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Feb 5, 2011 at 12:10 comment added Gottfried Helms upps - the "restriction" in the previous comment was lost. Initially I meant to restrict the lhs to be a repunit/q-analogue of the base-number like in the 13^3 example. For the analogy to the 5^3 example another meaningful might then be to use $ [base]^k + 1 $ but this is only a shot in the dark...
Feb 5, 2011 at 11:21 comment added Gottfried Helms @Aaron: the fact that 13 in $13^3-3^7=10$ is a q-binomial with q=3 gave me the idea to write the both solutions in your "later" statement in the base 3 and base 2 numbersystem respective and also put the minus-term to the rhs. then we have $ [base 3]: 111^3=10000101$ and $[base 2]: 101^3 = 10000011 $ and the left number 111 and 101 are greater than the nonzero tail of the rhs-number. Maybe this view adds a useful restriction to your last idea and might make it solvable with this.(I remember vaguely there is something with max. consecutive zeros in binomial expansions)
Jan 29, 2011 at 9:19 vote accept Asterios Gkantzounis
Jan 29, 2011 at 9:19
Jan 22, 2011 at 23:21 history edited Aaron Meyerowitz CC BY-SA 2.5
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Jan 22, 2011 at 23:14 history edited Aaron Meyerowitz CC BY-SA 2.5
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Jan 22, 2011 at 22:56 history edited Aaron Meyerowitz CC BY-SA 2.5
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Jan 22, 2011 at 22:34 history edited Aaron Meyerowitz CC BY-SA 2.5
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Jan 22, 2011 at 22:22 history answered Aaron Meyerowitz CC BY-SA 2.5