# What real number is exactly one less than its cube?

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

• You've basically solved the equation $x^3-x-1=0$ right? – Maazul May 17 '13 at 2:28
• Wolfram says the number is approximately $1.32472$ – Maazul May 17 '13 at 2:31
• Also, thank you for upvoting! (I assume it was you, @Maazul.) I just discovered that it was acceptable and even encouraged to share knowledge on Stack Exchange, so I shared my little discovery of today only to receive two downvotes. – Lee Sleek May 17 '13 at 2:34
• Let $x$ be a real number. Its cube is $x^3$. $x$ should be $1$ less than its cube $x^3$. The equation should be $x=x^3 -1$. – Maazul May 17 '13 at 2:47
• I have understood my mistake and removed a previous erroneous comment because of it. The equation I gave in that comment was the one I actually solved. – Lee Sleek May 17 '13 at 2:55

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.

• The golden ratio and silver ratio belong to the special case $J(n^2+2)=\dfrac{\sqrt{n^2+4}+n}2$, which is the set of real numbers $x \geq 1$ such that $x-\dfrac{1}x$ is an integer (one of the more well-known properties of Phi), as can be seen from the relation $\dfrac{\sqrt{n^2+4}+n}2=\dfrac{2}{\sqrt{n^2+4}-n}$. Specifically they satisfy $x^2=nx+1$. I'm sure you know this, but for OP and others... – Jaycob Coleman Oct 1 '13 at 4:52
• The particular cases like $J(n^2+2)$ are generally grouped under pseudosquares, solutions to Pell's equation that are comeasurable with their square. – wendy.krieger Oct 1 '13 at 11:46

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.

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.

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.

• In systems like the heptagonal numbers, or those that derive from any higher-class system (like the polygon of 13 sides), one can find PV numbers as small as one choose. The thing for example, is a lattice of units in 5D space, where the PV condition is that the point lies below five planes, and above the sixth. Because it can not lie above all planes, it is thus quite possible to find a lattice point that satisfies any particular condition. – wendy.krieger May 17 '13 at 9:33
• So @wendy.krieger can you tell me a PV number smaller than $1.3$ – P.. May 17 '13 at 13:03
• What i was actually thinking of was PVs that converge really slowly, not PVs that were incredibly small. My bad. – wendy.krieger May 18 '13 at 8:43

I would expand on the previous answers as follows. In addition to the fact that the golden ratio is one less than its square, it is one more than its inverse. The plastic number (or constant) has a similar relationship in that it is one more than its inverse to the $4^{th}$ power.

In fact, the golden ratio and the plastic number are the only two $morphic$ numbers. (Reference: J. Aarts, R. Fokkink, and G. Kruijtzer, “Morphic Numbers,” $Nieuw \ Arch. Wiskd.$, 5 (2) (2001) 56–58.)

By definition, a real number $p>1$ is a morphic number if there exists natural numbers $k$ and $l$ such that

$$p+1=p^k \ \ \text{and} \ \ p-1=p^{-l}$$

The values of $[k,l]$ for the golden ratio and plastic number are $[2,1]$ and $[3,4]$ , respectively.