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
53.6k23890
$endgroup$
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
14.3k22438
$endgroup$
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
52.9k340115
$endgroup$
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
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1367032%2fa-natural-number-that-is-both-a-perfect-square-and-a-perfect-cube-is-a-perfect-s%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
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
53.6k23890
$endgroup$
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
53.6k23890
$endgroup$
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
53.6k23890
$endgroup$
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
53.6k23890
answered Jul 19 '15 at 23:41
ajotatxeajotatxe
53.6k23890
answered Jul 19 '15 at 23:41
ajotatxeajotatxe
53.6k23890
answered Jul 19 '15 at 23:41
ajotatxeajotatxe
53.6k23890
53.6k23890
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
14.3k22438
$endgroup$
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
14.3k22438
$endgroup$
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
14.3k22438
$endgroup$
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
14.3k22438
answered Jul 19 '15 at 23:42
hunterhunter
14.3k22438
answered Jul 19 '15 at 23:42
hunterhunter
14.3k22438
answered Jul 19 '15 at 23:42
hunterhunter
14.3k22438
14.3k22438
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
52.9k340115
$endgroup$
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
52.9k340115
$endgroup$
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
52.9k340115
$endgroup$
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
52.9k340115
answered Jul 20 '15 at 3:24
David KDavid K
52.9k340115
answered Jul 20 '15 at 3:24
David KDavid K
52.9k340115
answered Jul 20 '15 at 3:24
David KDavid K
52.9k340115
52.9k340115
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 |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1367032%2fa-natural-number-that-is-both-a-perfect-square-and-a-perfect-cube-is-a-perfect-s%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
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