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This question was in an oral entry exam for a university in Italy. I have tried to procede by contradiction but don't seem to get anywhere. How can a question like this one be approached?

Lorenzo Catani
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    What are $a,,b$: positive integers? Edit a clarification into your question. – J.G. Apr 30 '20 at 11:52
  • https://math.stackexchange.com/questions/457382/can-sqrtn-sqrtm-be-rational-if-neither-n-m-are-perfect-squares?rq=1 – jimjim Apr 30 '20 at 17:14
  • Let $F = \mathbf{Q}(\sqrt{a})$ and $E = \mathbf{Q}(\sqrt[3]{b})$. If $\sqrt{a} + \sqrt[3]{b} = r$ is rational then $\sqrt{a} = r - \sqrt[3]{b} \in E$ and $\sqrt[3]{b} = r - \sqrt{a} \in F$, so $E \subset F$ and $F \subset E$. Can you finish from here? (Writing down an explicit polynomial like in the answer below is completely unnecessary.) – user760870 Apr 30 '20 at 18:21
  • I guess it would follow that $E = F$ from which one can derive a contradiction. Could you please clarify what the notation $\pmb{Q}(\sqrt{a})$ indicates? Does it mean that $\sqrt{a} \in \pmb{Q}$? – Lorenzo Catani May 01 '20 at 10:09

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Assuming $a,\,b\in\Bbb N$:

Let $q:=\sqrt{a}+\sqrt[3]{b}$ so $\sqrt[3]{b}$ is a root of both $x^2-2qx+q^2-a$ and $x^3-b$, so also of$$x^3-b-(x+2q)(x^2-2qx+q^2-a)=(a+3q^2)x+2qa-2a^3-b.$$So if $q\in\Bbb Q$, $\sqrt[3]{b}\in\Bbb Q$ and $b$ is a perfect cube, and $\sqrt{a}=q-\sqrt[3]{b}\in\Bbb Q$ so $a$ is a perfect square.

J.G.
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  • Thank you. How did you come up with this solution though? Is it standard practice in these types of questions to define polynomials whose roots are (at least partially) known? I don't see how one could come up with the polynomial $^2-2qx+q^2-a$ and the factor $(x+2q)$ which somehow seem to work so well... – Lorenzo Catani Apr 30 '20 at 12:12
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    @LorenzoCatani The question looks us to prove each surd is rational, i.e. of algebraic degree $1$, so we seek a linear solved by whichever surd has the highest apparent degree. We know one polynomial of that degree it solves by definition; we get another by rearranging the definition of $q$ and squaring; and then we take a linear combination of degree $1$. The $x+2q$ results from solving simultaneous equations to get rid of higher-order terms. – J.G. Apr 30 '20 at 12:31
  • (looks being a typo for wants) – J.G. Apr 30 '20 at 12:49