Showing two different moduli are same












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$begingroup$



  1. $t$ is an integer coprime to primes $p,a,b$.


  2. $max(pa,pb)<t<pab$ and $t'equiv tbmod p$ and $t'in[-p^{1/2},p^{1/2}]$ holds.


  3. $a,bin[p^{1/4},2p^{1/4}]$ holds.


  4. $t''equiv tabmod pb$ and $t''in[-p^{3/4},p^{3/4}]$


  5. Denote $r=tabbmod pb$.



There are two ways to interpret $r$.




  1. $tabbmod pbequiv (tabmod p)bequiv(((tbmod p)(abmod p))bmod p)b=(t'abmod p)b$ and so $r=t'ab$ holds over $mathbb Z$ since $t'a<p$.


  2. $tabbmod pbequiv((tabmod pb)(bbmod pb))bmod pbequiv((tabmod pb)b)bmod pbequiv((tabmod pb)bmod p)bequiv(t''bmod p)b$



How do we show $t''equiv(tabmod pb)$ gives $t''=t'a$ over $mathbb Z$? If they are not equal which is correct route to compute $r$?






elementary-number-theory modular-arithmetic







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    0












    $begingroup$



    1. $t$ is an integer coprime to primes $p,a,b$.


    2. $max(pa,pb)<t<pab$ and $t'equiv tbmod p$ and $t'in[-p^{1/2},p^{1/2}]$ holds.


    3. $a,bin[p^{1/4},2p^{1/4}]$ holds.


    4. $t''equiv tabmod pb$ and $t''in[-p^{3/4},p^{3/4}]$


    5. Denote $r=tabbmod pb$.



    There are two ways to interpret $r$.




    1. $tabbmod pbequiv (tabmod p)bequiv(((tbmod p)(abmod p))bmod p)b=(t'abmod p)b$ and so $r=t'ab$ holds over $mathbb Z$ since $t'a<p$.


    2. $tabbmod pbequiv((tabmod pb)(bbmod pb))bmod pbequiv((tabmod pb)b)bmod pbequiv((tabmod pb)bmod p)bequiv(t''bmod p)b$



    How do we show $t''equiv(tabmod pb)$ gives $t''=t'a$ over $mathbb Z$? If they are not equal which is correct route to compute $r$?






    elementary-number-theory modular-arithmetic







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$



      1. $t$ is an integer coprime to primes $p,a,b$.


      2. $max(pa,pb)<t<pab$ and $t'equiv tbmod p$ and $t'in[-p^{1/2},p^{1/2}]$ holds.


      3. $a,bin[p^{1/4},2p^{1/4}]$ holds.


      4. $t''equiv tabmod pb$ and $t''in[-p^{3/4},p^{3/4}]$


      5. Denote $r=tabbmod pb$.



      There are two ways to interpret $r$.




      1. $tabbmod pbequiv (tabmod p)bequiv(((tbmod p)(abmod p))bmod p)b=(t'abmod p)b$ and so $r=t'ab$ holds over $mathbb Z$ since $t'a<p$.


      2. $tabbmod pbequiv((tabmod pb)(bbmod pb))bmod pbequiv((tabmod pb)b)bmod pbequiv((tabmod pb)bmod p)bequiv(t''bmod p)b$



      How do we show $t''equiv(tabmod pb)$ gives $t''=t'a$ over $mathbb Z$? If they are not equal which is correct route to compute $r$?






      elementary-number-theory modular-arithmetic







      share|cite|improve this question











      $endgroup$





      1. $t$ is an integer coprime to primes $p,a,b$.


      2. $max(pa,pb)<t<pab$ and $t'equiv tbmod p$ and $t'in[-p^{1/2},p^{1/2}]$ holds.


      3. $a,bin[p^{1/4},2p^{1/4}]$ holds.


      4. $t''equiv tabmod pb$ and $t''in[-p^{3/4},p^{3/4}]$


      5. Denote $r=tabbmod pb$.



      There are two ways to interpret $r$.




      1. $tabbmod pbequiv (tabmod p)bequiv(((tbmod p)(abmod p))bmod p)b=(t'abmod p)b$ and so $r=t'ab$ holds over $mathbb Z$ since $t'a<p$.


      2. $tabbmod pbequiv((tabmod pb)(bbmod pb))bmod pbequiv((tabmod pb)b)bmod pbequiv((tabmod pb)bmod p)bequiv(t''bmod p)b$



      How do we show $t''equiv(tabmod pb)$ gives $t''=t'a$ over $mathbb Z$? If they are not equal which is correct route to compute $r$?






      elementary-number-theory modular-arithmetic




      elementary-number-theory modular-arithmetic






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 15 at 19:59







      Brout

















      asked Jan 15 at 19:27









      BroutBrout

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