Any product of primes in the form of 4n+1 is the sum of 2 relatively prime squares
$begingroup$
How can we prove the theorem that any product of primes in the form $4n+1$ is the sum of 2 relatively prime square numbers?
EDIT: My question is not the same as the responses that were linked. I'm asking specifically about the product of primes in the form $4n+1$ being the sum of $2$ relatively prime squares, not just primes in the form $4n+1$ being the sum of squares.
elementary-number-theory
asked Dec 25 '18 at 17:48
OmicronGammaOmicronGamma
536
$endgroup$
add a comment |
$begingroup$
How can we prove the theorem that any product of primes in the form $4n+1$ is the sum of 2 relatively prime square numbers?
EDIT: My question is not the same as the responses that were linked. I'm asking specifically about the product of primes in the form $4n+1$ being the sum of $2$ relatively prime squares, not just primes in the form $4n+1$ being the sum of squares.
elementary-number-theory
asked Dec 25 '18 at 17:48
OmicronGammaOmicronGamma
536
$endgroup$
$begingroup$
See also this question. The "unique" squares must be coprime, because $p$ is prime.
$endgroup$
– Dietrich Burde
Dec 25 '18 at 17:53
1
$begingroup$
Why is it a duplicate? The link refers to a prime, but the question is about a product of primes.
$endgroup$
– Michael Behrend
Dec 25 '18 at 18:08
$begingroup$
Dear Dietrich Burde, thank you for the link to proving that all primes in the form $4k+1$ are the sum of two squares. However, I was wondering about the product of several primes in the form $4k+1$ being the sum of 2 relatively prime square numbers. I don't have quite the same question as the response you linked.
$endgroup$
– OmicronGamma
Dec 25 '18 at 18:13
$begingroup$
@DietrichBurde The question may well be a dupe, but the link you gave does not suffice, Please be more careful.
$endgroup$
– Bill Dubuque
Dec 25 '18 at 19:59
add a comment |
$begingroup$
How can we prove the theorem that any product of primes in the form $4n+1$ is the sum of 2 relatively prime square numbers?
EDIT: My question is not the same as the responses that were linked. I'm asking specifically about the product of primes in the form $4n+1$ being the sum of $2$ relatively prime squares, not just primes in the form $4n+1$ being the sum of squares.
elementary-number-theory
asked Dec 25 '18 at 17:48
OmicronGammaOmicronGamma
536
$endgroup$
How can we prove the theorem that any product of primes in the form $4n+1$ is the sum of 2 relatively prime square numbers?
EDIT: My question is not the same as the responses that were linked. I'm asking specifically about the product of primes in the form $4n+1$ being the sum of $2$ relatively prime squares, not just primes in the form $4n+1$ being the sum of squares.
elementary-number-theory
elementary-number-theory
asked Dec 25 '18 at 17:48
OmicronGammaOmicronGamma
536
asked Dec 25 '18 at 17:48
OmicronGammaOmicronGamma
536
edited Dec 25 '18 at 18:26
OmicronGamma
asked Dec 25 '18 at 17:48
OmicronGammaOmicronGamma
536
asked Dec 25 '18 at 17:48
OmicronGammaOmicronGamma
536
asked Dec 25 '18 at 17:48
OmicronGammaOmicronGamma
536
536
$begingroup$
See also this question. The "unique" squares must be coprime, because $p$ is prime.
$endgroup$
– Dietrich Burde
Dec 25 '18 at 17:53
1
$begingroup$
Why is it a duplicate? The link refers to a prime, but the question is about a product of primes.
$endgroup$
– Michael Behrend
Dec 25 '18 at 18:08
$begingroup$
Dear Dietrich Burde, thank you for the link to proving that all primes in the form $4k+1$ are the sum of two squares. However, I was wondering about the product of several primes in the form $4k+1$ being the sum of 2 relatively prime square numbers. I don't have quite the same question as the response you linked.
$endgroup$
– OmicronGamma
Dec 25 '18 at 18:13
$begingroup$
@DietrichBurde The question may well be a dupe, but the link you gave does not suffice, Please be more careful.
