What real number is exactly one less than its cube?

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And does it have any of the special properties that the golden ratio (one less than its square) has?

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The numbers that have the sorts of features that $\phi$ does, are to be found in the series of $J(n) = \frac{\sqrt{n+2}+\sqrt{n-2})}2$, where $J(3) = \phi$.

The exotic-looking formula reflects the use of isoseries, in the form that $t(a+1) = n t(a) - t(a-1)$. Applied to n=3, we get alternating lucas numbers, eg 2 (1) 3 (4) 7 (11) 18 (29) ...

The value for $J(4)$ gives the heron triangles, and is used to hunt down fermat-like primes. It turns up in the dodecagon, for example. The series 2, 4, 14, 52, 194, ..., are the middle side of a triangle, whose edges are n-1, n, n+1, and the area is integral.

The value for $J(6)$ is involved in the octagon, as well as the various approximations to the square-root of 2, Also, the silver ratio is $1:J(6)$.

On the other hand $J(0)$ and $J(1)$ are complex numbers, which produce ultimately the gaussian and eisenstein integers.

All of the $J(n)$ eventually turn up in polygons: $J(5)$ can be constructed from the edges of a polygon of 21 sides, and $J(11)$ turns up as in the polygon of 13 sides.

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If you multiply everything out and solve the resulting cubic function, you should get

$$\frac{\sqrt[3]{12\sqrt{69} + 108} - \sqrt[3]{12\sqrt{69} - 108}}{6}$$

But I don't know if this number has any special properties yet.

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Actually, this is fairly well studied, although not as extensively as the Golden Ratio. As such, it does indeed have a name, the Plastic number. It's also called the silver ratio but as the linked page says, this is somewhat improper.

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A Pisot–Vijayaraghavan number is an algebraic integer greater than 1 where all its conjucates have modulus strictly less than 1.
The number you are describing is the smallest Pisot–Vijayaraghavan number (proved by Carl Ludwig Siegel here) called plastic number.
The golden ratio $\phi$ is the smallest limit point of the set of Pisot–Vijayaraghavan numbers.

For more see this.