Pagina:Werkecarlf02gausrich.djvu/100

denotando per ${\textstyle P}$ coëfficientem medium

$${\frac {{\frac {1}{2}}(p-1)\cdot {\frac {1}{2}}(p-3)\cdot {\frac {1}{2}}(p-5)\ldots {\frac {1}{2}}(p+3)}{1\cdot 2\cdot 3\ldots {\frac {1}{4}}(p-1)}}$$

Substituendo itaque pro ${\textstyle x}$ deinceps numeros ${\textstyle 1}$, ${\textstyle 2}$, ${\textstyle 3\ldots p-1}$, obtinebimus per lemma art. 19

$$\Sigma (x^{4}+1)^{{\frac {1}{2}}(p-1)}\equiv -2-P$$

At perpendendo ea quae in art. 19 exposuimus, insuperque, quod numeri complexuum ${\textstyle A}$, ${\textstyle B}$, ${\textstyle C}$, ${\textstyle D}$, ad potestatem exponentis ${\textstyle {\frac {1}{2}}(p-1)}$ evecti congrui sunt, secundum modulum ${\textstyle p}$, numeris ${\textstyle +1}$, ${\textstyle -1}$, ${\textstyle +1}$, ${\textstyle -1}$ resp., facile intelligitur fieri

$$\Sigma (x^{4}+1)^{{\frac {1}{2}}(p-1)}\equiv 4(00)-4(01)+4(02)-4(03)$$

adeoque per schemata in fine artt. 18, 20 tradita

$$\Sigma (x^{4}+1)^{{\frac {1}{2}}(p-1)}\equiv -2a-2$$

Comparatio horum duorum valorum suppeditat elegantissimum theorema: scilicet habemus

$$P\equiv 2a{\pmod {p}}$$

Denotando quatuor producta

$${\begin{aligned}&1.2.3\ldots {\frac {1}{4}}(p-1)\\&{\frac {1}{4}}(p+3).{\frac {1}{4}}(p+7).{\frac {1}{4}}(p+11)\ldots {\frac {1}{2}}(p-1)\\&{\frac {1}{2}}(p+1).{\frac {1}{2}}(p+3).{\frac {1}{2}}(p+5)\ldots {\frac {3}{4}}(p-1)\\&{\frac {1}{4}}(3p+1).{\frac {1}{4}}(3p+5).{\frac {1}{4}}(3p+9)\ldots (p-1)\end{aligned}}$$

resp. per ${\textstyle q}$, ${\textstyle r}$, ${\textstyle s}$, ${\textstyle t}$, theorema praecedens ita exhibetur:

$$2a\equiv {\frac {r}{q}}{\pmod {p}}$$

Quum quilibet factorum ipsius ${\textstyle q}$ complementum suum ad ${\textstyle p}$ habeat in ${\textstyle t}$, erit ${\textstyle q\equiv t{\pmod {p}}}$, quoties multitudo factorum par est, i.e. quoties ${\textstyle p}$ est formae ${\textstyle 8n+1}$, contra ${\textstyle q\equiv -t}$, quoties multitudo factorum impar est, sive ${\textstyle p}$ formae ${\textstyle 8n+5}$. Perinde in casu priori erit ${\textstyle r\equiv s}$, in posteriori ${\textstyle r\equiv -s}$. In utroque casu erit ${\textstyle qr\equiv st}$, et quum constet, haberi ${\textstyle qrst\equiv -1}$, erit ${\textstyle qqrr\equiv -1}$,