Which rationals can be written as the sum of two rational squares?
$begingroup$
Which rational numbers can be written as the sum of two rational squares?
That is, for which rational numbers $a$, are there rational numbers $x$ and $y$ such that $a = x^2 + y^2$.
It is a famous theorem that if an integer can be written as the sum of two rational squares then it can be written as the sum of two integral squares, and then the solution is the famous one by Fermat, but I didn't find anything about the general case.
number-theory
asked Sep 21 '15 at 13:50
lhflhf
168k11172404
$endgroup$
|
show 3 more comments
$begingroup$
Which rational numbers can be written as the sum of two rational squares?
That is, for which rational numbers $a$, are there rational numbers $x$ and $y$ such that $a = x^2 + y^2$.
It is a famous theorem that if an integer can be written as the sum of two rational squares then it can be written as the sum of two integral squares, and then the solution is the famous one by Fermat, but I didn't find anything about the general case.
number-theory
asked Sep 21 '15 at 13:50
lhflhf
168k11172404
$endgroup$
$begingroup$
math.stackexchange.com/questions/1200794/… mentions a solution but gives no references.
$endgroup$
– lhf
Sep 21 '15 at 13:53
$begingroup$
the last part of that question states the result asked for here.
$endgroup$
– Ross Millikan
Sep 21 '15 at 13:55
$begingroup$
What about $x=frac35,y=frac45$?
$endgroup$
– abiessu
Sep 21 '15 at 13:56
2
$begingroup$
Basically an Pythagorean triple?
$endgroup$
– Gummy bears
Sep 21 '15 at 14:02
4
$begingroup$
can you not clear the denominators and apply the standard results?
$endgroup$
– Matt B
Sep 21 '15 at 14:19
|
show 3 more comments
$begingroup$
Which rational numbers can be written as the sum of two rational squares?
That is, for which rational numbers $a$, are there rational numbers $x$ and $y$ such that $a = x^2 + y^2$.
It is a famous theorem that if an integer can be written as the sum of two rational squares then it can be written as the sum of two integral squares, and then the solution is the famous one by Fermat, but I didn't find anything about the general case.
number-theory
asked Sep 21 '15 at 13:50
lhflhf
168k11172404
$endgroup$
Which rational numbers can be written as the sum of two rational squares?
That is, for which rational numbers $a$, are there rational numbers $x$ and $y$ such that $a = x^2 + y^2$.
It is a famous theorem that if an integer can be written as the sum of two rational squares then it can be written as the sum of two integral squares, and then the solution is the famous one by Fermat, but I didn't find anything about the general case.
number-theory
number-theory
asked Sep 21 '15 at 13:50
lhflhf
168k11172404
asked Sep 21 '15 at 13:50
lhflhf
168k11172404
edited Sep 21 '15 at 14:00
lhf
asked Sep 21 '15 at 13:50
lhflhf
168k11172404
asked Sep 21 '15 at 13:50
lhflhf
168k11172404
asked Sep 21 '15 at 13:50
lhflhf
168k11172404
168k11172404
$begingroup$
math.stackexchange.com/questions/1200794/… mentions a solution but gives no references.
$endgroup$
– lhf
Sep 21 '15 at 13:53
$begingroup$
the last part of that question states the result asked for here.
$endgroup$
– Ross Millikan
Sep 21 '15 at 13:55
$begingroup$
What about $x=frac35,y=frac45$?
$endgroup$
– abiessu
Sep 21 '15 at 13:56
2
$begingroup$
Basically an Pythagorean triple?
$endgroup$
– Gummy bears
Sep 21 '15 at 14:02
4
$begingroup$
can you not clear the denominators and apply the standard results?
$endgroup$
– Matt B
Sep 21 '15 at 14:19
|
show 3 more comments
$begingroup$
math.stackexchange.com/questions/1200794/… mentions a solution but gives no references.
$endgroup$
– lhf
Sep 21 '15 at 13:53
$begingroup$
the last part of that question states the result asked for here.
