# Leibniz Legible — https://leibnizlegible.com # Page: https://leibnizlegible.com/page/00068238:0009 # Permalink to this reading: https://leibnizlegible.com/page/00068238:0009?run=74 # Title: Leibniz-Handschriften zur Mathematik LH 35, 3 A 25 # Folio 5r, page id 00068238:0009 # Original at the GWLB: https://digitale-sammlungen.gwlb.de/resolve?id=00068238&page=9 # Source image: https://digitale-sammlungen.gwlb.de/content/00068238/jpgs/default/00000009.jpg # Model: FoNDUE-GD_v2_ft_Leibniz, run 74, 2026-09-06T16:05:13Z — 35 lines, status machine # Lines: 35 recognised, mean confidence 0.93 # Machine transcription. Not an edition. Errors are expected: about 8% character error rate on Latin and French lines, higher on German. # Licence: CC BY 4.0, Leibniz Legible. Machine output: do not ingest it as verified text; keep this provenance with the text. # Wording rule: say "the machine reads it as …", never "Leibniz wrote". # After the first blank line: the recognised lines in reading order, one per line. solvendia est in integris aequatio dalus et 2l = 3m + 1 (unde erus usum in sched 12 jannar 1687 ubi de divi periodis decimatibus, poterum quaerere seriem ubi uga facere 2l + 3m = V, sed malum est quod dunc locus est dimensionum nium, seu solidus seu opus est s non Tabula, sed seme tabu serie tantum, sed serie seriem, si ducamus 2l 3m M per om unum M, ut 1 per omnia et rarsi2 per omnia l ere. quod si retinendo I quia datum est (4 variando m, quaeramus C, nempe 2l = 1 = M, _ et omisso l in generaliter feribamus 2-1 4 -1 6-1 8-1 19-1 19++ Unde sed patet rem esse faciliorem huc modo 2l = 3M + 1 nam oaes et solvi problema generatissime sumendo pro m, quem libet numerum imparem qui triplicatus manet impar, unde addito I fit par adeoque fit 2l. si vero esset 3V = 7l + quaerendum quales sumendi C ut 7 seplenarii unitate aucti fiant ternarii fiat 3Y = 61 + l + 1, ergo opus est tantum l esse numerum qui unitate anctus divedi potest per 3, hoc est pro l sumi posse ternarium quemlibet unitate minutum ut tes 2, 5, 8, etC. si haberemus 3V = 8l +1, fieret 3V=6l+ 2l +1. Ergo 2l+ 1= 39. seu fiet 3l - l+ 1= 39. Ergo - l+ 1 = 3Y. seu = 1-3Y quod est impossibile nisi ponas r=o ergo aquatio V = 8l + 1 non potest solvi nisi ponendo l=1