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 enzocreti 2020-02-15 09:26

Numbers sum of two cubes and product of two numbers of the form 6^j+7^k

344 and 559 are numbers that are sum of two positive cubes and product of two numbers of the form 6^j+7^k with j, k >=0. FOR EXAMPLE 344=43*8=7^3+1
Are there infinitely many such numbers?

IS 16 THE ONLY perfect POWER SUM OF TWO CUBES AND PRODUCT OF TWO NUMERS OF THE FORM 6^J+7^K J, K NONNEGATIVE?

 CRGreathouse 2020-02-16 03:22

[QUOTE=enzocreti;537634]344 and 559 are numbers that are sum of two positive cubes and product of two numbers of the form 6^j+7^k with j, k >=0. FOR EXAMPLE 344=43*8=7^3+1
Are there infinitely many such numbers?[/QUOTE]

Probably. Here's a bunch:
[code]16, 91, 344, 559, 1736, 2752, 4472, 8029, 9331, 12913, 14023, 20683, 71665, 74648, 207145, 326599, 373256, 375992, 941200, 942920, 1314440, 1688911, 4797295, 8456552, 12365695, 16283293, 23588209, 66926791, 80621576, 80624312, 81562760, 322828864, 322830584, 323202104, 362851489, 403450424, 17414258696, 17414261432, 17415199880, 17737087544, 110730297616, 110730299336, 110730670856, 110810919176, 128144556296, 3761479876616, 3761479879352, 3761480817800, 3761802705464, 3872210174216, 37980492079552, 37980492081272, 37980492452792, 37980572701112, 37997906338232, 41741971956152, 812479653347336, 812479653350072, 812479654288520, 812479976176184, 812590383644936, 850460145426872, 13027308783283600, 13027308783285320, 13027308783656840, 13027308863905160, 13027326197542280, 13031070263160200, 13839788436630920, 175495605123022856, 175495605123025592, 175495605123964040, 175495605445851704, 175495715853320456, 175533585615102392, 188522913906306440, 789831783010279009, 4468366912666272064, 4468366912666273784, 4468366912666645304, 4468366912746893624, 4468366930080530744, 4468370674146148664, 4469179392319619384, 4643862517789294904, 37907050706572935176, 37907050706572937912, 37907050706573876360, 37907050706895764024, 37907050817303232776, 37907088687065014712, 37920078015356218760, 42375417619239207224, 1532649851044531315216, 1532649851044531316936, 1532649851044531688456, 1532649851044611936776, 1532649851061945573896, 1532649854806011191816, 1532650663524184662536, 1532825346649654338056, 1570556901751104250376, 8187922952619753996296, 8187922952619753999032, 8187922952619754937480, 8187922952620076825144, 8187922952730484293896, 8187922990600246075832, 8187935979928537279880, 8192391319532420268344, 9720572803664285311496, 525698898908274241116352, 525698898908274241118072, 525698898908274241489592, 525698898908274321737912, 525698898908291655375032, 525698898912035720992952, 525698899720753894463672, 525699074403879364139192, 525736805958980814051512, 533886821860893995112632, 1768591357765866863198216, 1768591357765866863200952, 1768591357765866864139400, 1768591357765867186027064, 1768591357765977593495816, 1768591357803847355277752, 1768591370793175646481800, 1768595826132779529470264, 1770124007616911394513416, 2294290256674141104314552[/code]

If you want more, you should probably look into fast systems for solving cubic Thue equations. I could share my PARI/GP code but it's not particularly performant.

[QUOTE=enzocreti;537634]IS 16 THE ONLY perfect POWER SUM OF TWO CUBES AND PRODUCT OF TWO NUMERS OF THE FORM 6^J+7^K J, K NONNEGATIVE?[/QUOTE]

If this is just the above problem, but asking for the numbers to be powers as well, there should be only finitely many, with 16 being presumably the only one.

 CRGreathouse 2020-02-16 03:24

Broughan has an alternate approach if you don't like modern Thue methods:
[url]https://cs.uwaterloo.ca/journals/JIS/VOL6/Broughan/broughan25.html[/url]

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