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$a, b$ are two naturals such that,

$$a^{2}-b^{2}=k^{3}$$ and $$a^{3}-b^{3}=c^{2}$$

where $k^{3}$ and $c^{2}$ are perfect cube and square respectively.

What can be the least possible pair of naturals $(a, b)$ for the above to hold true?

This link suggests $(10, 6)$ but I'm not satisfied with the answer. Please help.

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Not clear what makes one pair of reals smaller than another, but $a=b$ can be made arbitrarily small, with $c=k=0$. –  Gerry Myerson Aug 9 '12 at 5:53
@TheApe, unless a≠b, a=b=0 or 1 should be the least possible solution according as you consider 0 as natural number or not. –  lab bhattacharjee Aug 9 '12 at 6:00
Well, $(10,6)$ works, so to prove it's the least possible pair, all you have to do is check all the smaller pairs - assuming you have decided what makes one pair of naturals smaller than another. –  Gerry Myerson Aug 9 '12 at 6:05
But I got $(10, 6)$'s idea through a site, not by myself. How to prove it'll be the smallest one analytically? –  TheApe Aug 9 '12 at 6:14
Not sure what "analytically" means. You could have tried $(1,1)$, then $(1,2)$, then $(2,1)$, etc., etc., until you found one that works - the first one you found would be the answer. –  Gerry Myerson Aug 9 '12 at 7:10
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