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# Title: Leibniz-Handschriften zur Mathematik LH 35, 13, 3 (shelfmark LH 13, 3)
# Folio 57r, page id 00068200:0115
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quaeritur an numemi 7 possit divedi in duos quadratos Tentemus
in quisitionem dy per dyadicos
= 11I = 1X + 1X + 1 posito X = 2L
siat quadrati B  29 2Z, yy et vv: gy, fiet 7 = vr + 2Z,: yy
seu  ex 7yy = vN + ZZ. sit v = 10 +UX + M2 XX, 3  et Z = 20 + 21X + 22 AX + 23 X 1 2+X1
et 7 = 30 + 31X + 32 XX fiet
23X + 3X+ 31+ 30
3X +32X + 31X +30
 +1 30. 5 X f 303+ b30. 3 X 30
+243 Z. F  +31X ea
3Ss 32X
142715X
+2. 10BX + Z.BIX F AIX EB 2BIXF1 - 00
EHI. 15X ,2. 1I14 2. 11. B 21. 12 lH. X
2.12 9L 2. 12. 14 2. 12. 13 12
21. 5X 21314 13
13X
24
idem repesatur, secundum 20, 2, 2e
CE v t2UB +2MX + 31 X
32
+29033 +250 9rX + 2.0g X OX
2.313) 2. 31. 31. 31
229.3 32
2.30. B 2. 30. 32 230 H + 30X
2.31.33 2.31.32 31
 23033, 230. 3t 23031 130
(ag 147F +abX +4SF +44R +43Y +tia uN 14cl 2o
2 - 248 - 247 -246 -45 24 , 13 42 A1L
40 = - 10820 +30, :2 41 = 8. 30( 4+ 2. 30. 31  -1 - 20l - 2. 10.11 = - 4. 10. 11 + ∞3. 30l
+30l
=4. 10. 11 = 2 10 l
2t
Meterodum hic quae sivi utum possit pervenm ad aequandam sorumtam du adicam quadrato, vel quadurti
multiplo ut hic per 7, vut alium nundrm, essi non fiant additiones seu res non peduaetur ad
membra haud excioentia 0 vel 1. Viditur tamam additiae illae nolae 40,41 et. esse quidem integri
sed non ost necesse ut sint ovel 1, etsi 10, 11, 12, etc. item 20, 21, 22, etc. item 3031 3r ea
huit c vel 1 sed ii fallor hinc con datur Tota ergas res hac radit, ut his ipsis sumtis utcaque
0, _ 42, et c. fiant jage integri
incipiendo a summo levitaturdvis 1o et prodeunt integri per se
Ovel 1, ipsi
49215 +25 48 = Z2154325 33 A9 47= 14 +2. 1451
2. le. 1
24 +223. 25
49 = 15+25 48 = 2. 14. 15 + 2. 15 -35722713= 14 2. 23. 25 4. 445 4. 13
Be(svilalis c cusu tentemus ponendo 32, 33; 22, 23, 2e, _ 12, 13, 14, sc = 0
2. 24. 25 42. 25
fint 44 = - 31 43 = -2.31 - 31 42 = - 6. 31 - 1. 31 - 2. 30. 31 = -7. 31 = 2. 30. 31
41 = = 14. 31 - 4. 30.31 - 30-2. 30.3] - 2.30. 32 4 2. 10. 12. + 11
49 = - 14.. 31 - 2. 30 = 12. 30. 31 - 4. 30.32 + 4. 10. 12 +2. 11 -30 -2. 30. 31 + 2.10. 11
2. 29, 22 + 21
4. 2022 +2. 21
2. 20. 21
= 28. 31 - 6. 30 = 28. 30. 31 -8.30.32 + 8. 10. 12 + 4. 11 + 4. 10. 11 = 10
8. 20. 22 +4. 21 4. 20. 21 20
quae aequatio est solvenda non solum in dyadicis, sed etiam in 0 et 1. seu ∞,t,
sed 10 et I1 non possunt simul esse O
 10. 1(1
ut
sunt 1 vet L
seu 10, U1; 20, 21, 30, 3) debent esse 0 vel 1
idem est de 20 et 21, 30+ 31 Ergo seu
X = 10r1 10 + H1 X ~Y X - 201X - 1 = XX - 3X, f 2 ∞10+ -21041
+1 f. (19 (+ 11) + 2. 10. 11 - β3)10-(3, 14 12 =0. ita 10 et U possunt esse ambo, a ,
non amba vel unum 1, alterum 0, sod non possunt esse ambo D
Habemus ergu aeqq. 1,2, 3, 4, et ope agy. 23A, tollamus  10, 2D, 3D ex aeg. 1.
semulis asquatio inveniatur pro 4,21, et pro 3031.
superetunt 1424,31 in aequatione ubi omnes literaee possunt esse 1 vel 0.
nulla ab aliis dependentia; cum ipsae 10, 20, 30, non possint essie 0, nisi cn 1, vel 21,
porro haeaquationes si haec aequatio si quadretur, et cubetur
manent eaedem literae nec altius assumnat, Ergo s habemus tres aeqgu. securdum
quas  haben potemul tres valores; sed  quomodo quaeso efficietur, ut quivis valor sit 1velo
vel 31 sunt 1.
Resus invenietur ope fractinis, cujus si numemior sit denominatur sit, o vel nomi
denoninatur et numerator simul sit 0, ipsa ert 1 vel 0. si e numerator sit
nominatur quicunque erit 0, si. Numator et nominator sint aliquid id aequales fiet 1.
sed hoa. hoc inquovis gradu tentare minis prolixem foret
ut habeatur
d+
n 10.1 - 10-11 + 1 =0
s
33 X +2. 5134
+33
2. 3237
ex