I was thinking that I could try to make some sort of substitution to convert $\sqrt[3]{5n}+\sqrt{10n}$ into a polynomial with integer coefficients then use the Rational Roots Theorem to find a rational root. I don't really know if that's going to get me anywhere other than $n=0$.

I would really appreciate some help, or some hints. I don't necessarily want a full solution, but a nudge in the right direction.


You should focus on the prime factorization of $n$. To have a number be a cube, all the primes in its factorization need to come with a power that is a multiple of $3$. Similarly, to be a square the primes need to have an even power.

In your example, all primes except $2$ and $5$ must be sixth powers. Think about what the leading coefficients do for $2$ and $5$.

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  • $\begingroup$ by leading coefficients you mean in the prime factorization? $\endgroup$ – Trevor Mason Dec 9 '18 at 15:31
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    $\begingroup$ By leading coefficients I mean the $5$ and $10$ under the radical signs. They mean you don't want the powers of $2$ and $5$ in $n$ to be multiples of $6$. $\endgroup$ – Ross Millikan Dec 9 '18 at 15:38
  • $\begingroup$ Ah, yes. That makes much more sense. I think I figured it out. Thanks so much for your help! $\endgroup$ – Trevor Mason Dec 9 '18 at 15:39
  • $\begingroup$ I think the limit of $30,000$ is unfortunately low. I think it should have been chosen large enough for there to be half a dozen solutions or so. $\endgroup$ – Ross Millikan Dec 9 '18 at 15:56
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    $\begingroup$ I just would like to have people recognize that once they find the smallest solution, which is $2^35^5=25,000$ they can multiply it by the sixth power of any number, including $2$ and $5$ and get another solution. $\endgroup$ – Ross Millikan Dec 9 '18 at 16:16

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