$endgroup$
– Bill Dubuque
Dec 25 '18 at 19:59
add a comment |
$begingroup$
See also this question. The "unique" squares must be coprime, because $p$ is prime.
$endgroup$
– Dietrich Burde
Dec 25 '18 at 17:53
1
$begingroup$
Why is it a duplicate? The link refers to a prime, but the question is about a product of primes.
$endgroup$
– Michael Behrend
Dec 25 '18 at 18:08
$begingroup$
Dear Dietrich Burde, thank you for the link to proving that all primes in the form $4k+1$ are the sum of two squares. However, I was wondering about the product of several primes in the form $4k+1$ being the sum of 2 relatively prime square numbers. I don't have quite the same question as the response you linked.
$endgroup$
– OmicronGamma
Dec 25 '18 at 18:13
$begingroup$
@DietrichBurde The question may well be a dupe, but the link you gave does not suffice, Please be more careful.
$endgroup$
– Bill Dubuque
Dec 25 '18 at 19:59
$begingroup$
See also this question. The "unique" squares must be coprime, because $p$ is prime.
$endgroup$
– Dietrich Burde
Dec 25 '18 at 17:53
$begingroup$
See also this question. The "unique" squares must be coprime, because $p$ is prime.
$endgroup$
– Dietrich Burde
Dec 25 '18 at 17:53
1
1
$begingroup$
Why is it a duplicate? The link refers to a prime, but the question is about a product of primes.
$endgroup$
– Michael Behrend
Dec 25 '18 at 18:08
$begingroup$
Why is it a duplicate? The link refers to a prime, but the question is about a product of primes.
$endgroup$
– Michael Behrend
Dec 25 '18 at 18:08
$begingroup$
Dear Dietrich Burde, thank you for the link to proving that all primes in the form $4k+1$ are the sum of two squares. However, I was wondering about the product of several primes in the form $4k+1$ being the sum of 2 relatively prime square numbers. I don't have quite the same question as the response you linked.
$endgroup$
– OmicronGamma
Dec 25 '18 at 18:13
$begingroup$
Dear Dietrich Burde, thank you for the link to proving that all primes in the form $4k+1$ are the sum of two squares. However, I was wondering about the product of several primes in the form $4k+1$ being the sum of 2 relatively prime square numbers. I don't have quite the same question as the response you linked.
$endgroup$
– OmicronGamma
Dec 25 '18 at 18:13
$begingroup$
@DietrichBurde The question may well be a dupe, but the link you gave does not suffice, Please be more careful.
$endgroup$
– Bill Dubuque
Dec 25 '18 at 19:59
$begingroup$
@DietrichBurde The question may well be a dupe, but the link you gave does not suffice, Please be more careful.
$endgroup$
– Bill Dubuque
Dec 25 '18 at 19:59
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Since every prime
of the form 4n+1
is the sum of two squares,
and
$begin{array}\
(a^2+b^2)(c^2+d^2)
&=a^2c^2+a^2d^2+b^2c^2+b^2d^2\
&=a^2c^2pm 2abcd+b^2d^2+a^2d^2mp 2abcd+b^2c^2\
&=(acpm bd)^2+(admp bc)^2\
end{array}
$
the product of
any number of primes
of the form 4n+1
is the sum of two squares.
If the two squares
are not relatively prime,
some prime divides both of them,
and so the square of that prime
divides each square and so divides
the product of the primes.
But the product of the primes
(unless two of them are the same)
is only divisible by
primes to the first power,
a contradiction.
answered Dec 25 '18 at 22:25
marty cohenmarty cohen
73.8k549128
$endgroup$
add a comment |
$begingroup$
I think more work is needed to show that the squares can be chosen to be coprime. The above answer assumes that the primes are all distinct, but that isn't stated in the question.
