a natural number that is both a perfect square and a perfect cube is a perfect sixth power?
$begingroup$
I really can't get a grasp on how to prove this, because if $x$ = $sqrt[6] n$ for some $n$, then $x^2$ = $a$ and $x^3$ = $b$, with $a$ and $b$ being different natural numbers right? Any help?
discrete-mathematics
asked Jul 19 '15 at 23:37
user3467433user3467433
4514
$endgroup$
add a comment |
$begingroup$
I really can't get a grasp on how to prove this, because if $x$ = $sqrt[6] n$ for some $n$, then $x^2$ = $a$ and $x^3$ = $b$, with $a$ and $b$ being different natural numbers right? Any help?
discrete-mathematics
asked Jul 19 '15 at 23:37
user3467433user3467433
4514
$endgroup$
$begingroup$
You can show that if $n$ is a perfect cube, then so is $sqrt n$.
$endgroup$
– MJD
Jul 20 '15 at 0:04
$begingroup$
related: math.stackexchange.com/questions/538801/…
$endgroup$
– Henry
Jul 20 '15 at 7:41
$begingroup$
It may be helpful to look at the first examples. The number $64$ is both a square and a cube because $64=8^2$ and $64=4^3$. And for $729$ we have $729=27^2$ and $729=9^3$. The next two are $4096=64^2, 4096=16^3$ and $15625=125^2, 15625=25^3$. Can you see where you confused yourself?
$endgroup$
– Jeppe Stig Nielsen
Jul 20 '15 at 8:56
add a comment |
$begingroup$
I really can't get a grasp on how to prove this, because if $x$ = $sqrt[6] n$ for some $n$, then $x^2$ = $a$ and $x^3$ = $b$, with $a$ and $b$ being different natural numbers right? Any help?
discrete-mathematics
asked Jul 19 '15 at 23:37
user3467433user3467433
4514
$endgroup$
I really can't get a grasp on how to prove this, because if $x$ = $sqrt[6] n$ for some $n$, then $x^2$ = $a$ and $x^3$ = $b$, with $a$ and $b$ being different natural numbers right? Any help?
discrete-mathematics
discrete-mathematics
asked Jul 19 '15 at 23:37
user3467433user3467433
4514
asked Jul 19 '15 at 23:37
user3467433user3467433
4514
asked Jul 19 '15 at 23:37
user3467433user3467433
4514
asked Jul 19 '15 at 23:37
user3467433user3467433
4514
asked Jul 19 '15 at 23:37
user3467433user3467433
4514
4514
$begingroup$
You can show that if $n$ is a perfect cube, then so is $sqrt n$.
$endgroup$
– MJD
Jul 20 '15 at 0:04
$begingroup$
related: math.stackexchange.com/questions/538801/…
$endgroup$
– Henry
Jul 20 '15 at 7:41
$begingroup$
It may be helpful to look at the first examples. The number $64$ is both a square and a cube because $64=8^2$ and $64=4^3$. And for $729$ we have $729=27^2$ and $729=9^3$. The next two are $4096=64^2, 4096=16^3$ and $15625=125^2, 15625=25^3$. Can you see where you confused yourself?
$endgroup$
– Jeppe Stig Nielsen
Jul 20 '15 at 8:56
add a comment |
$begingroup$
You can show that if $n$ is a perfect cube, then so is $sqrt n$.
$endgroup$
– MJD
Jul 20 '15 at 0:04
$begingroup$
related: math.stackexchange.com/questions/538801/…
$endgroup$
– Henry
Jul 20 '15 at 7:41
$begingroup$
It may be helpful to look at the first examples. The number $64$ is both a square and a cube because $64=8^2$ and $64=4^3$. And for $729$ we have $729=27^2$ and $729=9^3$. The next two are $4096=64^2, 4096=16^3$ and $15625=125^2, 15625=25^3$. Can you see where you confused yourself?
