Pagina:Werkecarlf02gausrich.djvu/31

cum termino ${\textstyle m+1^{\text{to}}}$, qui est ${\textstyle =x^{{\frac {1}{2}}m}(m,m)}$. Quum sit

$$(m,m)=1,\quad (m,m-1)=(m,1),\quad (m,m-2)=(m,2){\text{ etc.}}$$

progressio ita quoque exhiberi poterit:

$$F(x,m)=x^{{\frac {1}{2}}m}+x^{{\frac {1}{2}}(m-1)}(m,1)+x^{{\frac {1}{2}}(m-2)}(m,2)+x^{{\frac {1}{2}}(m-3)}(m,3)+{\text{etc.}}$$

Hinc fit

$${\begin{aligned}(1+x^{{\frac {1}{2}}m+{\frac {1}{2}}})F(x,m)=1&+x^{\frac {1}{2}}(m,1)+x(m,2)+x^{\frac {3}{2}}(m,3)+{\text{etc.}}\\&+x^{\frac {1}{2}}.x^{m}+x.x^{m-1}(m,1)+x^{\frac {3}{2}}.x^{m-2}(m,2)+{\text{etc.}}\end{aligned}}$$

Quare quum habeatur (art. 5. II)

$${\begin{aligned}(m,1)+x^{m}&=(m+1,1)\\(m,2)+x^{m-1}(m,1)&=(m+1,2)\\(m,3)+x^{m-2}(m,2)&=(m+1,3){\text{ etc.,}}\end{aligned}}$$

provenit

$$(1+x^{{\frac {1}{2}}m+{\frac {1}{2}}})F(x,m)=F(x,m+1)\dots \dots \dots [3]$$

Sed fit ${\textstyle F(x,0)=1}$: quamobrem erit

$${\begin{aligned}&F(x,1)=1+x^{\frac {1}{2}}\\&F(x,2)=(1+x^{\frac {1}{2}})(1+x)\\&F(x,3)=(1+x^{\frac {1}{2}})(1+x)(1+x^{3}){\text{ etc.,}}\end{aligned}}$$

sive generaliter

$$F(x,m)=(1+x^{\frac {1}{2}})(1+x)(1+x^{\frac {3}{2}})\ldots (1+x^{{\frac {1}{2}}m})\dots \dots \dots [4]$$

10.

Praemissis hisce disquisitionibus praeliminaribus iam propius ad propositum nostrum accedamus. Quum pro valore primo ipsius ${\textstyle n}$ quadrata ${\textstyle 1}$, ${\textstyle 4}$, ${\textstyle 9\ldots ({\frac {1}{2}}(n-1))^{2}}$ omnia inter se incongrua sint secundum modulum ${\textstyle n}$, patet, illorum residua minima secundum hunc modulum cum numeris ${\textstyle a}$ identica esse debere, adeoque

$${\begin{aligned}&\Sigma \cos ak\omega =\cos k\omega +\cos 4k\omega +\cos 9k\omega +{\text{etc.}}+\cos({\frac {1}{2}}(n-1))^{2}k\omega \\&\Sigma \sin ak\omega =\sin k\omega +\sin 4k\omega +\sin 9k\omega +{\text{etc.}}+\sin({\frac {1}{2}}(n-1))^{2}k\omega \end{aligned}}$$

Perinde quum eadem quadrata ${\textstyle 1,4,9\ldots ({\frac {1}{2}}(n-1))^{2}}$ ordine inverso congrua sint his ${\textstyle ({\frac {1}{2}}(n+1))^{2}}$,${\textstyle ({\frac {1}{2}}(n+3))^{2}}$,${\textstyle ({\frac {1}{2}}(n+5))^{2}\ldots (n-1)^{2}}$, etiam erit