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I am not able to make headway on solving the diophantine equation $x^m - y^n = 6.$ Are there any solutions to this? What about $x^m - y^n = 14,$ and $= 30$ (both $m$ and $n$ are at least $2$).

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$83^2-19^3 = 30$ – joro Dec 17 '12 at 8:32
What about others? – Salahuddin559 Dec 17 '12 at 8:37
See and references given there. – Robert Israel Dec 17 '12 at 9:08
Heck, the second sequence I originally came up with, that is in the OEIS again. I think that happens a lot to good people, like here. – Salahuddin559 Dec 17 '12 at 9:15
One more thing. What if we use 3 terms, like numbers not expressed like x^m +/- y^n +/- z^p. Looks like 70 does not fit the bill here also. – Salahuddin559 Dec 18 '12 at 7:52

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