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# Title: Leibniz-Handschriften zur Mathematik LH 35, 15, 5
# Folio 8r, page id 00068207:0011
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Cs
Problema Leibnitianu.
Esto Aequatio: bx +1 = vy. quaeritur ejus solutio in integris,
sosito b esse numerum datum, X et y quaesitos, y autem mino
rem quam b. Letitur autem solutio ea lege, ut dissimulentur
factores numeri dati b.
Solutio.
Cum petatur y esse minorem quam b.
ponat b -2 = y
Ergo b2- 2bZ +3 = y 2 = bx + 1
Unde bx = b - 2bZ +27- 1
et X = b- 2b7 +E=1
Proinde si Z sit unitas, evanescent +2 -1
Atque ita habetur solutio in integrus, quantus cunque detur B. b. dummodo
major quam 2. Frit enim b - Z = b = 1 = y
= 2 = X
atque ita bx + 1 = b = 2b + 1 = y2
cujus S. b - 1 = y. O. F. F.
Si vero Z assumatu major unitate, tunc quoque b datus debet esse a
e major ternario, et (ut habeatu solutio in integrii) Z ea cautione su
mendum est, ut ejus quadratu unitate minutu (22- 1) dividi possit
sine fractione per b.
Sista b = t 3. 2- 2. erit b-2 - b - 2 = 32 = 1 = y
b= 2bZ 1 E 1 = qu 12  4 ut
Esto b = 8. 2= 3. erit b - Z = 6-3 = 5 = y
b= 262 +X2 = 25 = y
b= 262 +22-1 = 24 =3 =x
Hic assumpsimus Z = 3. quia Z =1 = 9 - 1 = 8 dividi pot est per b = 8
Esto rursus b = 8. 7 =5. erit b -Z = 6-5 = y = 3
b - 2b2 + 2-1 = b
5
Nic 2= 1 = 25- 1 = 24
= 1 = x. dividi potest pb = 8
Atque ita. si b sit = 8. habentu tres solutiones in integris
1. P Z = 1. Quadit enim y = 7. X = 6
da
e
etia
c
2 2 3.
2 = 5.
y = 5. X = 3
y = 3. X = 1
Lterud sit b = 12. 2 - 5. erit b - 2 = 12 -5 = 3 = y
b=2b7 +7-1 = 4b = 4 = X divisibile p b = 12
2
Iem b = 12. Z-7. erit b - 2 = 12 -7 = 5 = y. 22- 1 = 49 -1 = 48 dividit
b2- 262 + 221
 24 = 2 = X
Iterud tres solutioneinintegrii posito b = 12
ta p
e
2daF Z  5
3ti a p2 =7
12
= - evadit y = 11. X = 10
y = 3. X = 5
Y = 5. X=2
Fb = 12
22 1 = 25 -( = 24
Vertatur.