# Writing $\sqrt[\large3]2+\sqrt[\large3]4$ with nested roots

Let $C\subset\Bbb R$ be the smallest set containing $0$ and closed under whole number addition/subtraction, whole number exponents, and whole number roots. That is, for all $c\in C$ and $n\in\Bbb N$, we have $c\pm n\in C$, $c^n\in C$, and $c^{1/n}\in C$.

We know that $\sqrt2+\sqrt3\in C$, since it's equal to $\sqrt{\sqrt{24}+5}$. Similarly, Ramanujan proved that: $$\sqrt[\large3]{\sqrt[\large3]{1458}-9}-1=\sqrt[\large3]4-\sqrt[\large3]2$$ so $\sqrt[\large3]4-\sqrt[\large3]2\in C$. (Well, Ramanujan proved something equivalent.)

I found that $\left(\sqrt[\large3]2-1\right)^{-1}-1=\sqrt[\large3]2+\sqrt[\large3]4$, but I'm not allowed to use negative exponents/roots, so this doesn't prove that $\sqrt[\large3]2+\sqrt[\large3]4\in C$. So:

Is it true that $\sqrt[\large3]2+\sqrt[\large3]4\in C$? If not, why not?

$\sqrt[\large3]2+\sqrt[\large3]4$ is indeed in $C$: $$\sqrt{(\sqrt[\large3]{2} + 1)^2+3}=\sqrt[\large3]2+\sqrt[\large3]4$$

First, let's think about the process the other way around, instead of starting at $0$ and working our way to the number, start at the number and get to a whole number. Our goal is to get an expression with only one surd. After some messing around you realize that just addition and exponentiation isn't cutting it like it did for the other examples, so we must take a root. For convenience let $x=\sqrt[\large3]2$, and observe that $ax^2 + bx + c$ can represent every result of the operations allowed. So if we can get to a quadratic with an integer single root, we found a way to get rid off one of the surds.

$$(x^2+x)^2=x^4+2x^3+x^2=2x+4+x^2=x^2+2x+1+3=(x+1)^2+3$$ $$x^2+x=\sqrt{(x+1)^2+3}$$

$\sqrt[\large3]{6+\sqrt{(\sqrt[\large3]{864}+3)^2+135}}$ was the original solution, based on the quadratic ($x^2 + x +\frac{1}{4}$) my friend suggested (He came up with the idea of thinking about it as a quadratic with a single root). By the way, we can use the same method to get another identity: $$\sqrt{(\sqrt[\large3]{2} + 1)^2-5}=\sqrt[\large3]4-\sqrt[\large3]2$$

My follow-up questions are:

1. $\sqrt[\large3]2+\sqrt[\large3]4$ can be expressed in similar way but without whole number exponentiation?
2. Is $\sqrt[\large3]2+\sqrt[\large3]3$ in $C$? Is $\sqrt2+\sqrt[\large3]3$ in $C$?
• Wow! Thank you! May I ask how you found this question (since it was written over a year ago)? – Akiva Weinberger Nov 11 '16 at 16:14
• From here (intresteing question) and through some of related questions links, I think. – i9Fn Nov 11 '16 at 16:24