$endgroup$
– Ross Millikan
Sep 21 '15 at 13:55
$begingroup$
What about $x=frac35,y=frac45$?
$endgroup$
– abiessu
Sep 21 '15 at 13:56
2
$begingroup$
Basically an Pythagorean triple?
$endgroup$
– Gummy bears
Sep 21 '15 at 14:02
4
$begingroup$
can you not clear the denominators and apply the standard results?
$endgroup$
– Matt B
Sep 21 '15 at 14:19
$begingroup$
math.stackexchange.com/questions/1200794/… mentions a solution but gives no references.
$endgroup$
– lhf
Sep 21 '15 at 13:53
$begingroup$
math.stackexchange.com/questions/1200794/… mentions a solution but gives no references.
$endgroup$
– lhf
Sep 21 '15 at 13:53
$begingroup$
the last part of that question states the result asked for here.
$endgroup$
– Ross Millikan
Sep 21 '15 at 13:55
$begingroup$
the last part of that question states the result asked for here.
$endgroup$
– Ross Millikan
Sep 21 '15 at 13:55
$begingroup$
What about $x=frac35,y=frac45$?
$endgroup$
– abiessu
Sep 21 '15 at 13:56
$begingroup$
What about $x=frac35,y=frac45$?
$endgroup$
– abiessu
Sep 21 '15 at 13:56
2
2
$begingroup$
Basically an Pythagorean triple?
$endgroup$
– Gummy bears
Sep 21 '15 at 14:02
$begingroup$
Basically an Pythagorean triple?
$endgroup$
– Gummy bears
Sep 21 '15 at 14:02
4
4
$begingroup$
can you not clear the denominators and apply the standard results?
$endgroup$
– Matt B
Sep 21 '15 at 14:19
$begingroup$
can you not clear the denominators and apply the standard results?
$endgroup$
– Matt B
Sep 21 '15 at 14:19
|
show 3 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Let $r=frac{p}{q}$, where $p$ and $q$ are integers. We will show that $r$ can be written as the sum of two rational squares if and only if $pq$ can be written as the sum of the squares of two integers.
Equivalently, if $pne 0$, then $frac{p}{q}$ is a sum of the squares of two rationals if and only if every prime divisor of $pq$ of the form $4k+3$ occurs to an even power.
Proof: If $pq$ is the sum of the squares of two integers, it is clear that $r$ is the sum of the squares of two rationals.
For the other direction, suppose that $r$ can be written as the sum of the squares of two rationals. Without loss of generality we may assume that the rationals are $frac{a}{c}$ and $frac{b}{c}$ for some integers $a,b,c$. Then
$$frac{a^2+b^2}{c^2}=frac{pq}{q^2}.$$
So $c^2pq$ is a sum of two squares. It follows that every prime of the form $4k+3$ occurs to an even degree in the prime power factorization of $c^2pq$, and hence of $pq$. It follows that $pq$ is a sum of two squares.
answered Sep 21 '15 at 14:23
André NicolasAndré Nicolas
455k36432821
$endgroup$
1
$begingroup$
The first observation is the clever trick!
$endgroup$
– lhf
Sep 21 '15 at 14:34
add a comment |
$begingroup$
This can be translated in a question about Hilbert symbols (or quaternion algebras). In fact you can check easily that a rational $a$ can be written as a sum of two squares if and only if the Hilbert symbol $(a,-1)_{mathbb Q}$ is trivial. Now write $a=p/q$ and notice that $(p/q,-1)=(pq,-1)$. You can forget about all squares in the factorization of $pq$, so you can assume that $p$ and $q$ are both squarefree. Now look at the prime factorization of $p$ and $q$ and use the fact that $(x,-1)(y,-1)=(xy,-1)$ for all $x,yinmathbb Q$, toghether with the fact that for a prime $l$ you have that $(l,-1)$ is trivial iff $lequiv 1,2bmod 4$ while if $lequiv 3bmod 4$ then $(l,-1)$ ramifies at $2$ and $l$ to get your answer: $a$ can be written as a sum of two squares iff it is non-negative and is of the form $b^2frac{p}{q}$ where $binmathbb Q$ and $p,q$ are coprime integers divided only by primes $equiv 1,2bmod 4$.
