Legendre (p/q) = 1 what can we conclude about the solution of $n^2 equiv p$ mod q? [on hold]












-4














Let p and q be odd primes and Legendre (p/q)=1. So we know that there exists (at least) one whole number n for which $n^2 equiv p$ mod q holds.
Is it possible to deduce that there must be an odd n from these facts, without having a solution procedure for $n^2 equiv p$ mod q, where we can possibly see wether even and/or odd solutions occur.
I've calculated some examples, so I'v seen there might be always two solutions, one n even, one n odd.
Please for now only yes or no as answer and perhaps a little hint.
Thank you.










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put on hold as off-topic by Eevee Trainer, Lord Shark the Unknown, Saad, Lord_Farin, Cesareo Dec 26 at 10:24


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." – Eevee Trainer, Saad, Lord_Farin

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  • $n^2equiv ppmod qRightarrow (n+q)^2equiv ppmod q$
    – Minz
    Dec 26 at 9:22


















-4














Let p and q be odd primes and Legendre (p/q)=1. So we know that there exists (at least) one whole number n for which $n^2 equiv p$ mod q holds.
Is it possible to deduce that there must be an odd n from these facts, without having a solution procedure for $n^2 equiv p$ mod q, where we can possibly see wether even and/or odd solutions occur.
I've calculated some examples, so I'v seen there might be always two solutions, one n even, one n odd.
Please for now only yes or no as answer and perhaps a little hint.
Thank you.










share|cite|improve this question









New contributor




i-have-no-clue 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 Eevee Trainer, Lord Shark the Unknown, Saad, Lord_Farin, Cesareo Dec 26 at 10:24


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." – Eevee Trainer, Saad, Lord_Farin

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













  • $n^2equiv ppmod qRightarrow (n+q)^2equiv ppmod q$
    – Minz
    Dec 26 at 9:22
















-4












-4








-4







Let p and q be odd primes and Legendre (p/q)=1. So we know that there exists (at least) one whole number n for which $n^2 equiv p$ mod q holds.
Is it possible to deduce that there must be an odd n from these facts, without having a solution procedure for $n^2 equiv p$ mod q, where we can possibly see wether even and/or odd solutions occur.
I've calculated some examples, so I'v seen there might be always two solutions, one n even, one n odd.
Please for now only yes or no as answer and perhaps a little hint.
Thank you.










share|cite|improve this question









New contributor




i-have-no-clue is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











Let p and q be odd primes and Legendre (p/q)=1. So we know that there exists (at least) one whole number n for which $n^2 equiv p$ mod q holds.
Is it possible to deduce that there must be an odd n from these facts, without having a solution procedure for $n^2 equiv p$ mod q, where we can possibly see wether even and/or odd solutions occur.
I've calculated some examples, so I'v seen there might be always two solutions, one n even, one n odd.
Please for now only yes or no as answer and perhaps a little hint.
Thank you.







modular-arithmetic






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edited Dec 26 at 8:54





















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asked Dec 26 at 8:06









i-have-no-clue

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put on hold as off-topic by Eevee Trainer, Lord Shark the Unknown, Saad, Lord_Farin, Cesareo Dec 26 at 10:24


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." – Eevee Trainer, Saad, Lord_Farin

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 Eevee Trainer, Lord Shark the Unknown, Saad, Lord_Farin, Cesareo Dec 26 at 10:24


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." – Eevee Trainer, Saad, Lord_Farin

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












  • $n^2equiv ppmod qRightarrow (n+q)^2equiv ppmod q$
    – Minz
    Dec 26 at 9:22




















  • $n^2equiv ppmod qRightarrow (n+q)^2equiv ppmod q$
    – Minz
    Dec 26 at 9:22


















$n^2equiv ppmod qRightarrow (n+q)^2equiv ppmod q$
– Minz
Dec 26 at 9:22






$n^2equiv ppmod qRightarrow (n+q)^2equiv ppmod q$
– Minz
Dec 26 at 9:22

















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