$endgroup$
– Jeppe Stig Nielsen
Jul 20 '15 at 8:56
$begingroup$
You can show that if $n$ is a perfect cube, then so is $sqrt n$.
$endgroup$
– MJD
Jul 20 '15 at 0:04
$begingroup$
You can show that if $n$ is a perfect cube, then so is $sqrt n$.
$endgroup$
– MJD
Jul 20 '15 at 0:04
$begingroup$
related: math.stackexchange.com/questions/538801/…
$endgroup$
– Henry
Jul 20 '15 at 7:41
$begingroup$
related: math.stackexchange.com/questions/538801/…
$endgroup$
– Henry
Jul 20 '15 at 7:41
$begingroup$
It may be helpful to look at the first examples. The number $64$ is both a square and a cube because $64=8^2$ and $64=4^3$. And for $729$ we have $729=27^2$ and $729=9^3$. The next two are $4096=64^2, 4096=16^3$ and $15625=125^2, 15625=25^3$. Can you see where you confused yourself?
$endgroup$
– Jeppe Stig Nielsen
Jul 20 '15 at 8:56
$begingroup$
It may be helpful to look at the first examples. The number $64$ is both a square and a cube because $64=8^2$ and $64=4^3$. And for $729$ we have $729=27^2$ and $729=9^3$. The next two are $4096=64^2, 4096=16^3$ and $15625=125^2, 15625=25^3$. Can you see where you confused yourself?
$endgroup$
– Jeppe Stig Nielsen
Jul 20 '15 at 8:56
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Every positive integer $>1$ can be factorized into a product of prime numbers. If the number is a perfect square, the exponents of these primes are even. And if the number is a perfect cube, the exponents are multiple of three.
If both things happen for the same number, the exponents are multiple of $2$ and $3$, that is, multiple of $6$, so the number is a perfect sixth power.
answered Jul 19 '15 at 23:41
ajotatxeajotatxe
53.8k23890
$endgroup$
add a comment |
$begingroup$
This is a consequence of unique prime factorization. Let $p$ be a prime dividing $n$. Since $n$ is a perfect square, the power of $p$ that divides $n$ is even. Similarly, since $n$ is a perfect cube, the power of $p$ that divides $n$ is divisible by 3. Thus, the power of $p$ that divides $n$ is divisible by $6$. So therefore if we write
$$n = p_1^{e_1}p_2^{e_2} ldots p_m^{e_m}$$
then each $e_i$ is divisible by $6$. Now you can conclude $n = a^6$ where
$$a = p_1^{e_1/6} ldots p_m^{e_m/6}.$$
By the way, if you know some abstract algebra, you might ask if we really need to use unique prime factorization; the answer is yes, you can construct a commutative ring where the theorem isn't true, e.g. $mathbb{C}[t^2, t^3]$.
answered Jul 19 '15 at 23:42
hunterhunter
14.6k22438
$endgroup$
add a comment |
$begingroup$
The title statement doesn't say to take the square and cube of a single number $x$. It says you take a square of something and a cube of something and the square and the cube are equal. The two "somethings" are not necessarily equal.
That is, you're not looking at $a = x^2$ and $b = x^3$,
you're looking at a single number $n$ such that $n = q^2 = r^3$
where $q$ and $r$ are integers.
The statement says that whenever you have such a number, there is
an integer $s$ such that $n = s^6$.
One possible step toward proving this is
to show that if $n = q^2$ (so $n$ is a square)
and if $q^3 = n$, then $q$ also is a square.
answered Jul 20 '15 at 3:24
David KDavid K
53.9k342116
$endgroup$
add a comment |
$begingroup$
The sequence of numbers that are both perfect squares and perfect cubes is:
$$
1, 64, 729, cdots, n^6
$$
for $ninmathbb{Z}^+$.
