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The question is: if $P$ is prime is $P^{1/3}$ rational?

I have been able to prove that if $P$ is prime then the square root of $P$ isn't rational (by contradiction) how would I go about the cube root?

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If $P$ is prime then neither $\sqrt{P}$ or $\sqrt[3]{P}$ is an integer, so the notion of primality makes no sense for them (tiny caveat: over $\mathbb{Z}$) and the simplest sensible answer is 'no'. –  Steven Stadnicki Feb 13 at 17:47
    
How can I rigorously prove that? –  IndividualThinker Feb 13 at 17:49
    
If $\sqrt{P} = a$ is a prime integer, then $P = a\cdot a$ is a prime factorization of $P$... –  BaronVT Feb 13 at 17:50
    
Omid: try searching for 'irrationality of square root proof' in your favorite search engine; there are several standard proofs, starting from various different sets of core principles, and you're sure to find one to your tastes somewhere in the stack. –  Steven Stadnicki Feb 13 at 17:51
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my apologies, I meant to type rational not prime. I have corrected the question. –  IndividualThinker Feb 13 at 17:57

3 Answers 3

Suppose $\sqrt[3]{P} = \dfrac{a}{b}$ where $a$ and $b$ have no common factors (i.e. the fraction is in reduced form). Then you have

$$ b^3 P = a^3. $$

Both sides must be divisible by $a$ (if they're both equal to $a^3$). We already know that $a$ does not divide $b$ (when we assumed the fraction is reduced). So then $a$ must divide $P$.

EDIT: and if $a = 1$, then $P = \dfrac{1}{b^3}$. How many integers are of the form $\dfrac{1}{B}$ for some $B$?

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To be pedantic, you need to add a line saying why $a = 1$ isn't possible either. :-) –  ShreevatsaR Feb 13 at 18:15
    
That would be very pedantic. P is prime, therefore greater than 1, therefore its roots are greater than 1, therefore a > b, but b is an integer greater than 0, therefore 0 < b < 1. No integers exist (as far as I know) between 0 and 1. –  Malvolio Feb 14 at 0:20

The main point is: The cube root of a natural number is rational iff it is infact an integer. More generally, any rational root of a monic polynomial with integer coefficients (such as $X^3-n$) is in fact integer. So if $\sqrt[3] n$ is rational then $n$ is a cube (and cannot be prime).

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Hint $ $ Any rational root of $\,x^3-p\,$ is an integer, by the Rational Root Test.

Alternatively $\, a^3 = pb^3\,$ contradicts the uniqueness of prime factorizations, since the prime $\,p\,$ occurs to power a multiple of $\,3\,$ on the lhs, but a nonmultiple $\,1\!+\!3n\,$ on rhs, i.e. $\,0\not\equiv 1\pmod 3.\,$ This is a generalization of the analogous proof of irrationality of square-roots by comparing the parity of exponents of $\,p,\,$ i.e. $\,0\not\equiv 1\pmod 2,\,$ i.e. even $\ne $ odd. Precisely the same proof works for $k$'th roots, by employing that $\ 0\not\equiv 1\pmod{\! k}$

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