For example having got $13 times 5 = 8^2 + 1^2$, we can extend to
$$13 times 5 times 5 = (8^2 + 1^2)(1^2 + 2^2) = 10^2 + 15^2 = 17^2 + 6^2,$$
where the first solution is invalid because $10$ and $15$ aren't coprime. We need to show that at least one of the two solutions is valid.
Suppose $a^2 + b^2$ is a product of primes $4n + 1$, and $c^2 + d^2$ is a prime of that form. If the first solution is invalid then some prime $q$ divides both $ac + bd$ and $ad - bc$. Hence
$$q vert (ac + bd)c + (ad - bc)d = a(c^2 + d^2),quad q vert (ac + bd)d - (ad - bc)c = b(c^2 + d^2).$$
Since $q$ doesn't divide both $a$ and $b$, we have $q vert c^2 + d^2$, hence $q = c^2 + d^2$.
If both solutions are invalid, then $c^2 + d^2$ divides $ac - bd$ as well as $ac + bd$, hence divides $2ac$ and $2bd$. Since $c^2 + d^2$ doesn't divide $2c$ or $2d$ it must divide both $a$ and $b$, contrary to hypothesis.
answered Dec 26 '18 at 10:04
Michael BehrendMichael Behrend
1,22746
$endgroup$
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%2f3052309%2fany-product-of-primes-in-the-form-of-4n1-is-the-sum-of-2-relatively-prime-squar%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
$begingroup$
Since every prime
of the form 4n+1
is the sum of two squares,
and
$begin{array}\
(a^2+b^2)(c^2+d^2)
&=a^2c^2+a^2d^2+b^2c^2+b^2d^2\
&=a^2c^2pm 2abcd+b^2d^2+a^2d^2mp 2abcd+b^2c^2\
&=(acpm bd)^2+(admp bc)^2\
end{array}
$
the product of
any number of primes
of the form 4n+1
is the sum of two squares.
If the two squares
are not relatively prime,
some prime divides both of them,
and so the square of that prime
divides each square and so divides
the product of the primes.
But the product of the primes
(unless two of them are the same)
is only divisible by
primes to the first power,
a contradiction.
answered Dec 25 '18 at 22:25
marty cohenmarty cohen
73.8k549128
$endgroup$
add a comment |
$begingroup$
Since every prime
of the form 4n+1
is the sum of two squares,
and
$begin{array}\
(a^2+b^2)(c^2+d^2)
&=a^2c^2+a^2d^2+b^2c^2+b^2d^2\
&=a^2c^2pm 2abcd+b^2d^2+a^2d^2mp 2abcd+b^2c^2\
&=(acpm bd)^2+(admp bc)^2\
end{array}
$
the product of
any number of primes
of the form 4n+1
is the sum of two squares.
If the two squares
are not relatively prime,
some prime divides both of them,
and so the square of that prime
divides each square and so divides
the product of the primes.
But the product of the primes
(unless two of them are the same)
is only divisible by
primes to the first power,
a contradiction.
answered Dec 25 '18 at 22:25
marty cohenmarty cohen
73.8k549128
$endgroup$
add a comment |
$begingroup$
Since every prime
of the form 4n+1
is the sum of two squares,
and
$begin{array}\
(a^2+b^2)(c^2+d^2)
&=a^2c^2+a^2d^2+b^2c^2+b^2d^2\
&=a^2c^2pm 2abcd+b^2d^2+a^2d^2mp 2abcd+b^2c^2\
&=(acpm bd)^2+(admp bc)^2\
end{array}
$
the product of
any number of primes
of the form 4n+1
is the sum of two squares.
If the two squares
are not relatively prime,
some prime divides both of them,
and so the square of that prime
divides each square and so divides
the product of the primes.
But the product of the primes
(unless two of them are the same)
is only divisible by
primes to the first power,
a contradiction.
answered Dec 25 '18 at 22:25
marty cohenmarty cohen
73.8k549128
$endgroup$
Since every prime
of the form 4n+1
is the sum of two squares,
and
$begin{array}\
(a^2+b^2)(c^2+d^2)
&=a^2c^2+a^2d^2+b^2c^2+b^2d^2\
&=a^2c^2pm 2abcd+b^2d^2+a^2d^2mp 2abcd+b^2c^2\
&=(acpm bd)^2+(admp bc)^2\
end{array}
$
the product of
any number of primes
of the form 4n+1
is the sum of two squares.
