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The Golden Ratio is calculated by assuming that a line segment is divided in to two subsegments so that the ratio between the entire segment and the larger subsegment is the same as the ratio between the larger subsegment and the smaller. So, if the entire line segment is defined to be unit length, with the larger subsegment length $a$, and the smaller length $b$ the Golden Ratio can be calculate by solving this system of equations:

$$ a^2 - b = 0 $$ $$ a + b = 1 $$

in which the $b$s cancel, and (assuming $a>0$) we are left with:

$$ a^2 + a - 1 = 0 \implies \boxed{a = \frac{\sqrt{5} - 1}{2}} \implies \boxed{b =\frac{3 - \sqrt{5}}{2}} $$

$$ \boxed{\phi = \frac{a}{b} = \frac{\sqrt{5}-1}{3-\sqrt{5}} \approx 1.618...} $$


I'm interested in what happens when the segment is divided in to larger number of subsegments. For instance, if the segment is divided in to three subsegments ($a$, the largest, $b$ the second largest, and $c$ the smallest), if we assume the ratio between the entire segment and $a$ is the same as the ratio between $a$ and $b$, which is the same as the ratio between $b$ and $c$, the problem comes down to solving a system of three equations:

$$ a^3 - c = 0 $$ $$ a^2 - b = 0 $$ $$ a + b + c = 1, $$

which reduces to:

$$ a^3 + a^2 + a - 1 = 0. $$

My questions are these:

  1. Is there a clever way to solve the above system of equations for a closed-form solution? Wolfram Alpha can give me the numerical approximation, but seems to struggle with finding the closed-form.

  2. One is tempted to think that this pattern will continue -- i.e. that for a line segment subdivided in to $n$ subsegments, the ratio will be determined by finding the roots of the polynomial:$$ a^n + a^{n-1} + \cdots + a - 1 = 0. $$ Is there any non-messy way to prove this (perhaps by induction)?

  3. If the answers are yes to (1) and (2), is there a closed form solution to the Golden Ratio for $n$ subsegments?

  • $\begingroup$ Well there is a formula for the roots of a cubic equation. Quartic too.None for general equations of higher degree but possibly for this one. $\endgroup$ Commented Nov 27, 2017 at 23:59
  • $\begingroup$ For an unexpected occurrence of the "tribonacci constant" see researchgate.net/publication/…. The authors describe the construction of a regular hendecagon with a marked straightedge and compasses; the reciprocal of their $a$ parameter is actually the tribonacci constant! $\endgroup$ Commented Nov 28, 2017 at 1:01

3 Answers 3


According to Wikipedia, the root between 0 and 1 of $x^3+x^2+x-1=0$ is $$\bigl(\root3\of{17+3\sqrt{33}}-\root3\of{-17+3\sqrt{33}}-1\bigr)/3=0.543689012$$ to nine decimals. The root between 0 and 1 of $x^4+x^3+x^2+x-1=0$ is the reciprocal of $$p_1+(1/4)+\sqrt{(p_1+(1/4))^2-(2\lambda_1/p_1)(p_1+(1/4))+(7/(24p_1))+(1/6)}$$ where $p_1=\sqrt{\lambda_1+(11/48)}$ and $$\lambda_1={\root3\of{3\sqrt{1689}-65}-\root3\of{3\sqrt{1689}+65}\over12\root3\of2}$$ According to D. A. Wolfram, "Solving Generalized Fibonacci Recurrences", Fib. Quart. May 1998 129-145, (see in particular page 136), for $5\le n\le11$ the Galois group is the symmetric group on $n$ letters, which is not a solvable group, from which it follows that for these values of $n$ there is no solution in radicals. Wolfram conjectures the Galois group is $S_n$ for all $n\ge5$, from which it would follow that there is no solution in radicals for any $n\ge5$.

I would not be surprised to learn that in the meantime someone has proved this conjecture. If I find it, I'll get back to you. Progress is made in the paper Paulo A. Martin, The Galois group of $x^n-x^{n-1}-\cdots-x-1$, J. Pure Appl. Algebr. 190 (2004) 213-223. Martin proves the conjecture if $n$ is even, and if $n$ is prime. The paper may be available at https://www.sciencedirect.com/science/article/pii/S0022404903002457 (or it may be behind a paywall).

  • $\begingroup$ Any chance I could get more detail on the way to arrive at the closed for solutions for the order 3 and 4 cases? You don't have to spell it all out for me, but any references that show the method for solving those would be greatly appreciate. Really happy to see this is an open question, $\endgroup$
    – D. W.
    Commented Dec 2, 2017 at 20:13
  • $\begingroup$ I expect that if you type "cubic formula" and "quartic formula" into your favorite search engine, any number of references will come up. Other useful search terms would be Cardano, Scipio del Ferro, Tartaglia. $\endgroup$ Commented Dec 2, 2017 at 21:12

2) The answer is yes. Indeed, if we assume that the initial segment has unit length and the first (the largest) subdivision segment has length $a$ then the second subdivision segment has length $a\cdot a=a^2$, the third subdivision segment has length $a^2\cdot a=a^3$, and so forth. Since the sum of lengths of the subdivision segments is $1$, we obtain the respective equality. It can be simplified to

$$\frac {1-a^{n+1}}{1-a}-2=0$$

$$a^{n+1}-2a+1=0, 0<a< 1.$$

I guess for (almost) all $n>4$ this equation has no closed form solution in radicals, but I don’t know Galois theory sufficiently well to show this. Another approach to look for a closed form is to put $a=\cos t$, but I don’t see how we can proceed farther from this.


Here are all 3 exact solutions in Wolfram Alpha. I don't think you will find it very helpful.


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