This can be proven using the identity $(b^m)^n=b^{mn}$.
$$
(n^3)^2=(n^2)^3=n^6
$$
answered Dec 6 '18 at 6:57
lukejanickelukejanicke
22719
$endgroup$
add a comment |
Your Answer
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4 Answers
4
active
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votes
4 Answers
4
active
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votes
$begingroup$
Every positive integer $>1$ can be factorized into a product of prime numbers. If the number is a perfect square, the exponents of these primes are even. And if the number is a perfect cube, the exponents are multiple of three.
If both things happen for the same number, the exponents are multiple of $2$ and $3$, that is, multiple of $6$, so the number is a perfect sixth power.
answered Jul 19 '15 at 23:41
ajotatxeajotatxe
53.8k23890
$endgroup$
add a comment |
$begingroup$
Every positive integer $>1$ can be factorized into a product of prime numbers. If the number is a perfect square, the exponents of these primes are even. And if the number is a perfect cube, the exponents are multiple of three.
If both things happen for the same number, the exponents are multiple of $2$ and $3$, that is, multiple of $6$, so the number is a perfect sixth power.
answered Jul 19 '15 at 23:41
ajotatxeajotatxe
53.8k23890
$endgroup$
add a comment |
$begingroup$
Every positive integer $>1$ can be factorized into a product of prime numbers. If the number is a perfect square, the exponents of these primes are even. And if the number is a perfect cube, the exponents are multiple of three.
If both things happen for the same number, the exponents are multiple of $2$ and $3$, that is, multiple of $6$, so the number is a perfect sixth power.
answered Jul 19 '15 at 23:41
ajotatxeajotatxe
53.8k23890
$endgroup$
Every positive integer $>1$ can be factorized into a product of prime numbers. If the number is a perfect square, the exponents of these primes are even. And if the number is a perfect cube, the exponents are multiple of three.
If both things happen for the same number, the exponents are multiple of $2$ and $3$, that is, multiple of $6$, so the number is a perfect sixth power.
answered Jul 19 '15 at 23:41
ajotatxeajotatxe
53.8k23890
answered Jul 19 '15 at 23:41
ajotatxeajotatxe
53.8k23890
answered Jul 19 '15 at 23:41
ajotatxeajotatxe
53.8k23890
answered Jul 19 '15 at 23:41
ajotatxeajotatxe
53.8k23890
53.8k23890
add a comment |
add a comment |
$begingroup$
This is a consequence of unique prime factorization. Let $p$ be a prime dividing $n$. Since $n$ is a perfect square, the power of $p$ that divides $n$ is even. Similarly, since $n$ is a perfect cube, the power of $p$ that divides $n$ is divisible by 3. Thus, the power of $p$ that divides $n$ is divisible by $6$. So therefore if we write
$$n = p_1^{e_1}p_2^{e_2} ldots p_m^{e_m}$$
then each $e_i$ is divisible by $6$. Now you can conclude $n = a^6$ where
$$a = p_1^{e_1/6} ldots p_m^{e_m/6}.$$
By the way, if you know some abstract algebra, you might ask if we really need to use unique prime factorization; the answer is yes, you can construct a commutative ring where the theorem isn't true, e.g. $mathbb{C}[t^2, t^3]$.
answered Jul 19 '15 at 23:42
hunterhunter
14.6k22438
$endgroup$
add a comment |
$begingroup$
This is a consequence of unique prime factorization. Let $p$ be a prime dividing $n$. Since $n$ is a perfect square, the power of $p$ that divides $n$ is even. Similarly, since $n$ is a perfect cube, the power of $p$ that divides $n$ is divisible by 3. Thus, the power of $p$ that divides $n$ is divisible by $6$. So therefore if we write
$$n = p_1^{e_1}p_2^{e_2} ldots p_m^{e_m}$$
then each $e_i$ is divisible by $6$. Now you can conclude $n = a^6$ where
$$a = p_1^{e_1/6} ldots p_m^{e_m/6}.$$
By the way, if you know some abstract algebra, you might ask if we really need to use unique prime factorization; the answer is yes, you can construct a commutative ring where the theorem isn't true, e.g. $mathbb{C}[t^2, t^3]$.