answered Sep 21 '15 at 14:19
FerraFerra
3,82111024
$endgroup$
add a comment |
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%2f1445058%2fwhich-rationals-can-be-written-as-the-sum-of-two-rational-squares%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $r=frac{p}{q}$, where $p$ and $q$ are integers. We will show that $r$ can be written as the sum of two rational squares if and only if $pq$ can be written as the sum of the squares of two integers.
Equivalently, if $pne 0$, then $frac{p}{q}$ is a sum of the squares of two rationals if and only if every prime divisor of $pq$ of the form $4k+3$ occurs to an even power.
Proof: If $pq$ is the sum of the squares of two integers, it is clear that $r$ is the sum of the squares of two rationals.
For the other direction, suppose that $r$ can be written as the sum of the squares of two rationals. Without loss of generality we may assume that the rationals are $frac{a}{c}$ and $frac{b}{c}$ for some integers $a,b,c$. Then
$$frac{a^2+b^2}{c^2}=frac{pq}{q^2}.$$
So $c^2pq$ is a sum of two squares. It follows that every prime of the form $4k+3$ occurs to an even degree in the prime power factorization of $c^2pq$, and hence of $pq$. It follows that $pq$ is a sum of two squares.
answered Sep 21 '15 at 14:23
André NicolasAndré Nicolas
455k36432821
$endgroup$
1
$begingroup$
The first observation is the clever trick!
$endgroup$
– lhf
Sep 21 '15 at 14:34
add a comment |
$begingroup$
Let $r=frac{p}{q}$, where $p$ and $q$ are integers. We will show that $r$ can be written as the sum of two rational squares if and only if $pq$ can be written as the sum of the squares of two integers.
Equivalently, if $pne 0$, then $frac{p}{q}$ is a sum of the squares of two rationals if and only if every prime divisor of $pq$ of the form $4k+3$ occurs to an even power.
Proof: If $pq$ is the sum of the squares of two integers, it is clear that $r$ is the sum of the squares of two rationals.
For the other direction, suppose that $r$ can be written as the sum of the squares of two rationals. Without loss of generality we may assume that the rationals are $frac{a}{c}$ and $frac{b}{c}$ for some integers $a,b,c$. Then
$$frac{a^2+b^2}{c^2}=frac{pq}{q^2}.$$
So $c^2pq$ is a sum of two squares. It follows that every prime of the form $4k+3$ occurs to an even degree in the prime power factorization of $c^2pq$, and hence of $pq$. It follows that $pq$ is a sum of two squares.
answered Sep 21 '15 at 14:23
André NicolasAndré Nicolas
455k36432821
$endgroup$
1
$begingroup$
The first observation is the clever trick!
$endgroup$
– lhf
Sep 21 '15 at 14:34
add a comment |
$begingroup$
Let $r=frac{p}{q}$, where $p$ and $q$ are integers. We will show that $r$ can be written as the sum of two rational squares if and only if $pq$ can be written as the sum of the squares of two integers.
Equivalently, if $pne 0$, then $frac{p}{q}$ is a sum of the squares of two rationals if and only if every prime divisor of $pq$ of the form $4k+3$ occurs to an even power.
Proof: If $pq$ is the sum of the squares of two integers, it is clear that $r$ is the sum of the squares of two rationals.
