$ est formae ${\textstyle 8n+3}$ vel ${\textstyle 8n+5}$.