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Find $x\in \mathbb{Z}$ such that $54x^3+1$ is a cube.

I found $x=0$, any others ?

Álvaro Lozano-Robledo
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Xeing
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  • You found the easiest. –  Mar 08 '13 at 09:49
  • :P So that I need stackexchane; I tried $x = 0,1,2,3,...,10000; x=0$ is only the solution; I can prove $3 \mid x$ too – Xeing Mar 08 '13 at 10:01
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    3|x may be a good start.. cubic diophantine can be extremely difficult. –  Mar 08 '13 at 10:12
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    Since $54=2\cdot3^3$ this equation may be written $2x^3+1=y^3$ where $x$ stands for your $3x$. And this equation has been discussed here, with an answer giving a reference that shows in your case that $0$ is the only $x$. http://math.stackexchange.com/questions/133754/diophantine-equations-x3-2y3-1 – coffeemath Mar 08 '13 at 10:44
  • @D3r0X4: If you've tried till $1000$, then $0$ is the only solution. – Inceptio Mar 09 '13 at 05:40

2 Answers2

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I think the trivial solution you got is the only solution. We need $54x^3 + 1 = y^3$. Let $v =3x$ and $u = y$. Then the equation is: $$1 - u^3 = 2v^3$$

However, there is an early result by Euler (I think) which says that the sum or difference of two cubes cannot be double another cube, unless it is the trivial case of both the cubes being the same.

EDIT: I notice someone had already commented on very similar lines. The Euler reference can be obtained here GoogleBook Theorem 247.

Aang
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Macavity
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The curve $54x^3+1=y^3$ is the equation of an elliptic curve. Let $C: 54U^3+V^3=W^3$ be the projectivization of the original curve, and let us declare $[0,1,1]$ to be the origin for the addition law on the elliptic curve.

The change of variables $$\begin{cases} U = X/18,\\ V=-Y,\\ W=-Y+Z/324,\\ \end{cases}$$ gives a birational equivalence of $C$ and $E: X^3 - Y^2Z + 1/324YZ^2 - 1/314928 Z^3=0$, which we can de-homogenize to be $E: y^2 - 1/324 y = x^3 - 1/314928$ (which in turn can be further simplified to be $y^2=x^3-27$, but we will not use this last model here). Using the method of descent, one can show that the rank of $E$ is zero, and the torsion subgroup is of order $2$. Hence, there are only two rational points on $E$, namely $[0,1,0]$ and $[1/108,1/648,1]$. These correspond to the only two rational points on the curve $C$, namely $$(0,1) \quad \text{ and } \quad (-1/3,-1).$$ In particular, the only rational numbers $x\in\mathbb{Q}$ such that $54x^3+1$ is also a cube are $x=0$ and $x=-1/3$.

Álvaro Lozano-Robledo
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  • Excellent, an even stronger result! – Dylan Yott Mar 18 '13 at 03:12
  • Just as a general question, I'm used to seeing elliptic curves in the form $y^{2}=x^{3}+bx+c$. Given that initial equation, how would you know it was an elliptic curve? – Dylan Yott Mar 18 '13 at 18:06
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    A model of the form $y^2=x^3+bx+c$ is a so-called (short) Weierstrass model, and we tend to write elliptic curves this way for convenience, but elliptic curves come in many different flavors (models). Any curve given by a cubic, non-singular as a projective curve, and with a rational point, is an elliptic curve. – Álvaro Lozano-Robledo Mar 18 '13 at 18:16
  • Also, how did you know to make those particular substitutions? Is there a general method at work here? – Dylan Yott Mar 18 '13 at 19:55
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    Yes, there is a method to know what substitution to use. Look at Silverman and Tate's "Rational points on elliptic curves", Chapter I, section 3. Some software, such as Magma, can do this for you, and that's what I used. – Álvaro Lozano-Robledo Mar 18 '13 at 20:28