Mixing
Exercise on Fundamentals of Divisibility
$begingroup$
(This question is on page 236 of Falko Lorenz's Algebra Volume 1: Fields and Galois Theory, exercise 4.5)
Prove that $Y^2 = X^3 - 2$ has exactly one solution in the natural numbers.
Hint: use the fact that $mathbb{Z}[sqrt{-2}]$ is a Euclidean domain, and that in any unique factorization domain $R$, if $alpha_1,dots ,alpha_n$ are pairwise rel. prime in $R$ and their product is an $m$-th power in $R$, each $alpha_i$ is associated to an $m$-th power in $R$.
Here's what I know. Since $mathbb{Z}[sqrt{-2}]$ is a Euclidean domain, then it is also a unique factorization domain. As well, $X^3=Y^2+2=Y^2-sqrt{-2}^2=(Y-sqrt{-2})(Y+sqrt{-2})$. Clues for the clueless?
abstract-algebra number-theory algebraic-number-theory diophantine-equations
edited Feb 3 at 6:22
darij grinberg
11.5k33168
asked Oct 3 '12 at 3:39
ChrisChris
256110
$endgroup$
add a comment |
$begingroup$
(This question is on page 236 of Falko Lorenz's Algebra Volume 1: Fields and Galois Theory, exercise 4.5)
Prove that $Y^2 = X^3 - 2$ has exactly one solution in the natural numbers.
Hint: use the fact that $mathbb{Z}[sqrt{-2}]$ is a Euclidean domain, and that in any unique factorization domain $R$, if $alpha_1,dots ,alpha_n$ are pairwise rel. prime in $R$ and their product is an $m$-th power in $R$, each $alpha_i$ is associated to an $m$-th power in $R$.
Here's what I know. Since $mathbb{Z}[sqrt{-2}]$ is a Euclidean domain, then it is also a unique factorization domain. As well, $X^3=Y^2+2=Y^2-sqrt{-2}^2=(Y-sqrt{-2})(Y+sqrt{-2})$. Clues for the clueless?
abstract-algebra number-theory algebraic-number-theory diophantine-equations
edited Feb 3 at 6:22
darij grinberg
11.5k33168
asked Oct 3 '12 at 3:39
ChrisChris
256110
$endgroup$
add a comment |
$begingroup$
(This question is on page 236 of Falko Lorenz's Algebra Volume 1: Fields and Galois Theory, exercise 4.5)
Prove that $Y^2 = X^3 - 2$ has exactly one solution in the natural numbers.
Hint: use the fact that $mathbb{Z}[sqrt{-2}]$ is a Euclidean domain, and that in any unique factorization domain $R$, if $alpha_1,dots ,alpha_n$ are pairwise rel. prime in $R$ and their product is an $m$-th power in $R$, each $alpha_i$ is associated to an $m$-th power in $R$.
Here's what I know. Since $mathbb{Z}[sqrt{-2}]$ is a Euclidean domain, then it is also a unique factorization domain. As well, $X^3=Y^2+2=Y^2-sqrt{-2}^2=(Y-sqrt{-2})(Y+sqrt{-2})$. Clues for the clueless?
abstract-algebra number-theory algebraic-number-theory diophantine-equations
edited Feb 3 at 6:22
darij grinberg
11.5k33168
asked Oct 3 '12 at 3:39
ChrisChris
256110
$endgroup$
(This question is on page 236 of Falko Lorenz's Algebra Volume 1: Fields and Galois Theory, exercise 4.5)
Prove that $Y^2 = X^3 - 2$ has exactly one solution in the natural numbers.
Hint: use the fact that $mathbb{Z}[sqrt{-2}]$ is a Euclidean domain, and that in any unique factorization domain $R$, if $alpha_1,dots ,alpha_n$ are pairwise rel. prime in $R$ and their product is an $m$-th power in $R$, each $alpha_i$ is associated to an $m$-th power in $R$.
Here's what I know. Since $mathbb{Z}[sqrt{-2}]$ is a Euclidean domain, then it is also a unique factorization domain. As well, $X^3=Y^2+2=Y^2-sqrt{-2}^2=(Y-sqrt{-2})(Y+sqrt{-2})$. Clues for the clueless?
abstract-algebra number-theory algebraic-number-theory diophantine-equations
abstract-algebra number-theory algebraic-number-theory diophantine-equations
edited Feb 3 at 6:22
darij grinberg
11.5k33168
asked Oct 3 '12 at 3:39
ChrisChris
256110
edited Feb 3 at 6:22
darij grinberg
11.5k33168
asked Oct 3 '12 at 3:39
ChrisChris
256110
edited Feb 3 at 6:22
darij grinberg
11.5k33168
edited Feb 3 at 6:22
darij grinberg
11.5k33168
edited Feb 3 at 6:22
darij grinberg
11.5k33168
11.5k33168
asked Oct 3 '12 at 3:39
ChrisChris
256110
asked Oct 3 '12 at 3:39
ChrisChris
256110
asked Oct 3 '12 at 3:39
ChrisChris
256110
256110
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
An outline: The units are $pm 1$. Show that $X$ and $Y$ must be odd. Then show $Y+sqrt{-2}$ and $Y-sqrt{-2}$ are relatively prime. So each is a cube.