answered Jul 19 '15 at 23:42
hunterhunter
14.6k22438
$endgroup$
add a comment |
$begingroup$
This is a consequence of unique prime factorization. Let $p$ be a prime dividing $n$. Since $n$ is a perfect square, the power of $p$ that divides $n$ is even. Similarly, since $n$ is a perfect cube, the power of $p$ that divides $n$ is divisible by 3. Thus, the power of $p$ that divides $n$ is divisible by $6$. So therefore if we write
$$n = p_1^{e_1}p_2^{e_2} ldots p_m^{e_m}$$
then each $e_i$ is divisible by $6$. Now you can conclude $n = a^6$ where
$$a = p_1^{e_1/6} ldots p_m^{e_m/6}.$$
By the way, if you know some abstract algebra, you might ask if we really need to use unique prime factorization; the answer is yes, you can construct a commutative ring where the theorem isn't true, e.g. $mathbb{C}[t^2, t^3]$.
answered Jul 19 '15 at 23:42
hunterhunter
14.6k22438
$endgroup$
This is a consequence of unique prime factorization. Let $p$ be a prime dividing $n$. Since $n$ is a perfect square, the power of $p$ that divides $n$ is even. Similarly, since $n$ is a perfect cube, the power of $p$ that divides $n$ is divisible by 3. Thus, the power of $p$ that divides $n$ is divisible by $6$. So therefore if we write
$$n = p_1^{e_1}p_2^{e_2} ldots p_m^{e_m}$$
then each $e_i$ is divisible by $6$. Now you can conclude $n = a^6$ where
$$a = p_1^{e_1/6} ldots p_m^{e_m/6}.$$
By the way, if you know some abstract algebra, you might ask if we really need to use unique prime factorization; the answer is yes, you can construct a commutative ring where the theorem isn't true, e.g. $mathbb{C}[t^2, t^3]$.
answered Jul 19 '15 at 23:42
hunterhunter
14.6k22438
answered Jul 19 '15 at 23:42
hunterhunter
14.6k22438
answered Jul 19 '15 at 23:42
hunterhunter
14.6k22438
answered Jul 19 '15 at 23:42
hunterhunter
14.6k22438
14.6k22438
add a comment |
add a comment |
$begingroup$
The title statement doesn't say to take the square and cube of a single number $x$. It says you take a square of something and a cube of something and the square and the cube are equal. The two "somethings" are not necessarily equal.
That is, you're not looking at $a = x^2$ and $b = x^3$,
you're looking at a single number $n$ such that $n = q^2 = r^3$
where $q$ and $r$ are integers.
The statement says that whenever you have such a number, there is
an integer $s$ such that $n = s^6$.
One possible step toward proving this is
to show that if $n = q^2$ (so $n$ is a square)
and if $q^3 = n$, then $q$ also is a square.
answered Jul 20 '15 at 3:24
David KDavid K
53.9k342116
$endgroup$
add a comment |
$begingroup$
The title statement doesn't say to take the square and cube of a single number $x$. It says you take a square of something and a cube of something and the square and the cube are equal. The two "somethings" are not necessarily equal.
That is, you're not looking at $a = x^2$ and $b = x^3$,
you're looking at a single number $n$ such that $n = q^2 = r^3$
where $q$ and $r$ are integers.
The statement says that whenever you have such a number, there is
an integer $s$ such that $n = s^6$.
One possible step toward proving this is
to show that if $n = q^2$ (so $n$ is a square)
and if $q^3 = n$, then $q$ also is a square.
answered Jul 20 '15 at 3:24
David KDavid K
53.9k342116
$endgroup$
add a comment |
$begingroup$
The title statement doesn't say to take the square and cube of a single number $x$. It says you take a square of something and a cube of something and the square and the cube are equal. The two "somethings" are not necessarily equal.
That is, you're not looking at $a = x^2$ and $b = x^3$,
you're looking at a single number $n$ such that $n = q^2 = r^3$
where $q$ and $r$ are integers.
