<s>[INST] Proving $\sqrt[3]{2}$ is irrational without using prime factorization Prove that $\sqrt[3]{2}$ is irrational without using prime factorization.
The standard proof that $\sqrt[3]{2}$ is irrational uses prime factorization in an essential way. So I wondered if there is a proof that does not use it.
This was inspired by the fact that I know two proofs that $\sqrt{2}$ is irrational that do not use  prime factorization.
The first uses
$$\sqrt{2}=\sqrt{2}\frac{\sqrt{2}-1}{\sqrt{2}-1}=\frac{2-\sqrt{2}}{\sqrt{2}-1}
$$
to show that if $\sqrt{2} = \dfrac{a}{b}$ then
$$\sqrt{2}=\frac{2-\dfrac{a}{b}}{\dfrac{a}{b}-1}=\frac{2b-a}{a-b}
$$
is a rational $\sqrt{2}$ with a smaller denominator.
The second uses
$$(x^2-2y^2)^2=(x^2+2y^2)^2-2(2xy)^2
$$
and $3^2-2\cdot 2^2 = 1$ to show that $x^2-2y^2=1$ has arbitrarily large solutions and this contradicts $\sqrt{2}$ being rational.
I have not been able to extend either of these proofs to $\sqrt[3]{2}$. Results that  I do not consider "legal" in solving this problem include Fermat's Last Theorem (which definitely uses unique factorization) and the  rational root theorem (which uses unique factorization
in its proof).
 [/INST] If $\sqrt[3]2$ we rational, it would be $a/b$ for $a$, $b\in\Bbb N$. Then there is $n\in\Bbb N$ with $n\sqrt[3]2$ and $n\sqrt[3]4$ integers, say $n=b^2$. Let $n$ be the least positive integer with $n\sqrt[3]2$, $n\sqrt[3]4\in\Bbb N$. But $(\sqrt[3]2-1)n$ would be an even smaller one...
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