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 2020-02-15, 09:26 #1 enzocreti   Mar 2018 17×31 Posts 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? Last fiddled with by enzocreti on 2020-02-15 at 12:06
2020-02-16, 03:22   #2
CRGreathouse

Aug 2006

22×1,483 Posts

Quote:
 Originally Posted by enzocreti 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?
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
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:
 Originally Posted by enzocreti 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?
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.

 2020-02-16, 03:24 #3 CRGreathouse     Aug 2006 22×1,483 Posts Broughan has an alternate approach if you don't like modern Thue methods: https://cs.uwaterloo.ca/journals/JIS...roughan25.html

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