Track / Stage 4 / 210 of 340 #470 of 1964

Problem 470

AMC 12 late, AIME early

Algebra Difficulty 4.8 Prove it The 14th Thailand Mathematical Olympiad · Thailand

Let $p$ be a prime. Show that $\sqrt[3]{p} + \sqrt[3]{p^5}$ is irrational.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solutions — 2

Solution 1

Let $r = \sqrt[3]{p} + \sqrt[3]{p^5}$ . Then,
$r^3 = (\sqrt[3]{p} + \sqrt[3]{p^5})^3 = p + p^5 + 3p^2(\sqrt[3]{p} + \sqrt[3]{p^5}) = p + p^5 + 3p^2 r.$
Hence $r$ is a root of the polynomial $x^3 - 3p^2x - p^5 - p$ . Assume to the contrary that $r$ is rational. By the rational root theorem we have $r$ is an integer. From $r^3 = p + p^5 + 3p^2r$ we get $p \mid r$ . Hence $p^3 \mid r^3 - 3p^2r - p^5$ , implying $p^3 \mid p$ a contradiction. Thus, $r$ is irrational.

Solution 2

Assume to the contrary that $r = \sqrt[3]{p} + \sqrt[3]{p^5}$ is rational. Thus
$\frac{r + p^3}{1 + pr} = \frac{\sqrt[3]{p} + \sqrt[3]{p^5} + \sqrt[3]{p^9}}{1 + \sqrt[3]{p^4} + \sqrt[3]{p^8}} = \sqrt[3]{p}$
is rational. This is clearly false, thus $r$ is irrational.

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