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I'm trying to demostrate that $ \sqrt{x} + \sqrt[3]{y} $ is either an integer or an irrational $ \forall x,y\in \mathbb{N} $. I have simplified the problem in 4 cases:

• Both $\sqrt{x}$ and $\sqrt[3]{y}$ are rational.

• $\sqrt{x}$ is irrational and $\sqrt[3]{y}$ rational.

• $\sqrt{x}$ is rational and $\sqrt[3]{y}$ irrational.

• Both $\sqrt{x}$ and $\sqrt[3]{y}$ are irrational.

The only case I haven't solved yet is the last one, I have no idea what should I try.

DanyWose
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  • @learning123 I think at the end you are missing some cases, because one would need to consider $x = p^k \cdot t$ with $p \not| t$ in order to be able to obtain a contradiction. But regardless of that, the case of the sum does not follow from this in any obvious way. – Cristian Gratie Mar 22 '22 at 11:28

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Suppose that $\sqrt{x}+{y}^{1/3}=q$, where $q\in \mathbb{Q}$ then $y=(q-\sqrt{x})^{3}=q^3-3q^2\sqrt{x}+3qx-x\sqrt{x}\implies -y+q^3+3qx=(3q^2+x)\sqrt{x}$ or $(x+q'')\sqrt{x}=q'$, where $q', q'' \in \mathbb{Q}$ ($x$ is integer).But then this also implies that $\sqrt{x}$ is rational... Which is a contradiction. So $\sqrt{x}+y^{1/3}$ cannot be a rational number (neither of course, an integer) .

dmtri
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