Surprising Surds

Yesterday, I stumbled upon this very surprising identity while reading on nested radicals: $$ \sqrt[3]{2\pm\sqrt{5}} = \frac{1\pm\sqrt{5}}{2} $$ While it is easy to prove the fact after seeing it, I will try to prove this from the perspective of someone who is not sure if \(\sqrt[3]{2\pm\sqrt{5}}\) is the simplest form of the pair.


Let \(x=\sqrt[3]{2\pm\sqrt{5}}\) then \((x^3-2)^2=5\). Therefore, \(x\) is a root of the equation: $$ x^6 -4x^3-1=0 $$
To find the degree of \(\sqrt[3]{2\pm\sqrt{5}}\), we need to know the degree of its minimal irreducible polynomial \(p(x)\) which will divide \(x^6 -4x^3-1\). If the degree is less than \(6\) then maybe we could rewrite the number in simpler forms.

Clearly, if \(x^6 -4x^3-1\) factors then it factors into either two cubics or a quadratic and a quartic. (Note: It has no linear factor since the equation has no rational solutions). Furthermore, since it is a polynomial in integer coefficients, it will factor into polynomials of integer coefficients. This implies that the leading coefficients are \(1\) and the constant terms are \(\pm 1\) for the two factors respectively.1


Case 1: \( x^6 -4x^3-1 = (x^3+a_2 x^2+a_1x+1) (x^3+b_2 x^2+b_1x-1) \)


By multiplying out the factors and comparing coefficients, the following contradictory equations are observed:

$$ \begin{aligned} b_2 &= - a_2 & \qquad (x^5) \\ b_1&=a_1 &(x)\\ 0&=-4 &(x^3) \end{aligned} $$


Case 2: \( x^6 -4x^3-1 = (x^2+a x+1) (x^4+b_3 x^3+b_2 x^2+b_1x-1) \)


This implies another set of contradictory equations: $$ \begin{aligned} b_3 &= - a & \qquad (x^5) \\ b_2 &= a^2-1 & \qquad (x^4) \\ b_1 &= a & \qquad (x)\\ a^3 - a &= -4 & \qquad (x^3)\\ 2a^2-2&=0 &\qquad(x^2) \end{aligned} $$
Remember that \(a\) must be an integer and there are no integer solutions to the last two equations.


Case 3: \( x^6 -4x^3-1 = (x^2+a x-1) (x^4+b_3 x^3+b_2 x^2+b_1x+1) \)


This factorization is indeed possible: $$ \begin{aligned} b_3 &= - a & \qquad (x^5) \\ b_2 &= a^2+1 & \qquad (x^4) \\ b_1 &= a & \qquad (x)\\ a^3+3a &= -4 & \qquad (x^3)\\ 0&=0 &\qquad(x^2) \end{aligned} $$
\(a=-1\) is the only integer solution and it leads to this factorization, $$ x^6 -4x^3-1 = (x^2- x-1) (x^4+ x^3+2 x^2-x+1) $$ It is also true that $$ x^4+ x^3+2 x^2-x+1> 0 \text{ for all real }x\text{ since } \begin{cases} x^2-x \ge 0, & x\ge1\\ 1-x > 0, & 0 \le x < 1 \\ x^3-x>0, & -1 < x < 0 \\ x^4+x^3\ge 0, & x\le-1 \end{cases} $$ So when trying to solve \( x^6 -4x^3-1 =0\) in the reals, we can divide through by the positive factor.


Therefore, \(\sqrt[3]{2\pm\sqrt{5}}\) are roots of the equation $$ x^2-x-1=0 $$
By the quadratic formula \(x=\dfrac{1\pm \sqrt{5}}{2}\).


For this particular example, it would have sufficed to plug one of the values into a calculator to get \(x=1.618034 \text{ or } -0.618034\). These are the golden ratios and I know their quadratic form. But that feels like cheating since it relies on recognizing the numerical value. At least now I know how I would attempt to simplify an arbitrary surd of the form: \(\sqrt[3]{m \pm\sqrt{n}}\)


To generalize my steps: simplifying \(\sqrt[3]{m\pm\sqrt{m^2+1}}\) will be equivalent to solving this diophantine equation: $$ a(a^2+3)=2m $$
An integer solution \((a,m)\) implies $$ \sqrt[3]{m\pm\sqrt{m^2+1}} = \frac{a\pm\sqrt{a^2+4}}{2} \ \ \ \ \ (1) $$ Also since these are the roots of \(x^2-ax-1=0\), then this is their continued fraction $$ a + \cfrac{1}{a+\cfrac{1}{a+\cfrac{1}{a+\cfrac{1}{a+\cfrac{1}{\ddots}}}}} $$
Finally, due to this continued fraction, $$ a= \sqrt[3]{m\pm\sqrt{m^2+1}}-\dfrac{1}{\sqrt[3]{m\pm\sqrt{m^2+1}}} \text{ is an integer.} \iff (1) $$ In the same fashion, I proved for \(|m|\ge 1\) or \(|a|\ge 2\), $$ x = \sqrt[3]{m\pm\sqrt{m^2-1}} = \dfrac{a\pm\sqrt{a^2-4}}{2} \iff a = x+\frac{1}{x} \text{ is an integer} $$

  1. Update (19 Jan 2017) - This is definitely false: \( x^4 + 1 = (x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1) \). There is a polynomial with integer coefficients whose factors have some irrational coefficients. I must have been thinking of Gauss' Lemma when I typed it. I should have said something like: "..., it will factor into polynomials with integer coefficients or with some irrational coefficients." In fact even the note right before (about not having linear factors) is based on false logic. I'm using the rational root theorem but it only allows us to conclude it has no linear factor with only integer coefficients, but they could be irrational... Oh my, what a mess! ↩