Mixing
How to prove that all primes of the form $4k+1$ can be represented by the sum of two squares in only one way...
$begingroup$
I am reading a book about Number Theory as a new learner.
The book has proved that all primes of the form $4k+1$ can be represented by the sum of two squares.
This question is given as exercise and I truly have no idea for the solution.
The only thing that I found may be helpful is that one of the squares is even and one is odd.
Thanks for any help ^_^
Edit: About duplication, I suppose that question is about the process in a specified proof.
Will Jagy has given a complete proof.
Thanks so much for the answer!
number-theory prime-numbers sums-of-squares
asked Jul 18 '18 at 13:25
Mira from EarthMira from Earth
19510
$endgroup$
add a comment |
$begingroup$
I am reading a book about Number Theory as a new learner.
The book has proved that all primes of the form $4k+1$ can be represented by the sum of two squares.
This question is given as exercise and I truly have no idea for the solution.
The only thing that I found may be helpful is that one of the squares is even and one is odd.
Thanks for any help ^_^
Edit: About duplication, I suppose that question is about the process in a specified proof.
Will Jagy has given a complete proof.
Thanks so much for the answer!
number-theory prime-numbers sums-of-squares
asked Jul 18 '18 at 13:25
Mira from EarthMira from Earth
19510
$endgroup$
5
$begingroup$
Look here : proofwiki.org/wiki/Fermat%27s_Two_Squares_Theorem
$endgroup$
– Peter
Jul 18 '18 at 13:38
$begingroup$
It would help to know what sort of proof the book gave of the existence. For example, if it's based on the fact that the Gaussian integers $mathbb{Z}[i]$ are a PID, then the uniqueness just follows from the fact that if $a^2 + b^2 = p$ is prime then $a+bi$ is irreducible in $mathbb{Z}[i]$. So if $c^2 + d^2 = p$ also then $c+di$ must be a unit times either $a+bi$ or $a-bi$...
$endgroup$
– Daniel Schepler
Jul 18 '18 at 23:11
add a comment |
$begingroup$
I am reading a book about Number Theory as a new learner.
The book has proved that all primes of the form $4k+1$ can be represented by the sum of two squares.
This question is given as exercise and I truly have no idea for the solution.
The only thing that I found may be helpful is that one of the squares is even and one is odd.
Thanks for any help ^_^
Edit: About duplication, I suppose that question is about the process in a specified proof.
Will Jagy has given a complete proof.
Thanks so much for the answer!
number-theory prime-numbers sums-of-squares
asked Jul 18 '18 at 13:25
Mira from EarthMira from Earth
19510
$endgroup$
I am reading a book about Number Theory as a new learner.
The book has proved that all primes of the form $4k+1$ can be represented by the sum of two squares.
This question is given as exercise and I truly have no idea for the solution.
The only thing that I found may be helpful is that one of the squares is even and one is odd.
Thanks for any help ^_^
Edit: About duplication, I suppose that question is about the process in a specified proof.
Will Jagy has given a complete proof.
Thanks so much for the answer!
number-theory prime-numbers sums-of-squares
number-theory prime-numbers sums-of-squares
asked Jul 18 '18 at 13:25
Mira from EarthMira from Earth
19510
asked Jul 18 '18 at 13:25
Mira from EarthMira from Earth
19510
edited Jul 19 '18 at 1:34
Mira from Earth
asked Jul 18 '18 at 13:25
Mira from EarthMira from Earth
19510
asked Jul 18 '18 at 13:25
Mira from EarthMira from Earth
19510
asked Jul 18 '18 at 13:25
Mira from EarthMira from Earth
19510
19510
5
$begingroup$
Look here : proofwiki.org/wiki/Fermat%27s_Two_Squares_Theorem
$endgroup$
– Peter
Jul 18 '18 at 13:38
$begingroup$
It would help to know what sort of proof the book gave of the existence. For example, if it's based on the fact that the Gaussian integers $mathbb{Z}[i]$ are a PID, then the uniqueness just follows from the fact that if $a^2 + b^2 = p$ is prime then $a+bi$ is irreducible in $mathbb{Z}[i]$. So if $c^2 + d^2 = p$ also then $c+di$ must be a unit times either $a+bi$ or $a-bi$...