Let $Y+sqrt{-2}=(a+bsqrt{-2})^3$. Expand. Compute in particular the coefficient of $sqrt{-2}$. This must be $1$. That will tell you something very important about $b$. But then you will know what $a$ must be, more or less. That will give the only solutions, and the only positive one.
answered Oct 3 '12 at 3:59
André NicolasAndré Nicolas
455k36432822
$endgroup$
$begingroup$
Ok I got that $1=3a^2b-2b^3$. I'm confused about what this should be telling me?
$endgroup$
– Chris
Oct 3 '12 at 5:05
$begingroup$
That $b$ divides $1$!
$endgroup$
– André Nicolas
Oct 3 '12 at 5:06
$begingroup$
Oh grief, thank you : )
$endgroup$
– Chris
Oct 3 '12 at 5:07
$begingroup$
Well I got it, $X=3$ and $Y=5$!
$endgroup$
– Chris
Oct 3 '12 at 5:13
$begingroup$
OK, but remember there are (deliberately) little missing steps in my outline. Little proofs to fill in. Like what are the units? (Norm argument) Any solutions must be odd (easy). $X+sqrt{-2}$ and its conjugate are relatively prime in the ring. For that I used fact that solutions are odd.
$endgroup$
– André Nicolas
Oct 3 '12 at 5:20
|
show 2 more comments
Your Answer
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%2f206342%2fexercise-on-fundamentals-of-divisibility%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
An outline: The units are $pm 1$. Show that $X$ and $Y$ must be odd. Then show $Y+sqrt{-2}$ and $Y-sqrt{-2}$ are relatively prime. So each is a cube.
Let $Y+sqrt{-2}=(a+bsqrt{-2})^3$. Expand. Compute in particular the coefficient of $sqrt{-2}$. This must be $1$. That will tell you something very important about $b$. But then you will know what $a$ must be, more or less. That will give the only solutions, and the only positive one.
answered Oct 3 '12 at 3:59
André NicolasAndré Nicolas
455k36432822
$endgroup$
$begingroup$
Ok I got that $1=3a^2b-2b^3$. I'm confused about what this should be telling me?
$endgroup$
– Chris
Oct 3 '12 at 5:05
$begingroup$
That $b$ divides $1$!
$endgroup$
– André Nicolas
Oct 3 '12 at 5:06
$begingroup$
Oh grief, thank you : )
$endgroup$
– Chris
Oct 3 '12 at 5:07
$begingroup$
Well I got it, $X=3$ and $Y=5$!
$endgroup$
– Chris
Oct 3 '12 at 5:13
$begingroup$
OK, but remember there are (deliberately) little missing steps in my outline. Little proofs to fill in. Like what are the units? (Norm argument) Any solutions must be odd (easy). $X+sqrt{-2}$ and its conjugate are relatively prime in the ring. For that I used fact that solutions are odd.
$endgroup$
– André Nicolas
Oct 3 '12 at 5:20
|
show 2 more comments
$begingroup$
An outline: The units are $pm 1$. Show that $X$ and $Y$ must be odd. Then show $Y+sqrt{-2}$ and $Y-sqrt{-2}$ are relatively prime. So each is a cube.
Let $Y+sqrt{-2}=(a+bsqrt{-2})^3$. Expand. Compute in particular the coefficient of $sqrt{-2}$. This must be $1$. That will tell you something very important about $b$. But then you will know what $a$ must be, more or less. That will give the only solutions, and the only positive one.
answered Oct 3 '12 at 3:59
André NicolasAndré Nicolas
455k36432822
$endgroup$
$begingroup$
Ok I got that $1=3a^2b-2b^3$. I'm confused about what this should be telling me?
$endgroup$
– Chris
Oct 3 '12 at 5:05
$begingroup$
That $b$ divides $1$!
$endgroup$
– André Nicolas
Oct 3 '12 at 5:06
$begingroup$
Oh grief, thank you : )
$endgroup$
– Chris
Oct 3 '12 at 5:07
$begingroup$
Well I got it, $X=3$ and $Y=5$!
$endgroup$
– Chris
Oct 3 '12 at 5:13
$begingroup$
OK, but remember there are (deliberately) little missing steps in my outline. Little proofs to fill in. Like what are the units? (Norm argument) Any solutions must be odd (easy). $X+sqrt{-2}$ and its conjugate are relatively prime in the ring. For that I used fact that solutions are odd.
$endgroup$
– André Nicolas
Oct 3 '12 at 5:20
|
show 2 more comments
$begingroup$
An outline: The units are $pm 1$. Show that $X$ and $Y$ must be odd. Then show $Y+sqrt{-2}$ and $Y-sqrt{-2}$ are relatively prime. So each is a cube.
