If $\sqrt[3]{a} + \sqrt[3]{b}$ is rational then prove $\sqrt[3]{a}$ and $\sqrt[3]{b}$ are rational

Question

Assume there exist some rationals $a, b$ such that $\sqrt[3]{a}, \sqrt[3]{b}$ are irrationals, but:

$$\sqrt[3]{a} + \sqrt[3]{b} = \frac{m}{n}$$

for some integers $m, n$

$$\implies \left(\sqrt[3]{a} + \sqrt[3]{b}\right)^3 = \frac{m^3}{n^3}$$

$$\implies a + b + 3 \cdot \sqrt[3]{ab}\left(\sqrt[3]{a} + \sqrt[3]{b}\right) = \frac{m^3}{n^3}$$

Since $a +b$, $\sqrt[3]{a} + \sqrt[3]{b}$ are rational, $\sqrt[3]{ab}$ must be rational as well.

For convenience let us say $\sqrt[3]{a} = p, \sqrt[3]{b} = q \implies pq$ is rational. This means, for all $i$, $p^iq^i$ is rational.

$$\implies (p + q)^2 = \frac{m^2}{n^2}$$

$$ = p^2 + q^2 + 2pq = \frac{m^2}{n^2}$$

Since $pq$ is rational, $2pq$ is rational and so is $p^2 + q^2$.

Assume, for some $i$ that $p^i + q^i$ and $p^{i-1} + q^{i-1}$ is rational.

$$\implies (p^i + q^i)(p + q) - pq(p^{i - 1} + q^{i - 1}) = p^{i+1} + q^{i+1}$$

is rational as well.

So for all $i$,

$$a^{\frac{i}{3}} + b^{\frac{i}{3}}$$

and

$$a^{\frac{i}{3}}b^{\frac{i}{3}}$$

are rational.

I know I'm really close to the answer, but it somehow just keeps slipping through my fingers.

Answer 1

As you have proved, $p^2+q^2\in\Bbb Q$ and $pq\in \Bbb Q$, so $$p-q=\frac{a-b}{p^2+pq+q^2}\in \Bbb Q.$$ Combining this with $p+q\in \Bbb Q$, we are done.

Answer 2

Statement for the general case (for any $n>1$ in place of $3$)

Let $a,b$ be two positive rational numbers such that $\sqrt[n]{a}+\sqrt[n]{b}$ is a rational number for some natural number $n>1$. Then both $\sqrt[n]{a}$, $\sqrt[n]{b}$ are rational numbers.

Proof

Let $s=\sqrt[n]{a}+\sqrt[n]{b}\in\mathbb{Q}$. Then $\sqrt[n]{a}$ is a root of the polynomial $p(X)=X^n-a$ (If $\sqrt[n]{a}$ is not rational then $p$ is in fact the minimal polynomial). Again $\sqrt[n]{a}$ is also a root of $q(X)=(s-X)^n-b$. Now we show that $\sqrt[n]{a}$ is the unique common root of $p$ and $q$. Let $z$ be a common root of $p$ and $q$. Then $z=\sqrt[n]{a}\omega_1$ and $s-z=\sqrt[n]{b}\omega_2$ for some $n^{th}$ roots of unity $\omega_1,\omega_2$. Then $$\sqrt[n]{a}+\sqrt[n]{b}=\sqrt[n]{a}\omega_1+\sqrt[n]{b}\omega_2\tag{1}$$ Since $|\omega_1|=|\omega_2|=1$ therefore $|\mathrm{Re}(\omega_i)|\leq1$ with equality if and only if $\omega_i=\pm1$ for $i\in\{1,2\}$. Now comparing real parts in the equation $(1)$ we can easily conclude that $\omega_1=\omega_2=1$ implying that $\sqrt[n]{a}$ is the unique common root.

Then $$\gcd(p(X),q(X))=X-\sqrt[n]{a}$$ But $p(X),q(X)\in\mathbb{Q}[X]$ implies $X-\sqrt[n]{a}$ is in $\mathbb{Q}[X]$ then $\sqrt[n]{a}\in\mathbb{Q}$. Similarly one can prove that $\sqrt[n]{b}$ is also rational.