The statement says that whenever you have such a number, there is
an integer $s$ such that $n = s^6$.
One possible step toward proving this is
to show that if $n = q^2$ (so $n$ is a square)
and if $q^3 = n$, then $q$ also is a square.
answered Jul 20 '15 at 3:24
David KDavid K
53.9k342116
$endgroup$
The title statement doesn't say to take the square and cube of a single number $x$. It says you take a square of something and a cube of something and the square and the cube are equal. The two "somethings" are not necessarily equal.
That is, you're not looking at $a = x^2$ and $b = x^3$,
you're looking at a single number $n$ such that $n = q^2 = r^3$
where $q$ and $r$ are integers.
The statement says that whenever you have such a number, there is
an integer $s$ such that $n = s^6$.
One possible step toward proving this is
to show that if $n = q^2$ (so $n$ is a square)
and if $q^3 = n$, then $q$ also is a square.
answered Jul 20 '15 at 3:24
David KDavid K
53.9k342116
answered Jul 20 '15 at 3:24
David KDavid K
53.9k342116
answered Jul 20 '15 at 3:24
David KDavid K
53.9k342116
answered Jul 20 '15 at 3:24
David KDavid K
53.9k342116
53.9k342116
add a comment |
add a comment |
$begingroup$
The sequence of numbers that are both perfect squares and perfect cubes is:
$$
1, 64, 729, cdots, n^6
$$
for $ninmathbb{Z}^+$.
This can be proven using the identity $(b^m)^n=b^{mn}$.
$$
(n^3)^2=(n^2)^3=n^6
$$
answered Dec 6 '18 at 6:57
lukejanickelukejanicke
22719
$endgroup$
add a comment |
$begingroup$
The sequence of numbers that are both perfect squares and perfect cubes is:
$$
1, 64, 729, cdots, n^6
$$
for $ninmathbb{Z}^+$.
This can be proven using the identity $(b^m)^n=b^{mn}$.
$$
(n^3)^2=(n^2)^3=n^6
$$
answered Dec 6 '18 at 6:57
lukejanickelukejanicke
22719
$endgroup$
add a comment |
$begingroup$
The sequence of numbers that are both perfect squares and perfect cubes is:
$$
1, 64, 729, cdots, n^6
$$
for $ninmathbb{Z}^+$.
This can be proven using the identity $(b^m)^n=b^{mn}$.
$$
(n^3)^2=(n^2)^3=n^6
$$
answered Dec 6 '18 at 6:57
lukejanickelukejanicke
22719
$endgroup$
The sequence of numbers that are both perfect squares and perfect cubes is:
$$
1, 64, 729, cdots, n^6
$$
for $ninmathbb{Z}^+$.
This can be proven using the identity $(b^m)^n=b^{mn}$.
$$
(n^3)^2=(n^2)^3=n^6
$$
answered Dec 6 '18 at 6:57
lukejanickelukejanicke
22719
answered Dec 6 '18 at 6:57
lukejanickelukejanicke
22719
answered Dec 6 '18 at 6:57
lukejanickelukejanicke
22719
answered Dec 6 '18 at 6:57
lukejanickelukejanicke
22719
22719
add a comment |
add a comment |
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$begingroup$
You can show that if $n$ is a perfect cube, then so is $sqrt n$.
$endgroup$
– MJD
Jul 20 '15 at 0:04
$begingroup$
related: math.stackexchange.com/questions/538801/…
$endgroup$
– Henry
Jul 20 '15 at 7:41
$begingroup$
It may be helpful to look at the first examples. The number $64$ is both a square and a cube because $64=8^2$ and $64=4^3$. And for $729$ we have $729=27^2$ and $729=9^3$. The next two are $4096=64^2, 4096=16^3$ and $15625=125^2, 15625=25^3$. Can you see where you confused yourself?
$endgroup$
– Jeppe Stig Nielsen
Jul 20 '15 at 8:56