Find natural number $0 < n < 30,000$ such that $sqrt[3]{5n}+sqrt{10n}$ is rational
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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
asked Dec 9 '18 at 15:09
Trevor MasonTrevor Mason
288
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add a comment |
$begingroup$
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
asked Dec 9 '18 at 15:09
Trevor MasonTrevor Mason
288
$endgroup$
add a comment |
$begingroup$
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
asked Dec 9 '18 at 15:09
Trevor MasonTrevor Mason
288
$endgroup$
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
asked Dec 9 '18 at 15:09
Trevor MasonTrevor Mason
288
asked Dec 9 '18 at 15:09
Trevor MasonTrevor Mason
288
asked Dec 9 '18 at 15:09
Trevor MasonTrevor Mason
288
asked Dec 9 '18 at 15:09
Trevor MasonTrevor Mason
288
asked Dec 9 '18 at 15:09
Trevor MasonTrevor Mason
288
288
add a comment |
add a comment |
1 Answer
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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 $5$.
answered Dec 9 '18 at 15:14
Ross MillikanRoss Millikan
300k24200374
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$begingroup$
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
|
show 1 more comment
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1 Answer
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1 Answer
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oldest
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active
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votes
$begingroup$
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 $5$.
answered Dec 9 '18 at 15:14
Ross MillikanRoss Millikan
300k24200374
$endgroup$
$begingroup$
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
|
show 1 more comment
$begingroup$
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 $5$.
answered Dec 9 '18 at 15:14
Ross MillikanRoss Millikan
300k24200374
$endgroup$
$begingroup$
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
|
show 1 more comment
$begingroup$
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 $5$.
answered Dec 9 '18 at 15:14
Ross MillikanRoss Millikan
300k24200374
$endgroup$
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 $5$.
answered Dec 9 '18 at 15:14
Ross MillikanRoss Millikan
300k24200374
answered Dec 9 '18 at 15:14
Ross MillikanRoss Millikan
300k24200374
answered Dec 9 '18 at 15:14
Ross MillikanRoss Millikan
300k24200374
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
|
show 1 more comment
$begingroup$
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
|
show 1 more comment
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