Solvability of congruences in number theory.

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Prove the following lemma.



Let $p_1,q_1,....,p_r,q_r,m$ be numbers with $gcd(p_1,....,p_r,m)=1$. Then the congruences $$p_i B equiv q_i (textrmmod m),,, ,,, i=1,...,r$$ have a unique solution modulo $m$ if and only if for all $i,j=1,......,r$ we have $$p_i q_j equiv p_j q_i (textrmmod m)$$.



I have tried using the congruences formula from the elementary number theory but didn't succeed. I would really appreciate if someone can give me an explicit proof.




number-theory modular-arithmetic quadratic-forms




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    up vote
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    down vote

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    Prove the following lemma.



    Let $p_1,q_1,....,p_r,q_r,m$ be numbers with $gcd(p_1,....,p_r,m)=1$. Then the congruences $$p_i B equiv q_i (textrmmod m),,, ,,, i=1,...,r$$ have a unique solution modulo $m$ if and only if for all $i,j=1,......,r$ we have $$p_i q_j equiv p_j q_i (textrmmod m)$$.



    I have tried using the congruences formula from the elementary number theory but didn't succeed. I would really appreciate if someone can give me an explicit proof.




    number-theory modular-arithmetic quadratic-forms




    share|cite|improve this question





















      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Prove the following lemma.



      Let $p_1,q_1,....,p_r,q_r,m$ be numbers with $gcd(p_1,....,p_r,m)=1$. Then the congruences $$p_i B equiv q_i (textrmmod m),,, ,,, i=1,...,r$$ have a unique solution modulo $m$ if and only if for all $i,j=1,......,r$ we have $$p_i q_j equiv p_j q_i (textrmmod m)$$.



      I have tried using the congruences formula from the elementary number theory but didn't succeed. I would really appreciate if someone can give me an explicit proof.




      number-theory modular-arithmetic quadratic-forms




      share|cite|improve this question











      Prove the following lemma.



      Let $p_1,q_1,....,p_r,q_r,m$ be numbers with $gcd(p_1,....,p_r,m)=1$. Then the congruences $$p_i B equiv q_i (textrmmod m),,, ,,, i=1,...,r$$ have a unique solution modulo $m$ if and only if for all $i,j=1,......,r$ we have $$p_i q_j equiv p_j q_i (textrmmod m)$$.



      I have tried using the congruences formula from the elementary number theory but didn't succeed. I would really appreciate if someone can give me an explicit proof.





      number-theory modular-arithmetic quadratic-forms





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      share|cite|improve this question




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      asked 2 days ago









      antony james

      105




      105




















          1 Answer
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          $p_iBequiv q_ibmod m$ all $i$ implies $Bequiv p_i^-1q_ibmod m$ all $i$, so
          $p_i^-1q_iequiv p_j^-1q_jbmod m$ all $i$, $j$, and we're done.






          share|cite|improve this answer





















          • I think you have to prove $p^-1$ always exist.
            – Mira from Earth
            2 days ago










          • You can use Femet's little theorem to prove.
            – Mira from Earth
            2 days ago










          • Could you prove it explicitly? -@MirafromEarth
            – antony james
            2 days ago










          • By hypothesis, $gcd(p_i,m)=1$ for all $i$, therefore, $p_i^-1$ exists modulo m. This is in Chapter One of every intro Number Theory text, and certainly doesn't need Fermat.
            – Gerry Myerson
            2 days ago










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          1 Answer
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          active

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          active

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          up vote
          1
          down vote



          accepted










          $p_iBequiv q_ibmod m$ all $i$ implies $Bequiv p_i^-1q_ibmod m$ all $i$, so
          $p_i^-1q_iequiv p_j^-1q_jbmod m$ all $i$, $j$, and we're done.






          share|cite|improve this answer





















          • I think you have to prove $p^-1$ always exist.
            – Mira from Earth
            2 days ago










          • You can use Femet's little theorem to prove.
            – Mira from Earth
            2 days ago










