Prove that, if $pequiv1pmod4$ is a prime, then there exist $m$, $x$, and $y$ in $Bbb N$ s.t. $x^2+y^2=mp$, with $pnmid x$, $p nmid y$, $0<m<p$. [on hold]

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Prove that, if $pequiv1pmod4$ is a prime, then there exists positive integers $m$, $x$, and $y$ such that $x^2+y^2=mp$, with $pnmid x$, $p nmid y$, $0<m<p$.




elementary-number-theory




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put on hold as off-topic by José Carlos Santos, TheSimpliFire, user477343, Micah, Peter Aug 3 at 18:39


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – TheSimpliFire, user477343, Micah, Peter
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    What have you attempted?
    – TheSimpliFire
    Aug 3 at 9:46






  • 1




    This is in all the textbooks; it's one step on the proof that each prime congruent to $1$ modulo $4$ is the sum of two squares.
    – Lord Shark the Unknown
    Aug 3 at 9:47






  • 1




    @TheSimpliFire I would have asked the same thing, but this prevented me from doing so...
    – user477343
    Aug 3 at 9:55






  • 1




    @user477343 I think one such question suffices for every one of these posts. I was giving OP a chance to share their efforts, but just looking at their previous posts, it seems like they don't respond to comments/accept answers. I have now voted to close.
    – TheSimpliFire
    Aug 3 at 10:02







  • 2




    I was too quick to close as off-topic. This post is a duplicate of this one.
    – TheSimpliFire
    Aug 3 at 10:15















up vote
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Prove that, if $pequiv1pmod4$ is a prime, then there exists positive integers $m$, $x$, and $y$ such that $x^2+y^2=mp$, with $pnmid x$, $p nmid y$, $0<m<p$.




elementary-number-theory




share|cite|improve this question













put on hold as off-topic by José Carlos Santos, TheSimpliFire, user477343, Micah, Peter Aug 3 at 18:39


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – TheSimpliFire, user477343, Micah, Peter
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    What have you attempted?
    – TheSimpliFire
    Aug 3 at 9:46






  • 1




    This is in all the textbooks; it's one step on the proof that each prime congruent to $1$ modulo $4$ is the sum of two squares.
    – Lord Shark the Unknown
    Aug 3 at 9:47






  • 1




    @TheSimpliFire I would have asked the same thing, but this prevented me from doing so...
    – user477343
    Aug 3 at 9:55






  • 1




    @user477343 I think one such question suffices for every one of these posts. I was giving OP a chance to share their efforts, but just looking at their previous posts, it seems like they don't respond to comments/accept answers. I have now voted to close.
    – TheSimpliFire
    Aug 3 at 10:02







  • 2




    I was too quick to close as off-topic. This post is a duplicate of this one.
    – TheSimpliFire
    Aug 3 at 10:15













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Prove that, if $pequiv1pmod4$ is a prime, then there exists positive integers $m$, $x$, and $y$ such that $x^2+y^2=mp$, with $pnmid x$, $p nmid y$, $0<m<p$.




elementary-number-theory




share|cite|improve this question













Prove that, if $pequiv1pmod4$ is a prime, then there exists positive integers $m$, $x$, and $y$ such that $x^2+y^2=mp$, with $pnmid x$, $p nmid y$, $0<m<p$.





elementary-number-theory





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited 2 days ago









Shaun

7,31892970




7,31892970









asked Aug 3 at 9:42









user90533

212112




212112




put on hold as off-topic by José Carlos Santos, TheSimpliFire, user477343, Micah, Peter Aug 3 at 18:39


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – TheSimpliFire, user477343, Micah, Peter
If this question can be reworded to fit the rules in the help center, please edit the question.




put on hold as off-topic by José Carlos Santos, TheSimpliFire, user477343, Micah, Peter Aug 3 at 18:39


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – TheSimpliFire, user477343, Micah, Peter
If this question can be reworded to fit the rules in the help center, please edit the question.







  • 1




    What have you attempted?
    – TheSimpliFire
    Aug 3 at 9:46






  • 1




    This is in all the textbooks; it's one step on the proof that each prime congruent to $1$ modulo $4$ is the sum of two squares.
    – Lord Shark the Unknown
    Aug 3 at 9:47






  • 1




    @TheSimpliFire I would have asked the same thing, but this prevented me from doing so...
    – user477343
    Aug 3 at 9:55






  • 1




    @user477343 I think one such question suffices for every one of these posts. I was giving OP a chance to share their efforts, but just looking at their previous posts, it seems like they don't respond to comments/accept answers. I have now voted to close.
    – TheSimpliFire
    Aug 3 at 10:02







  • 2




    I was too quick to close as off-topic. This post is a duplicate of this one.
    – TheSimpliFire
    Aug 3 at 10:15













  • 1




    What have you attempted?
    – TheSimpliFire
    Aug 3 at 9:46






  • 1




    This is in all the textbooks; it's one step on the proof that each prime congruent to $1$ modulo $4$ is the sum of two squares.
    – Lord Shark the Unknown
    Aug 3 at 9:47






  • 1




    @TheSimpliFire I would have asked the same thing, but this prevented me from doing so...
    – user477343
    Aug 3 at 9:55






  • 1




    @user477343 I think one such question suffices for every one of these posts. I was giving OP a chance to share their efforts, but just looking at their previous posts, it seems like they don't respond to comments/accept answers. I have now voted to close.
    – TheSimpliFire
    Aug 3 at 10:02







  • 2




    I was too quick to close as off-topic. This post is a duplicate of this one.
    – TheSimpliFire
    Aug 3 at 10:15








1




1




What have you attempted?
– TheSimpliFire
Aug 3 at 9:46




What have you attempted?
– TheSimpliFire
Aug 3 at 9:46




1




1




This is in all the textbooks; it's one step on the proof that each prime congruent to $1$ modulo $4$ is the sum of two squares.
– Lord Shark the Unknown
Aug 3 at 9:47




This is in all the textbooks; it's one step on the proof that each prime congruent to $1$ modulo $4$ is the sum of two squares.
– Lord Shark the Unknown
Aug 3 at 9:47




1




1




@TheSimpliFire I would have asked the same thing, but this prevented me from doing so...
– user477343
Aug 3 at 9:55




@TheSimpliFire I would have asked the same thing, but this prevented me from doing so...
– user477343
Aug 3 at 9:55




1




1




@user477343 I think one such question suffices for every one of these posts. I was giving OP a chance to share their efforts, but just looking at their previous posts, it seems like they don't respond to comments/accept answers. I have now voted to close.
– TheSimpliFire
Aug 3 at 10:02





@user477343 I think one such question suffices for every one of these posts. I was giving OP a chance to share their efforts, but just looking at their previous posts, it seems like they don't respond to comments/accept answers. I have now voted to close.
– TheSimpliFire
Aug 3 at 10:02





2




2




I was too quick to close as off-topic. This post is a duplicate of this one.
– TheSimpliFire
Aug 3 at 10:15





I was too quick to close as off-topic. This post is a duplicate of this one.
– TheSimpliFire
Aug 3 at 10:15
















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