If the two squares
are not relatively prime,
some prime divides both of them,
and so the square of that prime
divides each square and so divides
the product of the primes.
But the product of the primes
(unless two of them are the same)
is only divisible by
primes to the first power,
a contradiction.
answered Dec 25 '18 at 22:25
marty cohenmarty cohen
73.8k549128
edited Dec 26 '18 at 21:27
answered Dec 25 '18 at 22:25
marty cohenmarty cohen
73.8k549128
answered Dec 25 '18 at 22:25
marty cohenmarty cohen
73.8k549128
answered Dec 25 '18 at 22:25
marty cohenmarty cohen
73.8k549128
73.8k549128
add a comment |
add a comment |
$begingroup$
I think more work is needed to show that the squares can be chosen to be coprime. The above answer assumes that the primes are all distinct, but that isn't stated in the question.
For example having got $13 times 5 = 8^2 + 1^2$, we can extend to
$$13 times 5 times 5 = (8^2 + 1^2)(1^2 + 2^2) = 10^2 + 15^2 = 17^2 + 6^2,$$
where the first solution is invalid because $10$ and $15$ aren't coprime. We need to show that at least one of the two solutions is valid.
Suppose $a^2 + b^2$ is a product of primes $4n + 1$, and $c^2 + d^2$ is a prime of that form. If the first solution is invalid then some prime $q$ divides both $ac + bd$ and $ad - bc$. Hence
$$q vert (ac + bd)c + (ad - bc)d = a(c^2 + d^2),quad q vert (ac + bd)d - (ad - bc)c = b(c^2 + d^2).$$
Since $q$ doesn't divide both $a$ and $b$, we have $q vert c^2 + d^2$, hence $q = c^2 + d^2$.
If both solutions are invalid, then $c^2 + d^2$ divides $ac - bd$ as well as $ac + bd$, hence divides $2ac$ and $2bd$. Since $c^2 + d^2$ doesn't divide $2c$ or $2d$ it must divide both $a$ and $b$, contrary to hypothesis.
answered Dec 26 '18 at 10:04
Michael BehrendMichael Behrend
1,22746
$endgroup$
add a comment |
$begingroup$
I think more work is needed to show that the squares can be chosen to be coprime. The above answer assumes that the primes are all distinct, but that isn't stated in the question.
For example having got $13 times 5 = 8^2 + 1^2$, we can extend to
$$13 times 5 times 5 = (8^2 + 1^2)(1^2 + 2^2) = 10^2 + 15^2 = 17^2 + 6^2,$$
where the first solution is invalid because $10$ and $15$ aren't coprime. We need to show that at least one of the two solutions is valid.
Suppose $a^2 + b^2$ is a product of primes $4n + 1$, and $c^2 + d^2$ is a prime of that form. If the first solution is invalid then some prime $q$ divides both $ac + bd$ and $ad - bc$. Hence
$$q vert (ac + bd)c + (ad - bc)d = a(c^2 + d^2),quad q vert (ac + bd)d - (ad - bc)c = b(c^2 + d^2).$$
Since $q$ doesn't divide both $a$ and $b$, we have $q vert c^2 + d^2$, hence $q = c^2 + d^2$.
If both solutions are invalid, then $c^2 + d^2$ divides $ac - bd$ as well as $ac + bd$, hence divides $2ac$ and $2bd$. Since $c^2 + d^2$ doesn't divide $2c$ or $2d$ it must divide both $a$ and $b$, contrary to hypothesis.
answered Dec 26 '18 at 10:04
Michael BehrendMichael Behrend
1,22746
$endgroup$
add a comment |
$begingroup$
I think more work is needed to show that the squares can be chosen to be coprime. The above answer assumes that the primes are all distinct, but that isn't stated in the question.