          • Could you prove it explicitly? -@MirafromEarth
            – antony james
            2 days ago










          • By hypothesis, $gcd(p_i,m)=1$ for all $i$, therefore, $p_i^-1$ exists modulo m. This is in Chapter One of every intro Number Theory text, and certainly doesn't need Fermat.
            – Gerry Myerson
            2 days ago














          up vote
          1
          down vote



          accepted










          $p_iBequiv q_ibmod m$ all $i$ implies $Bequiv p_i^-1q_ibmod m$ all $i$, so
          $p_i^-1q_iequiv p_j^-1q_jbmod m$ all $i$, $j$, and we're done.






          share|cite|improve this answer





















          • I think you have to prove $p^-1$ always exist.
            – Mira from Earth
            2 days ago










          • You can use Femet's little theorem to prove.
            – Mira from Earth
            2 days ago










          • Could you prove it explicitly? -@MirafromEarth
            – antony james
            2 days ago










          • By hypothesis, $gcd(p_i,m)=1$ for all $i$, therefore, $p_i^-1$ exists modulo m. This is in Chapter One of every intro Number Theory text, and certainly doesn't need Fermat.
            – Gerry Myerson
            2 days ago












          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          $p_iBequiv q_ibmod m$ all $i$ implies $Bequiv p_i^-1q_ibmod m$ all $i$, so
          $p_i^-1q_iequiv p_j^-1q_jbmod m$ all $i$, $j$, and we're done.






          share|cite|improve this answer













          $p_iBequiv q_ibmod m$ all $i$ implies $Bequiv p_i^-1q_ibmod m$ all $i$, so
          $p_i^-1q_iequiv p_j^-1q_jbmod m$ all $i$, $j$, and we're done.







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered 2 days ago









          Gerry Myerson

          142k7142292




          142k7142292











          • I think you have to prove $p^-1$ always exist.
            – Mira from Earth
            2 days ago










          • You can use Femet's little theorem to prove.
            – Mira from Earth
            2 days ago










          • Could you prove it explicitly? -@MirafromEarth
            – antony james
            2 days ago










          • By hypothesis, $gcd(p_i,m)=1$ for all $i$, therefore, $p_i^-1$ exists modulo m. This is in Chapter One of every intro Number Theory text, and certainly doesn't need Fermat.
            – Gerry Myerson
            2 days ago
















          • I think you have to prove $p^-1$ always exist.
            – Mira from Earth
            2 days ago










          • You can use Femet's little theorem to prove.
            – Mira from Earth
            2 days ago










          • Could you prove it explicitly? -@MirafromEarth
            – antony james
            2 days ago










          • By hypothesis, $gcd(p_i,m)=1$ for all $i$, therefore, $p_i^-1$ exists modulo m. This is in Chapter One of every intro Number Theory text, and certainly doesn't need Fermat.
            – Gerry Myerson
            2 days ago















          I think you have to prove $p^-1$ always exist.
          – Mira from Earth
          2 days ago




          I think you have to prove $p^-1$ always exist.
          – Mira from Earth
          2 days ago












          You can use Femet's little theorem to prove.
          – Mira from Earth
          2 days ago




          You can use Femet's little theorem to prove.
          – Mira from Earth
          2 days ago












          Could you prove it explicitly? -@MirafromEarth
          – antony james
          2 days ago




          Could you prove it explicitly? -@MirafromEarth
          – antony james
          2 days ago












          By hypothesis, $gcd(p_i,m)=1$ for all $i$, therefore, $p_i^-1$ exists modulo m. This is in Chapter One of every intro Number Theory text, and certainly doesn't need Fermat.
          – Gerry Myerson
          2 days ago




          By hypothesis, $gcd(p_i,m)=1$ for all $i$, therefore, $p_i^-1$ exists modulo m. This is in Chapter One of every intro Number Theory text, and certainly doesn't need Fermat.
          – Gerry Myerson
          2 days ago












           

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