$endgroup$
– Daniel Schepler
Jul 18 '18 at 23:11
add a comment |
5
$begingroup$
Look here : proofwiki.org/wiki/Fermat%27s_Two_Squares_Theorem
$endgroup$
– Peter
Jul 18 '18 at 13:38
$begingroup$
It would help to know what sort of proof the book gave of the existence. For example, if it's based on the fact that the Gaussian integers $mathbb{Z}[i]$ are a PID, then the uniqueness just follows from the fact that if $a^2 + b^2 = p$ is prime then $a+bi$ is irreducible in $mathbb{Z}[i]$. So if $c^2 + d^2 = p$ also then $c+di$ must be a unit times either $a+bi$ or $a-bi$...
$endgroup$
– Daniel Schepler
Jul 18 '18 at 23:11
5
5
$begingroup$
Look here : proofwiki.org/wiki/Fermat%27s_Two_Squares_Theorem
$endgroup$
– Peter
Jul 18 '18 at 13:38
$begingroup$
Look here : proofwiki.org/wiki/Fermat%27s_Two_Squares_Theorem
$endgroup$
– Peter
Jul 18 '18 at 13:38
$begingroup$
It would help to know what sort of proof the book gave of the existence. For example, if it's based on the fact that the Gaussian integers $mathbb{Z}[i]$ are a PID, then the uniqueness just follows from the fact that if $a^2 + b^2 = p$ is prime then $a+bi$ is irreducible in $mathbb{Z}[i]$. So if $c^2 + d^2 = p$ also then $c+di$ must be a unit times either $a+bi$ or $a-bi$...
$endgroup$
– Daniel Schepler
Jul 18 '18 at 23:11
$begingroup$
It would help to know what sort of proof the book gave of the existence. For example, if it's based on the fact that the Gaussian integers $mathbb{Z}[i]$ are a PID, then the uniqueness just follows from the fact that if $a^2 + b^2 = p$ is prime then $a+bi$ is irreducible in $mathbb{Z}[i]$. So if $c^2 + d^2 = p$ also then $c+di$ must be a unit times either $a+bi$ or $a-bi$...
$endgroup$
– Daniel Schepler
Jul 18 '18 at 23:11
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let us have positive integers with $$ n = a^2 + b^2 = c^2 + d^2 $$ with $a,c$ odd, then $b,d$ even, also $b < d$ and $a > c.$ So these are two genuinely distinct ways of writing $n.$
$$ (a-c)(a+c) = (d-b)(d+b). $$
Define $$ r = gcd(a-c, d-b) $$
Note that $r$ is even. Next we define
$$ a-c = rs ; , ; ; ; d-b = r t ; , $$ so that
$$ gcd(s,t) = 1. $$
Note that at least one of $s,t$ is odd. This gives us
$$ (a+c)s = (d+b)t. $$
The gcd property tells us that $t | (a+c).$ We define $u$ with
$$ a+c = t u. $$ We immediately conclude $d+b = s u.$ As $a+c, d+b$ are even, but at least one of $s,t$ is odd, we find $u$ is even.
In one line,
$$ color{magenta}{ a-c = rs ; , ; ; a+c = tu ; , ; ; d-b = rt ; , ; ; d+b = su ; }. $$
If we now solve for $a$ and $b$ and square and combine, we get
$$ a = frac{1}{2}(rs+tu) ; , ; ; ; b = frac{1}{2}(su-rt) ; , $$
$$ a^2 + b^2 = frac{1}{4}left( r^2 s^2 + r^2 t^2 + u^2 s^2 + u^2 t^2right) = frac{1}{4}left( r^2 + u^2 right) left( s^2 + t^2 right); , $$
$$ color{magenta}{ n = a^2 + b^2 = c^2 + d^2 = left( left(frac{r}{2}right)^2 +left(frac{u}{2}right)^2 right) left(s^2 + t^2 right) } . $$
That is to say, because $n$ had two distinct representations as the sum of two squares, it is composite.
The contrapositive is that a number $4k+1$ with just one expression as the sum of two nonzero squares is prime. It is probably worth pointing out that this is the contrapositive in the setting of having at least one representation. We are not making any conclusions about numbers that have no representation as the sum of two squares.
answered Jul 18 '18 at 22:59
Will JagyWill Jagy
103k5102200
$endgroup$
$begingroup$
Did you mean converse in the last sentence? Note that for example, 9 has essentially only $3^2 + 0^2$ as a representation as a sum of two squares, yet 9 is not prime.