For the other direction, suppose that $r$ can be written as the sum of the squares of two rationals. Without loss of generality we may assume that the rationals are $frac{a}{c}$ and $frac{b}{c}$ for some integers $a,b,c$. Then
$$frac{a^2+b^2}{c^2}=frac{pq}{q^2}.$$
So $c^2pq$ is a sum of two squares. It follows that every prime of the form $4k+3$ occurs to an even degree in the prime power factorization of $c^2pq$, and hence of $pq$. It follows that $pq$ is a sum of two squares.
answered Sep 21 '15 at 14:23
André NicolasAndré Nicolas
455k36432821
$endgroup$
Let $r=frac{p}{q}$, where $p$ and $q$ are integers. We will show that $r$ can be written as the sum of two rational squares if and only if $pq$ can be written as the sum of the squares of two integers.
Equivalently, if $pne 0$, then $frac{p}{q}$ is a sum of the squares of two rationals if and only if every prime divisor of $pq$ of the form $4k+3$ occurs to an even power.
Proof: If $pq$ is the sum of the squares of two integers, it is clear that $r$ is the sum of the squares of two rationals.
For the other direction, suppose that $r$ can be written as the sum of the squares of two rationals. Without loss of generality we may assume that the rationals are $frac{a}{c}$ and $frac{b}{c}$ for some integers $a,b,c$. Then
$$frac{a^2+b^2}{c^2}=frac{pq}{q^2}.$$
So $c^2pq$ is a sum of two squares. It follows that every prime of the form $4k+3$ occurs to an even degree in the prime power factorization of $c^2pq$, and hence of $pq$. It follows that $pq$ is a sum of two squares.
answered Sep 21 '15 at 14:23
André NicolasAndré Nicolas
455k36432821
edited Sep 21 '15 at 14:33
answered Sep 21 '15 at 14:23
André NicolasAndré Nicolas
455k36432821
answered Sep 21 '15 at 14:23
André NicolasAndré Nicolas
455k36432821
answered Sep 21 '15 at 14:23
André NicolasAndré Nicolas
455k36432821
455k36432821
1
$begingroup$
The first observation is the clever trick!
$endgroup$
– lhf
Sep 21 '15 at 14:34
add a comment |
1
$begingroup$
The first observation is the clever trick!
$endgroup$
– lhf
Sep 21 '15 at 14:34
1
1
$begingroup$
The first observation is the clever trick!
$endgroup$
– lhf
Sep 21 '15 at 14:34
$begingroup$
The first observation is the clever trick!
$endgroup$
– lhf
Sep 21 '15 at 14:34
add a comment |
$begingroup$
This can be translated in a question about Hilbert symbols (or quaternion algebras). In fact you can check easily that a rational $a$ can be written as a sum of two squares if and only if the Hilbert symbol $(a,-1)_{mathbb Q}$ is trivial. Now write $a=p/q$ and notice that $(p/q,-1)=(pq,-1)$. You can forget about all squares in the factorization of $pq$, so you can assume that $p$ and $q$ are both squarefree. Now look at the prime factorization of $p$ and $q$ and use the fact that $(x,-1)(y,-1)=(xy,-1)$ for all $x,yinmathbb Q$, toghether with the fact that for a prime $l$ you have that $(l,-1)$ is trivial iff $lequiv 1,2bmod 4$ while if $lequiv 3bmod 4$ then $(l,-1)$ ramifies at $2$ and $l$ to get your answer: $a$ can be written as a sum of two squares iff it is non-negative and is of the form $b^2frac{p}{q}$ where $binmathbb Q$ and $p,q$ are coprime integers divided only by primes $equiv 1,2bmod 4$.
answered Sep 21 '15 at 14:19
FerraFerra
3,82111024
$endgroup$
add a comment |
$begingroup$
This can be translated in a question about Hilbert symbols (or quaternion algebras). In fact you can check easily that a rational $a$ can be written as a sum of two squares if and only if the Hilbert symbol $(a,-1)_{mathbb Q}$ is trivial. Now write $a=p/q$ and notice that $(p/q,-1)=(pq,-1)$. You can forget about all squares in the factorization of $pq$, so you can assume that $p$ and $q$ are both squarefree. Now look at the prime factorization of $p$ and $q$ and use the fact that $(x,-1)(y,-1)=(xy,-1)$ for all $x,yinmathbb Q$, toghether with the fact that for a prime $l$ you have that $(l,-1)$ is trivial iff $lequiv 1,2bmod 4$ while if $lequiv 3bmod 4$ then $(l,-1)$ ramifies at $2$ and $l$ to get your answer: $a$ can be written as a sum of two squares iff it is non-negative and is of the form $b^2frac{p}{q}$ where $binmathbb Q$ and $p,q$ are coprime integers divided only by primes $equiv 1,2bmod 4$.
