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Let $f(x) = x^n - a$ be a polynomial with integer coefficients, when does $ f (x) $ have rational solutions? Is there a necessary and sufficient condition?

I understand this is equivalent to asking for integer solutions.

This is beacuse if $\frac{p}{q}$ is rational solution with $ gcd(p,q) =1$ then $ q \vert ( p^n - q^n a =0)$ and $q \vert q^n a$

So $q\vert (p^n -q^n a + q^n a = p^n)$. Moreover, since $p$ and $q$ are coprime, $q=1$.

Bysshed
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    Um, $x^n-a$ has a rational root exactly if each prime in the prime factorization of $a$ appears with an exponent that is a multiple of $n$. (That's for positive $a$). Did you have something more explicit than that in mind? For general polynomials, there's also the Rational Root Theorem, which implies that any rational root of a monic integer polynomial will be an integer. – hmakholm left over Monica Sep 04 '15 at 01:14
  • Yeah I realise it's quite simple really. Just needed to use the fundamental theorem of arithmetic – Bysshed Sep 04 '15 at 01:28

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It has rational solutions if and only if it has integer solutions, i.e. if and only if $a$ is an $n$-th power in $\mathbf Z$. This can be proved in an elementary way (I mean, without unique factorisation).

Bernard
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In terms of reals, there can be at most two real roots (depending on the parity of n). Let's assume $n$ is odd for convenience. The even case is similar. Then the real root is $\sqrt[n]{a}$. So you're just asking when is $\sqrt[n]{a}$ an integer. But that's easy. It's when $a$ is the $n^{\text{th}}$ power of an integer. It's clear that this is both necessary and sufficient. It's also about as nice a condition as you can expect.

Zach Stone
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