Let $Y+sqrt{-2}=(a+bsqrt{-2})^3$. Expand. Compute in particular the coefficient of $sqrt{-2}$. This must be $1$. That will tell you something very important about $b$. But then you will know what $a$ must be, more or less. That will give the only solutions, and the only positive one.
answered Oct 3 '12 at 3:59
André NicolasAndré Nicolas
455k36432822
$endgroup$
An outline: The units are $pm 1$. Show that $X$ and $Y$ must be odd. Then show $Y+sqrt{-2}$ and $Y-sqrt{-2}$ are relatively prime. So each is a cube.
Let $Y+sqrt{-2}=(a+bsqrt{-2})^3$. Expand. Compute in particular the coefficient of $sqrt{-2}$. This must be $1$. That will tell you something very important about $b$. But then you will know what $a$ must be, more or less. That will give the only solutions, and the only positive one.
answered Oct 3 '12 at 3:59
André NicolasAndré Nicolas
455k36432822
answered Oct 3 '12 at 3:59
André NicolasAndré Nicolas
455k36432822
answered Oct 3 '12 at 3:59
André NicolasAndré Nicolas
455k36432822
answered Oct 3 '12 at 3:59
André NicolasAndré Nicolas
455k36432822
455k36432822
$begingroup$
Ok I got that $1=3a^2b-2b^3$. I'm confused about what this should be telling me?
$endgroup$
– Chris
Oct 3 '12 at 5:05
$begingroup$
That $b$ divides $1$!
$endgroup$
– André Nicolas
Oct 3 '12 at 5:06
$begingroup$
Oh grief, thank you : )
$endgroup$
– Chris
Oct 3 '12 at 5:07
$begingroup$
Well I got it, $X=3$ and $Y=5$!
$endgroup$
– Chris
Oct 3 '12 at 5:13
$begingroup$
OK, but remember there are (deliberately) little missing steps in my outline. Little proofs to fill in. Like what are the units? (Norm argument) Any solutions must be odd (easy). $X+sqrt{-2}$ and its conjugate are relatively prime in the ring. For that I used fact that solutions are odd.
$endgroup$
– André Nicolas
Oct 3 '12 at 5:20
|
show 2 more comments
$begingroup$
Ok I got that $1=3a^2b-2b^3$. I'm confused about what this should be telling me?
$endgroup$
– Chris
Oct 3 '12 at 5:05
$begingroup$
That $b$ divides $1$!
$endgroup$
– André Nicolas
Oct 3 '12 at 5:06
$begingroup$
Oh grief, thank you : )
$endgroup$
– Chris
Oct 3 '12 at 5:07
$begingroup$
Well I got it, $X=3$ and $Y=5$!
$endgroup$
– Chris
Oct 3 '12 at 5:13
$begingroup$
OK, but remember there are (deliberately) little missing steps in my outline. Little proofs to fill in. Like what are the units? (Norm argument) Any solutions must be odd (easy). $X+sqrt{-2}$ and its conjugate are relatively prime in the ring. For that I used fact that solutions are odd.
$endgroup$
– André Nicolas
Oct 3 '12 at 5:20
$begingroup$
Ok I got that $1=3a^2b-2b^3$. I'm confused about what this should be telling me?
$endgroup$
– Chris
Oct 3 '12 at 5:05
$begingroup$
Ok I got that $1=3a^2b-2b^3$. I'm confused about what this should be telling me?
$endgroup$
– Chris
Oct 3 '12 at 5:05
$begingroup$
That $b$ divides $1$!
$endgroup$
– André Nicolas
Oct 3 '12 at 5:06
$begingroup$
That $b$ divides $1$!
$endgroup$
– André Nicolas
Oct 3 '12 at 5:06
$begingroup$
Oh grief, thank you : )
$endgroup$
– Chris
Oct 3 '12 at 5:07
$begingroup$
Oh grief, thank you : )
$endgroup$
– Chris
Oct 3 '12 at 5:07
$begingroup$
Well I got it, $X=3$ and $Y=5$!
$endgroup$
– Chris
Oct 3 '12 at 5:13
$begingroup$
Well I got it, $X=3$ and $Y=5$!
$endgroup$
– Chris
Oct 3 '12 at 5:13
$begingroup$
OK, but remember there are (deliberately) little missing steps in my outline. Little proofs to fill in. Like what are the units? (Norm argument) Any solutions must be odd (easy). $X+sqrt{-2}$ and its conjugate are relatively prime in the ring. For that I used fact that solutions are odd.
$endgroup$
– André Nicolas
Oct 3 '12 at 5:20
$begingroup$
OK, but remember there are (deliberately) little missing steps in my outline. Little proofs to fill in. Like what are the units? (Norm argument) Any solutions must be odd (easy). $X+sqrt{-2}$ and its conjugate are relatively prime in the ring. For that I used fact that solutions are odd.
$endgroup$
– André Nicolas
Oct 3 '12 at 5:20
|
show 2 more comments
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%2f206342%2fexercise-on-fundamentals-of-divisibility%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
