Full text: [Höhere Arithmetik] Theorematis arithmetici (2. Band)

riften 
= —2 
-2 = 0 
folglich 
0 + 0' + 0"4-0'" = —¿r+G4“-|-(aö»M.B) + *(ö) — i(6)+i(JB) 
= -,?+<?+« 
ah + + 
K 
— — 
+ 
— 
A B a ö 
8 
a b 
8 
a ß 
8 
a b 
8 
4—h 4—h 
+1 
- + 
+1 
— 
+l 
+ - 
+1 
—l 
+ + 
0 
- + 
4-1 
— — 
0 
-+ +- 
0 
4- -}- 
h 
+ 2 
— 
+1 
— 
— 1 
+ - 
— 1 
4— 
+ 1 
- + 
+1 
1 
+ - 
— 1 
+ + 
h 
-j-l 
- + 
—1 
— 
— 2 
+ - 
— 1 
+ + 
0 
+ — + 
0 
— 
— 1 
4— 
0 
+ + 
4-i 
— 1 1 
- + 
— 1 
— 
— 1 
+ - 
—i 
Hiernach bekommt nun die erste Regel folgende Gestalt: 
4 Dec. = I. —42s von denen, wo y ganz, \ps\ gerade 
II. -1-42e von denen, wo [oc] gerade, [y] ungerade 
III. —2e von allen 
IV. —j— 4 2s von denen, wo nicht zugleich \oo] und [y] gerade 
-f- Q-f- 8 
In unserm Beispiel 
0 
— 4 
+ 1 
0 
— 5 
0 
I 
II 
III 
IV 
Q 
8 
Man denke sich nun in III diejenigen besonders bemerkt, wo y ganz, M gerade, 
so ist 
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