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My question concerns the diophantine equation $a^3 + b^3 = c^2$. I know one solution: $1^3 + 2^3 = 3^2$. But this is special in (at least) two ways: the $a$ and $b$ are not coprime; the solution is a special case of the identity: sum of the first $n$ cubes $=$ the square of the $n$th triangular number.

Are there other solutions?

If not, is there an elementary proof of the fact?


marked as duplicate by Namaste, Adrian Keister, Cesareo, Nosrati, Brahadeesh Aug 8 '18 at 13:41

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There are, in fact, a lot!

For example, $(2, 2, 4), (7, 21, 98), (65, 56, 671),$ etc.

  • $\begingroup$ Do you know a parametrization for the solutions? $\endgroup$ – lhf Aug 8 '18 at 10:30
  • $\begingroup$ see the discussion here math.stackexchange.com/questions/369846/…. $\endgroup$ – pointguard0 Aug 8 '18 at 10:32
  • $\begingroup$ Thank you pointguardo: a bit dozy of me! The factorisation of your second example suggests a way to generate a particular class of solutions. $\endgroup$ – Paul Stephenson Aug 9 '18 at 12:22

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