For example having got $13 times 5 = 8^2 + 1^2$, we can extend to
$$13 times 5 times 5 = (8^2 + 1^2)(1^2 + 2^2) = 10^2 + 15^2 = 17^2 + 6^2,$$
where the first solution is invalid because $10$ and $15$ aren't coprime. We need to show that at least one of the two solutions is valid.
Suppose $a^2 + b^2$ is a product of primes $4n + 1$, and $c^2 + d^2$ is a prime of that form. If the first solution is invalid then some prime $q$ divides both $ac + bd$ and $ad - bc$. Hence
$$q vert (ac + bd)c + (ad - bc)d = a(c^2 + d^2),quad q vert (ac + bd)d - (ad - bc)c = b(c^2 + d^2).$$
Since $q$ doesn't divide both $a$ and $b$, we have $q vert c^2 + d^2$, hence $q = c^2 + d^2$.
If both solutions are invalid, then $c^2 + d^2$ divides $ac - bd$ as well as $ac + bd$, hence divides $2ac$ and $2bd$. Since $c^2 + d^2$ doesn't divide $2c$ or $2d$ it must divide both $a$ and $b$, contrary to hypothesis.
answered Dec 26 '18 at 10:04
Michael BehrendMichael Behrend
1,22746
$endgroup$
I think more work is needed to show that the squares can be chosen to be coprime. The above answer assumes that the primes are all distinct, but that isn't stated in the question.
For example having got $13 times 5 = 8^2 + 1^2$, we can extend to
$$13 times 5 times 5 = (8^2 + 1^2)(1^2 + 2^2) = 10^2 + 15^2 = 17^2 + 6^2,$$
where the first solution is invalid because $10$ and $15$ aren't coprime. We need to show that at least one of the two solutions is valid.
Suppose $a^2 + b^2$ is a product of primes $4n + 1$, and $c^2 + d^2$ is a prime of that form. If the first solution is invalid then some prime $q$ divides both $ac + bd$ and $ad - bc$. Hence
$$q vert (ac + bd)c + (ad - bc)d = a(c^2 + d^2),quad q vert (ac + bd)d - (ad - bc)c = b(c^2 + d^2).$$
Since $q$ doesn't divide both $a$ and $b$, we have $q vert c^2 + d^2$, hence $q = c^2 + d^2$.
If both solutions are invalid, then $c^2 + d^2$ divides $ac - bd$ as well as $ac + bd$, hence divides $2ac$ and $2bd$. Since $c^2 + d^2$ doesn't divide $2c$ or $2d$ it must divide both $a$ and $b$, contrary to hypothesis.
answered Dec 26 '18 at 10:04
Michael BehrendMichael Behrend
1,22746
answered Dec 26 '18 at 10:04
Michael BehrendMichael Behrend
1,22746
answered Dec 26 '18 at 10:04
Michael BehrendMichael Behrend
1,22746
answered Dec 26 '18 at 10:04
Michael BehrendMichael Behrend
1,22746
1,22746
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%2f3052309%2fany-product-of-primes-in-the-form-of-4n1-is-the-sum-of-2-relatively-prime-squar%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
$begingroup$
See also this question. The "unique" squares must be coprime, because $p$ is prime.
$endgroup$
– Dietrich Burde
Dec 25 '18 at 17:53
1
$begingroup$
Why is it a duplicate? The link refers to a prime, but the question is about a product of primes.
$endgroup$
– Michael Behrend
Dec 25 '18 at 18:08
$begingroup$
Dear Dietrich Burde, thank you for the link to proving that all primes in the form $4k+1$ are the sum of two squares. However, I was wondering about the product of several primes in the form $4k+1$ being the sum of 2 relatively prime square numbers. I don't have quite the same question as the response you linked.
$endgroup$
– OmicronGamma
Dec 25 '18 at 18:13
$begingroup$
@DietrichBurde The question may well be a dupe, but the link you gave does not suffice, Please be more careful.
$endgroup$
– Bill Dubuque
Dec 25 '18 at 19:59