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# Title: Leibniz-Handschriften zur Mathematik LH 35, 12, 2
# Folio 80r, page id 00068195:0159
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invenire Triangulum Rectangulum in Numeris
cujus cuborum a lateribus summa sit cubus.
Ex natura Friangali rectanguli latera semper possunt intelligi
X- y
2Xy X+ y
quorum cubi X - 3XXy Y, +9X + X +3Xy1 3y Ly.
sit s. XX 154M X
rexy = 4X
3γ+ 4 =5R st
Et cuborum summa 2X + 6X7 + 8Xy, aequanda cubo, is sit sex
(2
et divisca aequatione per 2X, fiet X + 3y + 4Xy = 4 ex.
Reducamus omnia ad y, et sit x = 10 +uy et e= 20+ 21y fiet
X = 10+ 4.13y + 610y+ 4 10y - 9 Et fiet
+ X = 10 +4. 10. 11y+6. 16. 1y + 9. 101y+ u y
+31 - . . . . ....
+D g. +3
+qX= . . . . . . . . . . .. + 4.10 V + 4. 11 y
4Cx = o28 - 11. 204 - 3. 126. 27 -3 11. 2 51 ) - 4 1 2y
-3,10. 20. 97 -340. 20. 25y 8, 0. 2(3 y
si e= m. X piet 4X M: X
X 13y V- xh +xh + x  4 ex
L-
20+3. 28. 119 + 3 20 2191 
20 411y
+1. 28y +3112027 13. 112 219 11 215
+ 10.28 +3 Wtry 4 3. 10201 1 102
Si lit 10 220, vanesit terninus ultimus ubi y et  terminoys destmendo evanescit 10 vel 20, adeoque fit
(3. 112= 11. 21 + 21, ubi pro 21: 11 ponendo 23 fit 25 + 23 4+ Z = + +2 fit 23+ 1 = + 3
seu 23 = +3 -1 (2) 21. ubi debet + signisicore  - nam si esset + suret A - T1
unde cum o jam fecerimus 10 = 2o fieret X ubi si faciamus + = + fit C = X
quia jam feciuus 10L 20 per 6, et nunc per 12 fieret 21 = 11. Nil tamen video quid prohibeat
sed retineamus generalitatem.
deuique pro termino y destruendo, fiet
102 11 : 4. 1 - 3. 21. Ergo tandem ob y, y fiet y=
si sit + = +, fiet 21 = 11 et 10 = 1, et 20 =1 et X = e, et y- 
_ 3+ 4. 11 +11 - 11. 213
== 3+ 4.1 - 4 74. 1 - 1
3 + 4. 11 3+ 4. 11
3. 1134 1 - 4. 1 - 2 = 5+ 4 1 et X = (i+ ny - 1- 3, 4
ubi 1 sic interpretanda est, ut vel  tam x quam y fiat tam el ffirmativa, qu
et y=
3+ 4. 11
6724 11 sit affirmativa fiet etam y negativa, ergo debet esse 4. 1 - 1 negativa, seu debet
begu esse 11 miuor quam . sit ergo 11 = h fiet 3= 34; seu =  et X = - 1. 2 = 
68
4C ergo X sed Ita X non fit major quam y, contra Hypothesin. Huic malo videamus an
1b9 alia expelicatione ipsius 11 liceat mederi debet autem 1 esse minor quam , in derctis
suppositombus. Ergo videamus an posito H esse minus quam L possit esse 4. H1 -1 majus quam 4
seu 4. 11 majus quam 5, s seu 11 mous quam E sed ita id quod est minus quam seret majus quas
1581
E quod absurdum. Ergo necesse est in Hypothesi praesenti cum + = +, fieri I quantitatem
negativam, seu 1 = -13. et fiet X =  E . -4. 13 et y =  41. ubi si 13 sit
quantitas major unitate, vel unitas fient ambae quantitates affirmativa, et X major quam yt.
+. 3 - 4. 13
sit ergo 13 =1 fiet X = 5 et y = 4 et 2Xy = 40. cujus cubus 640 et X  = 9 cujus ibus 729
8 et X + y = 41 cujus cubus 68921
a 4
g2 1L 168110
A
a
468921
9
640
20
40 45
erit enor
in calculo
70290
70290