# The Diophantine equations $X^3=DY^3+A^3$

Does anyone know for which values of $D$ the equation $X^3=DY^3+A^3$ has solutions? All numbers non-zero naturals.

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The answer is that nobody knows! An answer to your question is equivalent to asking what numbers $D$ can be expressed as the sum of two rational cubes $u^3+v^3=D$, where $u=X/Y$ and $v=-A/Y$, but the classification of such numbers $D$ is not known. (If one believes the Birch and Swinnerton-Dyer conjecture, then there is an analytic method to test that will check whether $D$ is a sum of rational cubes, though.)

This article by J. H. Silverman is a great introduction to the subject. I can't recommend it highly enough.

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I concur that Silverman's article is great. Slight caution: I don't think this is strictly equivalent to the question asked as the OP requires natural number solutions as opposed to integer solutions. If your $u$ and $v$ were both positive, it wouldn't correspond to a solution to the original problem. I'd be surprised if this changed the moral of your answer, though. – Cam McLeman Dec 16 '11 at 4:49
We can obtain infinite values of D by solving the equation 3m^2+3m+1-D=0. If this equation has a positive integer solution then this D gives a solution to the above equation but I do not know if that way we obtain all solutions. – Vassilis Parassidis Dec 16 '11 at 5:25
@CamMcLeman, actually, the two questions are strictly equivalent, thanks to the geometry of the problem (see Figure 1 in Silverman's article, p. 333). If $D=2E^3$ is twice a cube, then it is clear that $D=E^3+E^3$. If $D$ is not twice a cube, and $u,v$ are both positive (necessarily $u\neq v$), and $P=(u,v)$ is in the curve $X^3+Y^3=D$, then $2P=(s,t)$ and one of $s$ or $t$ is negative (but not both!), because $2P=(\frac{v(u^3+A)}{v^3-u^3},\frac{u(v^3+A)}{u^3-v^3})$, see p. 335 in Silverman's article. – Álvaro Lozano-Robledo Dec 16 '11 at 14:31
Aha! Very nice. – Cam McLeman Dec 16 '11 at 15:50

For the equation:

$X^3=DZ^3+Y^3$

If the number of $D$ can be represented in this form $D=\frac{3p^2+1}{4}$

Then primitive solution can be written:

$X=p^2+p$

$Y=p^2-p$

$Z=2p$

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Actually, since $p$ must be an odd integer, then the smallest integer solution is $$X = \frac{p+1}{2}$$ $$Y = \frac{p-1}{2}$$ $$Z=1$$ – Tito Piezas III Mar 17 '15 at 14:08