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Prove or disprove: For all positive integers $ n$ ,

$\sqrt[3]{n}+\sqrt[3]{n+1}$ are irrational numbers.

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Since $x=n^{1/3}$ and $y=(n+1)^{1/3}$ are algebraic integers then $x+y$ is an algebraic integer. Hence, if $x+y$ is rational then it is an integer. –  i. m. soloveichik Sep 29 '12 at 2:43
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2 Answers 2

up vote 7 down vote accepted

Note that $(x+y)^{3} = x^{3} +y^{3} + 3xy(x+y).$ If $n$ or $n+1$ is the cube of an integer, the result is clear by the uniqueness of prime factorization, so we assume that neither is the cube of an integer. Then neither is $n(n+1)$. Set $x = n^{\frac{1}{3}}$ and $y = (n+1)^{\frac{1}{3}}.$ If $x+y$ is rational, then so are $(x+y)^{3}$ and $3xy(x+y)$. Hence $3xy$ is rational. Thus $27n(n+1)$ must be the cube of an integer, a contradiction.

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Why 27n(n+1) not a cube?Can explain in detail Although n and n+1 not a cube, but their product may be ah, for example, 2 and 4 are not cubic number, but 8 cubic number. –  tianzhidaosunyouyu Sep 29 '12 at 8:26
    
But $n$ and $n+1$ are relatively prime. If $p$ is any prime, then if $n(n+1)$ is a cube, the power of $p$ dividing $n(n+1)$ is an integr divisible by $3$. If $p$ divides $n$, then $p$ does not divide $n+1,$ so the power of $p$ dividing $n$ would have to be an integer multiple of $3$. Hence $n$ must be a cube, as $p$ was an arbitrary prime divisor of $n.$ Similarly for $n+1$. –  Geoff Robinson Sep 29 '12 at 10:02
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Hint $\ $ If $\rm\:n,m\in\Bbb N$ then cubing $\rm\: \sqrt[3]{n}+\sqrt[3]{m}\in\Bbb Q\:\Rightarrow\: nm\in \Bbb N^3\ (\Rightarrow\ n,m\in \Bbb N^3\ if\ (n,m) = 1)$

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