Pagina:Werkecarlf02gausrich.djvu/97

Quibus valoribus in aequatione praecedente substitutis, prodit:

$$0=4mm-4m-1-8hm+4hh+(i-l)^{2}$$

Quodsi tandem pro ${\textstyle 4m}$ hic substituimus ${\textstyle 2(h+m)-2(h-m)}$ sive, propter aequationem ultimam in I, ${\textstyle 2n-2(h-m)}$, obtinemus:

$$0=4(h-m)^{2}-2n+2(h-m)-1+(i-l)^{2}$$

adeoque

$$8n+5=(4(h-m)+1)^{2}+4(i-l)^{2}$$

Statuendo itaque

$$4(h-m)+1=a,\quad 2i-2l=b$$

fiet

$$p=aa+bb$$

Iam quum in hoc quoque casu ${\textstyle p}$ unico tantum modo in duo quadrata, par alterum, alterum impar, discerpi possit, ${\textstyle aa}$ et ${\textstyle bb}$ erunt numeri prorsus determinati; manifesto enim ${\textstyle aa}$ quadrato impari, ${\textstyle bb}$ pari aequalis statui debet. Praeterea signum ipsius ${\textstyle a}$ ita erit stabiliendum, ut fiat ${\textstyle a\equiv 1{\pmod {4}}}$, signumque ipsius ${\textstyle b}$ ita, ut habeatur ${\textstyle b\equiv af{\pmod {p}}}$, uti per ratiocinia iis, quibus in art. praec. usi sumus, prorsus similia facile demonstratur.

His praemissis quinque numeri ${\textstyle h}$, ${\textstyle i}$, ${\textstyle k}$, ${\textstyle l}$, ${\textstyle m}$ per ${\textstyle a}$, ${\textstyle b}$ et ${\textstyle n}$ ita determinantur:

$${\begin{aligned}&8h=4n+a-1\\&8i=4n+a+2b+3\\&8k=4n-3a+3\\&8l=4n+a-2b+3\\&8m=4n-a+1\end{aligned}}$$

aut si expressiones per ${\textstyle p}$ praeferimus, termini schematis ${\textstyle S}$ per 16 multiplicati ita se habebunt:

$${\begin{array}{l|l|l|l}p+2a-7&p+2a+4b+1&p-6a+1&p+2a-4b+1\\p-2a-3&p-2a-3&p+2a-4b+1&p+2a+4b+1\\p+2a-7&p-2a-3&p+2a-7&p-2a-3\\p-2a-3&p+2a-4b+1&p+2a+4b+1&p-2a-3\end{array}}$$