$endgroup$
– Daniel Schepler
Jul 19 '18 at 0:33
$begingroup$
@DanielSchepler I will need to think about it. What I wrote is just about two representations with nonzero $a,b,c,d,$ so is silent about some issues.
$endgroup$
– Will Jagy
Jul 19 '18 at 0:36
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%2f2855583%2fhow-to-prove-that-all-primes-of-the-form-4k1-can-be-represented-by-the-sum-of%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
$begingroup$
Let us have positive integers with $$ n = a^2 + b^2 = c^2 + d^2 $$ with $a,c$ odd, then $b,d$ even, also $b < d$ and $a > c.$ So these are two genuinely distinct ways of writing $n.$
$$ (a-c)(a+c) = (d-b)(d+b). $$
Define $$ r = gcd(a-c, d-b) $$
Note that $r$ is even. Next we define
$$ a-c = rs ; , ; ; ; d-b = r t ; , $$ so that
$$ gcd(s,t) = 1. $$
Note that at least one of $s,t$ is odd. This gives us
$$ (a+c)s = (d+b)t. $$
The gcd property tells us that $t | (a+c).$ We define $u$ with
$$ a+c = t u. $$ We immediately conclude $d+b = s u.$ As $a+c, d+b$ are even, but at least one of $s,t$ is odd, we find $u$ is even.
In one line,
$$ color{magenta}{ a-c = rs ; , ; ; a+c = tu ; , ; ; d-b = rt ; , ; ; d+b = su ; }. $$
If we now solve for $a$ and $b$ and square and combine, we get
$$ a = frac{1}{2}(rs+tu) ; , ; ; ; b = frac{1}{2}(su-rt) ; , $$
$$ a^2 + b^2 = frac{1}{4}left( r^2 s^2 + r^2 t^2 + u^2 s^2 + u^2 t^2right) = frac{1}{4}left( r^2 + u^2 right) left( s^2 + t^2 right); , $$
$$ color{magenta}{ n = a^2 + b^2 = c^2 + d^2 = left( left(frac{r}{2}right)^2 +left(frac{u}{2}right)^2 right) left(s^2 + t^2 right) } . $$
That is to say, because $n$ had two distinct representations as the sum of two squares, it is composite.
The contrapositive is that a number $4k+1$ with just one expression as the sum of two nonzero squares is prime. It is probably worth pointing out that this is the contrapositive in the setting of having at least one representation. We are not making any conclusions about numbers that have no representation as the sum of two squares.
answered Jul 18 '18 at 22:59
Will JagyWill Jagy
103k5102200
$endgroup$
$begingroup$
Did you mean converse in the last sentence? Note that for example, 9 has essentially only $3^2 + 0^2$ as a representation as a sum of two squares, yet 9 is not prime.
$endgroup$
– Daniel Schepler
Jul 19 '18 at 0:33
$begingroup$
@DanielSchepler I will need to think about it. What I wrote is just about two representations with nonzero $a,b,c,d,$ so is silent about some issues.
$endgroup$
– Will Jagy
Jul 19 '18 at 0:36
add a comment |
$begingroup$
Let us have positive integers with $$ n = a^2 + b^2 = c^2 + d^2 $$ with $a,c$ odd, then $b,d$ even, also $b < d$ and $a > c.$ So these are two genuinely distinct ways of writing $n.$
$$ (a-c)(a+c) = (d-b)(d+b). $$
Define $$ r = gcd(a-c, d-b) $$
Note that $r$ is even. Next we define
$$ a-c = rs ; , ; ; ; d-b = r t ; , $$ so that
$$ gcd(s,t) = 1. $$
Note that at least one of $s,t$ is odd. This gives us
$$ (a+c)s = (d+b)t. $$
The gcd property tells us that $t | (a+c).$ We define $u$ with
$$ a+c = t u. $$ We immediately conclude $d+b = s u.$ As $a+c, d+b$ are even, but at least one of $s,t$ is odd, we find $u$ is even.