answered Sep 21 '15 at 14:19
FerraFerra
3,82111024
$endgroup$
add a comment |
$begingroup$
This can be translated in a question about Hilbert symbols (or quaternion algebras). In fact you can check easily that a rational $a$ can be written as a sum of two squares if and only if the Hilbert symbol $(a,-1)_{mathbb Q}$ is trivial. Now write $a=p/q$ and notice that $(p/q,-1)=(pq,-1)$. You can forget about all squares in the factorization of $pq$, so you can assume that $p$ and $q$ are both squarefree. Now look at the prime factorization of $p$ and $q$ and use the fact that $(x,-1)(y,-1)=(xy,-1)$ for all $x,yinmathbb Q$, toghether with the fact that for a prime $l$ you have that $(l,-1)$ is trivial iff $lequiv 1,2bmod 4$ while if $lequiv 3bmod 4$ then $(l,-1)$ ramifies at $2$ and $l$ to get your answer: $a$ can be written as a sum of two squares iff it is non-negative and is of the form $b^2frac{p}{q}$ where $binmathbb Q$ and $p,q$ are coprime integers divided only by primes $equiv 1,2bmod 4$.
answered Sep 21 '15 at 14:19
FerraFerra
3,82111024
$endgroup$
This can be translated in a question about Hilbert symbols (or quaternion algebras). In fact you can check easily that a rational $a$ can be written as a sum of two squares if and only if the Hilbert symbol $(a,-1)_{mathbb Q}$ is trivial. Now write $a=p/q$ and notice that $(p/q,-1)=(pq,-1)$. You can forget about all squares in the factorization of $pq$, so you can assume that $p$ and $q$ are both squarefree. Now look at the prime factorization of $p$ and $q$ and use the fact that $(x,-1)(y,-1)=(xy,-1)$ for all $x,yinmathbb Q$, toghether with the fact that for a prime $l$ you have that $(l,-1)$ is trivial iff $lequiv 1,2bmod 4$ while if $lequiv 3bmod 4$ then $(l,-1)$ ramifies at $2$ and $l$ to get your answer: $a$ can be written as a sum of two squares iff it is non-negative and is of the form $b^2frac{p}{q}$ where $binmathbb Q$ and $p,q$ are coprime integers divided only by primes $equiv 1,2bmod 4$.
answered Sep 21 '15 at 14:19
FerraFerra
3,82111024
answered Sep 21 '15 at 14:19
FerraFerra
3,82111024
answered Sep 21 '15 at 14:19
FerraFerra
3,82111024
answered Sep 21 '15 at 14:19
FerraFerra
3,82111024
3,82111024
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%2f1445058%2fwhich-rationals-can-be-written-as-the-sum-of-two-rational-squares%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
$begingroup$
math.stackexchange.com/questions/1200794/… mentions a solution but gives no references.
$endgroup$
– lhf
Sep 21 '15 at 13:53
$begingroup$
the last part of that question states the result asked for here.
$endgroup$
– Ross Millikan
Sep 21 '15 at 13:55
$begingroup$
What about $x=frac35,y=frac45$?
$endgroup$
– abiessu
Sep 21 '15 at 13:56
2
$begingroup$
Basically an Pythagorean triple?
$endgroup$
– Gummy bears
Sep 21 '15 at 14:02
4
$begingroup$
can you not clear the denominators and apply the standard results?
$endgroup$
– Matt B
Sep 21 '15 at 14:19