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, this trivially leads to $a \in \mathbb Q$ and therefore $\deg \Pi = 1$, which is a contradiction.
from Hot Weekly Questions - Mathematics Stack Exchange
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