$ x = \sqrt [3]{2 + \sqrt {a}} + \sqrt [3]{2 - \sqrt {a}}$ $ \Leftrightarrow x^3 = 4 + 3x\sqrt [3]{4 - a}$ So $ 4 - a = \frac {p^3}{q^3}$ where $ \gcd(q,p) = 1$ $ \Rightarrow x^3 = 4 + \frac {p}{q}x$ $ \Leftrightarrow qx^3 - 3px - 4q = 0$ $ \Rightarrow q|3px \Rightarrow q|3x$ Let $ 3x = \frac {mq}$ Let replace we has: $ q(mq)^3 - 27pmq - 108q = 0$ $ (mq)^3 - 27mp - 108 = 0$ $ m|108$ Case 1 $ m = 1$ $ q^3 - 27p - 108 = 0$ so $ q = 3k$ $ k^3 - p - 4 = 0$ $ x = k,p = k^3 - 4,q = 3k$ $ \Rightarrow 4 - a = (\frac {k^3 - 4}{3k})^3\Rightarrow a = 4 - (\frac {k^3 - 4}{3k})^3$ where $ k\in N$ $ a > 0$ so $ 4 - \frac {k^3 - 4}{3k}\geq 0$ Implies that $ k = 2$ so $ a = \frac {100}{27}$ Case 2 $ m = 2$ Implies that $ q = 3k$ $ 27(2k)^3 - 81kp - 108 = 0$ $ (2k)^3 - 3kp - 4 = 0$ So $ k|4$ $ p = 10,k = 2$ $ p = 10,q = 6$ but $ a < 0$ so it is not satisfy. Case 3 $ m = 3$ $ q^3 - 3p - 4 = 0$ Easy to check it not satisfty. Case 4...

euh... what have you proven now?

$\sqrt[3]{2-\sqrt{a}}=A$ $\sqrt[3]{2+\sqrt{a}}=B$ if $A+B$ is integer so $(A+B)^3$is integer with a little Calculation We conclude that $AB$ is integer so we have $\sqrt[3]{4-a}$ is integer so for $a\leq 4$ we should check 3 and 4 if a=3, $2<A+B<3 $ if a=4 ,$ A+B=\sqrt[3]{4}$ if a>4 an $A+B$ is integer $A+B=1$ so with a little Calculation We conclude that$AB=-1$so $\sqrt[3]{4-a}=-1$ so $a=5$ so a=5 is only answer

Peter wrote: For what positive numbers $a$ is \[\sqrt[3]{2+\sqrt{a}}+\sqrt[3]{2-\sqrt{a}}\] an integer? Property. If $a+b+c=0$ then $a^3+b^3+c^3=3abc$. Proof. We have $a^3+b^3+c^3-3abc= (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$. Solution. Let $\sqrt[3]{2+\sqrt{a}}+\sqrt[3]{2-\sqrt{a}}=b$. We have $b + \left( - \sqrt[3]{2+ \sqrt a} \right) + \left( - \sqrt[3]{2- \sqrt a} \right) =0$ Therefore $b^3 - \left( 2+ \sqrt a \right) - \left( 2- \sqrt a \right) = 3b \sqrt[3]{ (2+ \sqrt a)(2 - \sqrt a)}$. Or $b^3-4= 3b \sqrt[3]{4-a} \Leftrightarrow b^3-3xb-4=0 \qquad (1)$ (we let $\sqrt[3]{4-a}=x$). Thus, we need to find $a$ such that $\sqrt[3]{4-a}$ and $b$ is an integer. If $b$ is an integer then from $(1)$ we obtain $b|4$. +) $b=1$ then $x=-1$ implies $a=5$. +) $b=-1$ then $x= \dfrac{-5}{3}$ isn't integer. +) $b=2$ ...... So the only solution is $a=5$.

sahadian wrote: $\sqrt[3]{2-\sqrt{a}}=A$ $\sqrt[3]{2+\sqrt{a}}=B$ if $A+B$ is integer so $(A+B)^3$is integer with a little Calculation We conclude that $AB$ is integer so we have $\sqrt[3]{4-a}$ is integer so for $a\leq 4$ we should check 3 and 4 if a=3, $2<A+B<3 $ if a=4 ,$ A+B=\sqrt[3]{4}$ if a>4 an $A+B$ is integer $A+B=1$ so with a little Calculation We conclude that$AB=-1$so $\sqrt[3]{4-a}=-1$ so $a=5$ so a=5 is only answer it just gives AB is rational, doesn't it?????

shinichiman wrote: Peter wrote: For what positive numbers $a$ is \[\sqrt[3]{2+\sqrt{a}}+\sqrt[3]{2-\sqrt{a}}\]an integer? Property. If $a+b+c=0$ then $a^3+b^3+c^3=3abc$. Proof. We have $a^3+b^3+c^3-3abc= (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$. Solution. Let $\sqrt[3]{2+\sqrt{a}}+\sqrt[3]{2-\sqrt{a}}=b$. We have $b + \left( - \sqrt[3]{2+ \sqrt a} \right) + \left( - \sqrt[3]{2- \sqrt a} \right) =0$ Therefore $b^3 - \left( 2+ \sqrt a \right) - \left( 2- \sqrt a \right) = 3b \sqrt[3]{ (2+ \sqrt a)(2 - \sqrt a)}$. Or $b^3-4= 3b \sqrt[3]{4-a} \Leftrightarrow b^3-3xb-4=0 \qquad (1)$ (we let $\sqrt[3]{4-a}=x$). Thus, we need to find $a$ such that $\sqrt[3]{4-a}$ and $b$ is an integer. If $b$ is an integer then from $(1)$ we obtain $b|4$. +) $b=1$ then $x=-1$ implies $a=5$. +) $b=-1$ then $x= \dfrac{-5}{3}$ isn't integer. +) $b=2$ ...... So the only solution is $a=5$. it just gives you that $ \sqrt[3]{4-a}$ is rational!