How to prove a class of elementary algebra? [on hold]












-4














It is known that m+n=x, mn=y, m,n∈Q,Prove that m, n∈Z if and only if x, y∈Z






algebra-precalculus group-theory elementary-number-theory







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maks L is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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put on hold as off-topic by KReiser, Lord Shark the Unknown, max_zorn, Shaun, Hans Lundmark Jan 4 at 6:38


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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – KReiser, max_zorn, Shaun, Hans Lundmark

If this question can be reworded to fit the rules in the help center, please edit the question.









  • 6




    What about $m=sqrt2$, $n=-sqrt2$?
    – Lord Shark the Unknown
    Jan 4 at 5:29






  • 1




    What are the solutions to the quadratic $z^2-xz+y=(z-m)(z-n)=0$?
    – Mark Bennet
    Jan 4 at 5:52










  • z=m,n and m+n=x,mn=y
    – maks L
    Jan 4 at 6:13










  • Let $m,n=(1pmsqrt{5})/2$, then $x=m+n=1ne0$ and $y=mn=-1$.
    – Alexander Burstein
    Jan 4 at 6:13












  • Try $m,n=1pm sqrt{2}$. Then $x=2$ and $y=-1$
    – Sauhard Sharma
    Jan 4 at 6:23
















-4














It is known that m+n=x, mn=y, m,n∈Q,Prove that m, n∈Z if and only if x, y∈Z






algebra-precalculus group-theory elementary-number-theory







share|cite|improve this question









New contributor




maks L is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











put on hold as off-topic by KReiser, Lord Shark the Unknown, max_zorn, Shaun, Hans Lundmark Jan 4 at 6:38


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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – KReiser, max_zorn, Shaun, Hans Lundmark

If this question can be reworded to fit the rules in the help center, please edit the question.









  • 6




    What about $m=sqrt2$, $n=-sqrt2$?
    – Lord Shark the Unknown
    Jan 4 at 5:29






  • 1




    What are the solutions to the quadratic $z^2-xz+y=(z-m)(z-n)=0$?
    – Mark Bennet
    Jan 4 at 5:52










  • z=m,n and m+n=x,mn=y
    – maks L
    Jan 4 at 6:13










  • Let $m,n=(1pmsqrt{5})/2$, then $x=m+n=1ne0$ and $y=mn=-1$.
    – Alexander Burstein
    Jan 4 at 6:13












  • Try $m,n=1pm sqrt{2}$. Then $x=2$ and $y=-1$
    – Sauhard Sharma
    Jan 4 at 6:23














-4












-4








-4







It is known that m+n=x, mn=y, m,n∈Q,Prove that m, n∈Z if and only if x, y∈Z






algebra-precalculus group-theory elementary-number-theory







share|cite|improve this question









New contributor




maks L is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











It is known that m+n=x, mn=y, m,n∈Q,Prove that m, n∈Z if and only if x, y∈Z






algebra-precalculus group-theory elementary-number-theory




algebra-precalculus group-theory elementary-number-theory






share|cite|improve this question









New contributor




maks L is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




maks L is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 6:29







maks L













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maks L is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked Jan 4 at 5:22









maks Lmaks L

11




11




New contributor




maks L is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





maks L is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






maks L is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




put on hold as off-topic by KReiser, Lord Shark the Unknown, max_zorn, Shaun, Hans Lundmark Jan 4 at 6:38


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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – KReiser, max_zorn, Shaun, Hans Lundmark

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 KReiser, Lord Shark the Unknown, max_zorn, Shaun, Hans Lundmark Jan 4 at 6:38


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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – KReiser, max_zorn, Shaun, Hans Lundmark

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 6




    What about $m=sqrt2$, $n=-sqrt2$?
    – Lord Shark the Unknown
    Jan 4 at 5:29






  • 1




    What are the solutions to the quadratic $z^2-xz+y=(z-m)(z-n)=0$?
    – Mark Bennet
    Jan 4 at 5:52










  • z=m,n and m+n=x,mn=y
    – maks L
    Jan 4 at 6:13










  • Let $m,n=(1pmsqrt{5})/2$, then $x=m+n=1ne0$ and $y=mn=-1$.
    – Alexander Burstein
    Jan 4 at 6:13












  • Try $m,n=1pm sqrt{2}$. Then $x=2$ and $y=-1$
    – Sauhard Sharma
    Jan 4 at 6:23














  • 6




    What about $m=sqrt2$, $n=-sqrt2$?
    – Lord Shark the Unknown
    Jan 4 at 5:29






  • 1




    What are the solutions to the quadratic $z^2-xz+y=(z-m)(z-n)=0$?
    – Mark Bennet
    Jan 4 at 5:52










  • z=m,n and m+n=x,mn=y
    – maks L
    Jan 4 at 6:13










  • Let $m,n=(1pmsqrt{5})/2$, then $x=m+n=1ne0$ and $y=mn=-1$.
    – Alexander Burstein
    Jan 4 at 6:13












  • Try $m,n=1pm sqrt{2}$. Then $x=2$ and $y=-1$
    – Sauhard Sharma
    Jan 4 at 6:23








6




6




What about $m=sqrt2$, $n=-sqrt2$?
– Lord Shark the Unknown
Jan 4 at 5:29




What about $m=sqrt2$, $n=-sqrt2$?
– Lord Shark the Unknown
Jan 4 at 5:29




1




1




What are the solutions to the quadratic $z^2-xz+y=(z-m)(z-n)=0$?
– Mark Bennet
Jan 4 at 5:52




What are the solutions to the quadratic $z^2-xz+y=(z-m)(z-n)=0$?
– Mark Bennet
Jan 4 at 5:52












z=m,n and m+n=x,mn=y
– maks L
Jan 4 at 6:13




z=m,n and m+n=x,mn=y
– maks L
Jan 4 at 6:13












Let $m,n=(1pmsqrt{5})/2$, then $x=m+n=1ne0$ and $y=mn=-1$.
– Alexander Burstein
Jan 4 at 6:13






Let $m,n=(1pmsqrt{5})/2$, then $x=m+n=1ne0$ and $y=mn=-1$.
– Alexander Burstein
Jan 4 at 6:13














Try $m,n=1pm sqrt{2}$. Then $x=2$ and $y=-1$
– Sauhard Sharma
Jan 4 at 6:23




Try $m,n=1pm sqrt{2}$. Then $x=2$ and $y=-1$
– Sauhard Sharma
Jan 4 at 6:23










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