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
164k10170395
$endgroup$
|
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
164k10170395
$endgroup$
$begingroup$
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
164k10170395
$endgroup$
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
164k10170395
asked Sep 21 '15 at 13:50
lhflhf
164k10170395
edited Sep 21 '15 at 14:00
lhf
asked Sep 21 '15 at 13:50
lhflhf
164k10170395
asked Sep 21 '15 at 13:50
lhflhf
164k10170395
asked Sep 21 '15 at 13:50
lhflhf
164k10170395
164k10170395
$begingroup$
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
452k36425810
$endgroup$
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
3,80911024
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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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
452k36425810
$endgroup$
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
452k36425810
$endgroup$
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
452k36425810
$endgroup$
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
452k36425810
edited Sep 21 '15 at 14:33
answered Sep 21 '15 at 14:23
André NicolasAndré Nicolas
452k36425810
answered Sep 21 '15 at 14:23
André NicolasAndré Nicolas
452k36425810
answered Sep 21 '15 at 14:23
André NicolasAndré Nicolas
452k36425810
452k36425810
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
3,80911024
$endgroup$
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
3,80911024
$endgroup$
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
3,80911024
$endgroup$
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
3,80911024
answered Sep 21 '15 at 14:19
FerraFerra
3,80911024
answered Sep 21 '15 at 14:19
FerraFerra
3,80911024
answered Sep 21 '15 at 14:19
FerraFerra
3,80911024
3,80911024
add a comment |
add a comment |
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$begingroup$
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