In one line,
$$ color{magenta}{ a-c = rs ; , ; ; a+c = tu ; , ; ; d-b = rt ; , ; ; d+b = su ; }. $$
If we now solve for $a$ and $b$ and square and combine, we get
$$ a = frac{1}{2}(rs+tu) ; , ; ; ; b = frac{1}{2}(su-rt) ; , $$
$$ a^2 + b^2 = frac{1}{4}left( r^2 s^2 + r^2 t^2 + u^2 s^2 + u^2 t^2right) = frac{1}{4}left( r^2 + u^2 right) left( s^2 + t^2 right); , $$
$$ color{magenta}{ n = a^2 + b^2 = c^2 + d^2 = left( left(frac{r}{2}right)^2 +left(frac{u}{2}right)^2 right) left(s^2 + t^2 right) } . $$
That is to say, because $n$ had two distinct representations as the sum of two squares, it is composite.
The contrapositive is that a number $4k+1$ with just one expression as the sum of two nonzero squares is prime. It is probably worth pointing out that this is the contrapositive in the setting of having at least one representation. We are not making any conclusions about numbers that have no representation as the sum of two squares.
answered Jul 18 '18 at 22:59
Will JagyWill Jagy
103k5102200
$endgroup$
$begingroup$
Did you mean converse in the last sentence? Note that for example, 9 has essentially only $3^2 + 0^2$ as a representation as a sum of two squares, yet 9 is not prime.
$endgroup$
– Daniel Schepler
Jul 19 '18 at 0:33
$begingroup$
@DanielSchepler I will need to think about it. What I wrote is just about two representations with nonzero $a,b,c,d,$ so is silent about some issues.
$endgroup$
– Will Jagy
Jul 19 '18 at 0:36
add a comment |
$begingroup$
Let us have positive integers with $$ n = a^2 + b^2 = c^2 + d^2 $$ with $a,c$ odd, then $b,d$ even, also $b < d$ and $a > c.$ So these are two genuinely distinct ways of writing $n.$
$$ (a-c)(a+c) = (d-b)(d+b). $$
Define $$ r = gcd(a-c, d-b) $$
Note that $r$ is even. Next we define
$$ a-c = rs ; , ; ; ; d-b = r t ; , $$ so that
$$ gcd(s,t) = 1. $$
Note that at least one of $s,t$ is odd. This gives us
$$ (a+c)s = (d+b)t. $$
The gcd property tells us that $t | (a+c).$ We define $u$ with
$$ a+c = t u. $$ We immediately conclude $d+b = s u.$ As $a+c, d+b$ are even, but at least one of $s,t$ is odd, we find $u$ is even.
In one line,
$$ color{magenta}{ a-c = rs ; , ; ; a+c = tu ; , ; ; d-b = rt ; , ; ; d+b = su ; }. $$
If we now solve for $a$ and $b$ and square and combine, we get
$$ a = frac{1}{2}(rs+tu) ; , ; ; ; b = frac{1}{2}(su-rt) ; , $$
$$ a^2 + b^2 = frac{1}{4}left( r^2 s^2 + r^2 t^2 + u^2 s^2 + u^2 t^2right) = frac{1}{4}left( r^2 + u^2 right) left( s^2 + t^2 right); , $$
$$ color{magenta}{ n = a^2 + b^2 = c^2 + d^2 = left( left(frac{r}{2}right)^2 +left(frac{u}{2}right)^2 right) left(s^2 + t^2 right) } . $$
That is to say, because $n$ had two distinct representations as the sum of two squares, it is composite.
The contrapositive is that a number $4k+1$ with just one expression as the sum of two nonzero squares is prime. It is probably worth pointing out that this is the contrapositive in the setting of having at least one representation. We are not making any conclusions about numbers that have no representation as the sum of two squares.
answered Jul 18 '18 at 22:59
Will JagyWill Jagy
103k5102200
$endgroup$
Let us have positive integers with $$ n = a^2 + b^2 = c^2 + d^2 $$ with $a,c$ odd, then $b,d$ even, also $b < d$ and $a > c.$ So these are two genuinely distinct ways of writing $n.$
$$ (a-c)(a+c) = (d-b)(d+b). $$
Define $$ r = gcd(a-c, d-b) $$
Note that $r$ is even. Next we define
$$ a-c = rs ; , ; ; ; d-b = r t ; , $$ so that
$$ gcd(s,t) = 1. $$
Note that at least one of $s,t$ is odd. This gives us
$$ (a+c)s = (d+b)t. $$
The gcd property tells us that $t | (a+c).$ We define $u$ with
$$ a+c = t u. $$ We immediately conclude $d+b = s u.$ As $a+c, d+b$ are even, but at least one of $s,t$ is odd, we find $u$ is even.
