Find natural number $0 < n < 30,000$ such that $sqrt[3]{5n}+sqrt{10n}$ is rational












0












$begingroup$


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.






irrational-numbers rational-numbers







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    0












    $begingroup$


    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.






    irrational-numbers rational-numbers







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      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.






      irrational-numbers rational-numbers







      share|cite|improve this question









      $endgroup$




      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.






      irrational-numbers rational-numbers




      irrational-numbers rational-numbers






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 9 '18 at 15:09









      Trevor MasonTrevor Mason

      288




      288






















          1 Answer
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          3












          $begingroup$

          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 $5$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            by leading coefficients you mean in the prime factorization?
            $endgroup$
            – Trevor Mason
            Dec 9 '18 at 15:31






          • 1




            $begingroup$
            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






          • 1




            $begingroup$
            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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          1 Answer
          1






          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$

          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 $5$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            by leading coefficients you mean in the prime factorization?
            $endgroup$
            – Trevor Mason
            Dec 9 '18 at 15:31






          • 1




            $begingroup$
            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






          • 1




            $begingroup$
            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


















          3












          $begingroup$

          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 $5$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            by leading coefficients you mean in the prime factorization?
            $endgroup$
            – Trevor Mason
            Dec 9 '18 at 15:31






          • 1




            $begingroup$
            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






          • 1




            $begingroup$
            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
















          3












          3








          3





          $begingroup$

          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 $5$.






          share|cite|improve this answer









          $endgroup$



          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 $5$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 9 '18 at 15:14









          Ross MillikanRoss Millikan

          300k24200374




          300k24200374












          • $begingroup$
            by leading coefficients you mean in the prime factorization?
            $endgroup$
            – Trevor Mason
            Dec 9 '18 at 15:31






          • 1




            $begingroup$
            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






          • 1




            $begingroup$
            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




















          • $begingroup$
            by leading coefficients you mean in the prime factorization?
            $endgroup$
            – Trevor Mason
            Dec 9 '18 at 15:31






          • 1




            $begingroup$
            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






          • 1




            $begingroup$
            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


















          $begingroup$
          by leading coefficients you mean in the prime factorization?
          $endgroup$
          – Trevor Mason
          Dec 9 '18 at 15:31




          $begingroup$
          by leading coefficients you mean in the prime factorization?
          $endgroup$
          – Trevor Mason
          Dec 9 '18 at 15:31




          1




          1




          $begingroup$
          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




          $begingroup$
          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






          $begingroup$
          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




          $begingroup$
          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




          1




          1




          $begingroup$
          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






          $begingroup$
          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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