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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.
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    $\begingroup$ Are you familiar/comfortable with field theory? You are essentially asking if the nth roots are linearly independent over Q. $\endgroup$ – Calvin Lin Nov 29 '19 at 23:46
  • $\begingroup$ @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). $\endgroup$ – NieDzejkob Nov 30 '19 at 11:18
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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 $$\sqrt[n]{p} = R(\sqrt[n]{p} + \sqrt[n]{q})$$

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  • $\begingroup$ 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. $\endgroup$ – NieDzejkob Nov 30 '19 at 13:26
  • $\begingroup$ @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/… $\endgroup$ – orangeskid Dec 2 '19 at 0:45
  • $\begingroup$ 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. $\endgroup$ – NieDzejkob Dec 2 '19 at 14:45
  • $\begingroup$ @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. $\endgroup$ – orangeskid Dec 2 '19 at 18:25

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