In one line,
$$ color{magenta}{ a-c = rs ; , ; ; a+c = tu ; , ; ; d-b = rt ; , ; ; d+b = su ; }. $$
If we now solve for $a$ and $b$ and square and combine, we get
$$ a = frac{1}{2}(rs+tu) ; , ; ; ; b = frac{1}{2}(su-rt) ; , $$
$$ a^2 + b^2 = frac{1}{4}left( r^2 s^2 + r^2 t^2 + u^2 s^2 + u^2 t^2right) = frac{1}{4}left( r^2 + u^2 right) left( s^2 + t^2 right); , $$
$$ color{magenta}{ n = a^2 + b^2 = c^2 + d^2 = left( left(frac{r}{2}right)^2 +left(frac{u}{2}right)^2 right) left(s^2 + t^2 right) } . $$
That is to say, because $n$ had two distinct representations as the sum of two squares, it is composite.
The contrapositive is that a number $4k+1$ with just one expression as the sum of two nonzero squares is prime. It is probably worth pointing out that this is the contrapositive in the setting of having at least one representation. We are not making any conclusions about numbers that have no representation as the sum of two squares.
answered Jul 18 '18 at 22:59
Will JagyWill Jagy
103k5102200
edited Jul 19 '18 at 0:44
answered Jul 18 '18 at 22:59
Will JagyWill Jagy
103k5102200
answered Jul 18 '18 at 22:59
Will JagyWill Jagy
103k5102200
answered Jul 18 '18 at 22:59
Will JagyWill Jagy
103k5102200
103k5102200
$begingroup$
Did you mean converse in the last sentence? Note that for example, 9 has essentially only $3^2 + 0^2$ as a representation as a sum of two squares, yet 9 is not prime.
$endgroup$
– Daniel Schepler
Jul 19 '18 at 0:33
$begingroup$
@DanielSchepler I will need to think about it. What I wrote is just about two representations with nonzero $a,b,c,d,$ so is silent about some issues.
$endgroup$
– Will Jagy
Jul 19 '18 at 0:36
add a comment |
$begingroup$
Did you mean converse in the last sentence? Note that for example, 9 has essentially only $3^2 + 0^2$ as a representation as a sum of two squares, yet 9 is not prime.
$endgroup$
– Daniel Schepler
Jul 19 '18 at 0:33
$begingroup$
@DanielSchepler I will need to think about it. What I wrote is just about two representations with nonzero $a,b,c,d,$ so is silent about some issues.
$endgroup$
– Will Jagy
Jul 19 '18 at 0:36
$begingroup$
Did you mean converse in the last sentence? Note that for example, 9 has essentially only $3^2 + 0^2$ as a representation as a sum of two squares, yet 9 is not prime.
$endgroup$
– Daniel Schepler
Jul 19 '18 at 0:33
$begingroup$
Did you mean converse in the last sentence? Note that for example, 9 has essentially only $3^2 + 0^2$ as a representation as a sum of two squares, yet 9 is not prime.
$endgroup$
– Daniel Schepler
Jul 19 '18 at 0:33
$begingroup$
@DanielSchepler I will need to think about it. What I wrote is just about two representations with nonzero $a,b,c,d,$ so is silent about some issues.
$endgroup$
– Will Jagy
Jul 19 '18 at 0:36
$begingroup$
@DanielSchepler I will need to think about it. What I wrote is just about two representations with nonzero $a,b,c,d,$ so is silent about some issues.
$endgroup$
– Will Jagy
Jul 19 '18 at 0:36
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%2f2855583%2fhow-to-prove-that-all-primes-of-the-form-4k1-can-be-represented-by-the-sum-of%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
5
$begingroup$
Look here : proofwiki.org/wiki/Fermat%27s_Two_Squares_Theorem
$endgroup$
– Peter
Jul 18 '18 at 13:38
$begingroup$
It would help to know what sort of proof the book gave of the existence. For example, if it's based on the fact that the Gaussian integers $mathbb{Z}[i]$ are a PID, then the uniqueness just follows from the fact that if $a^2 + b^2 = p$ is prime then $a+bi$ is irreducible in $mathbb{Z}[i]$. So if $c^2 + d^2 = p$ also then $c+di$ must be a unit times either $a+bi$ or $a-bi$...
$endgroup$
– Daniel Schepler
Jul 18 '18 at 23:11