# $p$, $q$ and $\sqrt[n]{p} + \sqrt[n]{q}$ are rational, with the latter being non-zero. Are $\sqrt[n]{p}$ and $\sqrt[n]{q}$ rational?

Let $$p, q \in \mathbb Q$$, $$n \in \mathbb Z^+$$ and label $$a = \sqrt[n]p, b=\sqrt[n]q$$.

Conjecture: If $$a + b$$ is a non-zero rational, then both $$a$$ and $$b$$ are rational.

(Preliminary question: is this a known result I'm not aware of?)

I believe I have found a partial proof of the above. Namely,

• I first proved that $$ab$$ is rational for $$n = 1, 2, 3$$.
• For $$n = 2$$, $$(a + b)^2 = a^2 + 2ab + b^2$$, so $$ab = \frac{(a+b)^2-a^2-b^2}{2} \in \mathbb Q$$
• For $$n = 3$$, $$(a + b)^3 = a^3 + 3ab(a + b) + b^3$$, so $$ab = \frac{(a+b)^3-a^3-b^3}{3(a+b)} \in \mathbb Q$$
• As it turns out, an additional assumption that $$ab \in \mathbb Q$$ allows proving the conjecture.
• The question: how can I prove $$ab \in \mathbb Q$$ for $$n > 3$$? (or what numbers are a counterexample?)
• The details of the proof assuming $$ab \in \mathbb Q$$ follow.
• Consider the polynomial $$(x - a)(x - b) = x^2 - (a + b)x + ab$$. Note that its coefficients are rational.
• This means that $$\Pi \in \mathbb Q[x]$$, the minimal polynomial of $$a$$, is of degree at most 2.
• Hence, $$\deg \Pi \in \{1, 2\}$$. If $$\deg \Pi = 1$$, then $$a$$ is rational, which was to be proven, so let's assume that $$\deg \Pi = 2$$ and hope for a contradiction.
• As the minimal polynomial is unique, $$\Pi = (x - a)(x - b)$$. Moreover, $$\Pi$$ divides $$x^n - p$$, since the latter has a root at $$a$$.
• Hence, the roots of $$\Pi$$ are a subset of the roots of $$x^n - p$$.
• For odd $$n$$, we have $$\{a, b\} \subseteq \{a\}$$, so $$a = b$$.
• For even $$n$$, we have $$\{a, b\} \subseteq \{a, -a\}$$. $$b$$ can't be equal to $$-a$$, since roots of even degree are nonnegative (or imaginary, but in that case $$a + b$$ wouldn't be rational). Hence, $$a = b$$.
• In both cases, from the assumption $$a + b \in \mathbb{Q} \setminus \{0\}$$, we get $$2a \in \mathbb{Q} \implies a \in \mathbb Q$$ and therefore $$\deg \Pi = 1$$, which is a contradiction.
• Are you familiar/comfortable with field theory? You are essentially asking if the nth roots are linearly independent over Q. – Calvin Lin Nov 29 '19 at 23:46
• @Calvin Lin No, I've started studying abstract algebra a few days ago after someone pointed me to the concept of minimal polynomials. (In a hope to eventually learn about algebraic number theory in general). – NieDzejkob Nov 30 '19 at 11:18

## 1 Answer

It may not be true if $$n$$ is odd and $$q=-p$$. But otherwise your statement is correct. Let's prove it in the case $$p$$, $$q \ge 0$$. The following is a standard trick that avoids the Galois theory. It is enough to show that $$\sqrt[n]{p}$$ is a rational fraction in $$\sqrt[n]{p} + \sqrt[n]{q}$$ with rational coefficients. Consider the two polynomials $$X^n - p$$ and $$(\sqrt[n]{p} + \sqrt[n]{q}-X)^n - q$$. The first polynomial has the roots $$\omega \cdot \sqrt[n]{p}$$, while the second polynomial has the roots $$(\sqrt[n]{p} + \sqrt[n]{q}) -\omega' \cdot \sqrt[n]{q}$$. where $$\omega$$, $$\omega'$$ are $$n$$-th roots of $$1$$. These polynomials have exactly one common root $$\sqrt[n]{p}$$. Therefore, their gcd is $$(X-\sqrt[n]{p})$$. Now, the coefficients of the gcd of two polynomials can be expressed rationally in terms of the coefficients of the given polynomials. We conclude that there exists a rational function $$R(t) \in \mathbb{Q}(t)$$ such that $$\sqrt[n]{p} = R(\sqrt[n]{p} + \sqrt[n]{q})$$

• Note that $q = -p$ violates the assumption that $a + b$ is a non-zero rational. I'll try to find some information about the gcd of polynomials and understand your proof. Thank you. – NieDzejkob Nov 30 '19 at 13:26
• @NieDzejkob: You are welcome! The most general result would be that roots of positive rationals are linearly independent over $\mathbb{Q}$ if they are pairwise incomensurable. you can check for instance math.stackexchange.com/questions/158722/… – orangeskid Dec 2 '19 at 0:45
• I thought quite a bit about various properties of the gcd of polynomials and I think I understand now. Though, I can't see where the assumption that $p, q \geq 0$ is used. Does this proof break for negative $p$ or $q$? I haven't noticed any such moment. – NieDzejkob Dec 2 '19 at 14:45
• @NieDzejkob: the proof uses the fact that $\sqrt[n]{p}+ \sqrt[n]{q}= \omega\sqrt[n]{p}+ \omega' \sqrt[n]{q}$ implies $\omega = \omega' = 1$. Now this is straightforward if $p$,$q$ are positive , but not that simple if say $q$ is negative. – orangeskid Dec 2 '19 at 18:25