Pagina:Werkecarlf02gausrich.djvu/15

xum ${\textstyle {A}}$ explere. Habemus itaque

$$1.2.3\ldots {\frac {1}{2}}(p-1)=aa^{\prime }a^{\prime \prime }\ldots (p-b)(p-b^{\prime })(p-b^{\prime \prime })\ldots$$

Productum posterius autem manifesto fit

$${\begin{aligned}&\equiv (-1)^{\mu }aa^{\prime }a^{\prime \prime }\ldots bb^{\prime }b^{\prime \prime }\ldots \equiv (-1)^{\mu }k.2k.3k\ldots {\frac {1}{2}}(p-1)k\\&\equiv (-1)^{\mu }k^{{\frac {1}{2}}(p-1)}1.2.3\ldots {\frac {1}{2}}(p-1)\quad {\pmod {p}}\end{aligned}}$$

Hinc erit

$$1\equiv (-1)^{\mu }k^{{\frac {1}{2}}(p-1)}$$

sive ${\textstyle k^{{\frac {1}{2}}(p-1)}\equiv \pm 1}$, prout ${\textstyle \mu }$ par est vel impar, unde theorema nostrum protinus demanat.

4.

Ratiocinia sequentia magnopere abbreviare licebit per introductionem quarundam designationum idonearum. Exprimet igitur nobis character ${\textstyle (k,p)}$ multitudinem productorum ex his

$$k,2k,3k\ldots {\frac {1}{2}}(p-1)k,$$

quorum residua minima positiva secundum modulum ${\textstyle p}$ huius semissem superant. Porro existente ${\textstyle x}$ quantitate quacunque non integra, per signum ${\textstyle [x]}$ exprimemus integrum ipsa ${\textstyle x}$ proxime minorem, ita ut ${\textstyle x-[x]}$ semper fiat quantitas positiva intra limites ${\textstyle 0}$ et 1 sita. Levi iam negotio relationes sequentes evolventur:

I. ${\textstyle [x]+[-x]=-1}$.

II. ${\textstyle [x]+h=[x+h]}$, quoties ${\textstyle h}$ est integer.

III. ${\textstyle [x]+[h-x]=h-1}$

IV. Si ${\textstyle x-[x]}$ est fractio minor quam ${\textstyle {\frac {1}{2}}}$, erit ${\textstyle [2x]-2[x]=0}$; si vero ${\textstyle x-[x]}$ est maior quam ${\textstyle {\frac {1}{2}}}$, erit ${\textstyle [2x]-2[x]=1}$.

V. Iacente itaque residuo minimo positivo integri ${\textstyle h}$ secundum modulum ${\textstyle p}$ infra ${\textstyle {\frac {1}{2}}p}$, erit ${\textstyle \left[{\frac {2h}{p}}\right]-2\left[{\frac {h}{p}}\right]=0}$; iacente autem residuo illo ultra ${\textstyle {\frac {1}{2}}p}$, erit ${\textstyle \left[{\frac {2h}{p}}\right]-2\left[{\frac {h}{p}}\right]=1}$.