How to solve such a quadratic congruence equation?












1












$begingroup$


I have the following equation: $y^2 equiv r^2 pmod n $



I know the values of y and n, I just need to find the values of r.



Assuming that $y = 12654$ and $n = 79061$, my working is as follows:



$ 12654^2$ mod $79061 = r^2$ mod $79061$



$25191 = r^2$ mod $79061$



The prime factorization of 79061 is $173*457$



Hence,



$r^2 = 25191$ mod $173$ $=>$ $106$ mod $173$



$r^2 = 25191$ mod $457$ $=>$ $56$ mod $457$



So now I have two equations,



$r^2 = 106$ mod $173$ and $r^2 = 56$ mod $457$



I am stuck here, I would appreciate if someone can help me move forward.



I've stumbled upon other similar questions where the answers show that they get rid of the squared but I cannot understand how they do it.






elementary-number-theory modular-arithmetic chinese-remainder-theorem







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    I have the following equation: $y^2 equiv r^2 pmod n $



    I know the values of y and n, I just need to find the values of r.



    Assuming that $y = 12654$ and $n = 79061$, my working is as follows:



    $ 12654^2$ mod $79061 = r^2$ mod $79061$



    $25191 = r^2$ mod $79061$



    The prime factorization of 79061 is $173*457$



    Hence,



    $r^2 = 25191$ mod $173$ $=>$ $106$ mod $173$



    $r^2 = 25191$ mod $457$ $=>$ $56$ mod $457$



    So now I have two equations,



    $r^2 = 106$ mod $173$ and $r^2 = 56$ mod $457$



    I am stuck here, I would appreciate if someone can help me move forward.



    I've stumbled upon other similar questions where the answers show that they get rid of the squared but I cannot understand how they do it.






    elementary-number-theory modular-arithmetic chinese-remainder-theorem







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I have the following equation: $y^2 equiv r^2 pmod n $



      I know the values of y and n, I just need to find the values of r.



      Assuming that $y = 12654$ and $n = 79061$, my working is as follows:



      $ 12654^2$ mod $79061 = r^2$ mod $79061$



      $25191 = r^2$ mod $79061$



      The prime factorization of 79061 is $173*457$



      Hence,



      $r^2 = 25191$ mod $173$ $=>$ $106$ mod $173$



      $r^2 = 25191$ mod $457$ $=>$ $56$ mod $457$



      So now I have two equations,



      $r^2 = 106$ mod $173$ and $r^2 = 56$ mod $457$



      I am stuck here, I would appreciate if someone can help me move forward.



      I've stumbled upon other similar questions where the answers show that they get rid of the squared but I cannot understand how they do it.






      elementary-number-theory modular-arithmetic chinese-remainder-theorem







      share|cite|improve this question









      $endgroup$




      I have the following equation: $y^2 equiv r^2 pmod n $



      I know the values of y and n, I just need to find the values of r.



      Assuming that $y = 12654$ and $n = 79061$, my working is as follows:



      $ 12654^2$ mod $79061 = r^2$ mod $79061$



      $25191 = r^2$ mod $79061$



      The prime factorization of 79061 is $173*457$



      Hence,



      $r^2 = 25191$ mod $173$ $=>$ $106$ mod $173$



      $r^2 = 25191$ mod $457$ $=>$ $56$ mod $457$



      So now I have two equations,



      $r^2 = 106$ mod $173$ and $r^2 = 56$ mod $457$



      I am stuck here, I would appreciate if someone can help me move forward.



      I've stumbled upon other similar questions where the answers show that they get rid of the squared but I cannot understand how they do it.






      elementary-number-theory modular-arithmetic chinese-remainder-theorem




      elementary-number-theory modular-arithmetic chinese-remainder-theorem






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 22 '18 at 11:22









      MulishiaMulishia

      62




      62






















          4 Answers
          4






          active

          oldest

          votes


















          1












          $begingroup$

          You know $r^2$ modulo $p$ and $q$ (the prime factors). There we have exactly two solutions: $y$ and $-y$ modulo $p$ resp. $q$. (we have a field modulo a prime so no more then $2$).



          Now the CRT now allows us to combine the $4$ pairs of solutions (corresponding to the 4 possible choices of sign) to $4$ solution modulo $n=pq$.



          So e.g. solve the systems $xequiv -y pmod p$, $x equiv y pmod q$ using the CRT formula (e.g. see wiki, constructive proof) and
          $xequiv y pmod p$, $x equiv -y pmod q$ to get the two extra solutions besides the already known solutions $y$ and $-y pmod{n} equiv n-y$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks for your input, I am aware that I have to use CRT in the end but my issue is how to get rid of the squared. Unfortunately I didn't understand your first part very well and I also think I didn't explain my question well enough as well. Basically my equation is similar to the answer in the following link math.stackexchange.com/a/335181/628342. I am stuck at this part: i.imgur.com/D7fENFp.png. I have no idea how he got rid of the squared and how the right hand side number became +-3. That's what I'm trying to understand.
            $endgroup$
            – Mulishia
            Dec 22 '18 at 13:11












          • $begingroup$
            @Mulishia The difference is that you are already given a solution, and the quoted question started with no info. You already know that $pm y pmod{p}$ are the solutions to $x^2 equiv r^2 pmod{p}$ while otherwise you have to solve an equation $x^2 equiv a pmod{p}$ from scratch. A solution existence can be checked with Lagrange symbols, a concrete solution from trial and error or knowing a generator for $mathbb{Z}_p$ etc.
            $endgroup$
            – Henno Brandsma
            Dec 22 '18 at 22:50












          • $begingroup$
            @Mulishia some general theory on this can be found here e.g. But note that you don't need it, really, as you're given $y$ already.
            $endgroup$
            – Henno Brandsma
            Dec 22 '18 at 22:54










          • $begingroup$
            @Mulishia You can continue with this answer where I completely solve an analogous example.
            $endgroup$
            – Bill Dubuque
            Dec 25 '18 at 20:15





















          1












          $begingroup$

          Similar to 'random' comment we have:



          $(y-r)(y+r)≡0mod n$



          $n=79061=173times457$



          Following cases can be considered:



          a: $y-r=173$ ⇒ $r=y-173=12654-173=12421$



          b: $y+r=173$ ⇒ $r=173-12654=-12421$



          And we have:



          $ 12654^2-(±12421)^2=55times 79061$



          c: $y-r=457$⇒ $r=12654+457=13111$



          d: $y+r=457$⇒ $r=-13111$



          And we have:



          $12654^2-13111^2=148.93...times79061$



          So $r= ±12421$ is acceptable.






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            The equation can be rewritten as $(y-r)(y+r)=y^2 -r^2equiv 0 pmod n$. Now each of the two factors of $n$ must divide at least one of the factors of $y^2 -r^2$.






            share|cite|improve this answer









            $endgroup$





















              0












              $begingroup$

              $$ 12654^2 - (91542 - 79061 m)^2 = 79061 (-79061 m^2 + 183084 m - 103968) $$
              $$ r=91542 - 79061 m $$






              share|cite|improve this answer









              $endgroup$













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                4 Answers
                4






                active

                oldest

                votes








                4 Answers
                4






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                1












                $begingroup$

                You know $r^2$ modulo $p$ and $q$ (the prime factors). There we have exactly two solutions: $y$ and $-y$ modulo $p$ resp. $q$. (we have a field modulo a prime so no more then $2$).



                Now the CRT now allows us to combine the $4$ pairs of solutions (corresponding to the 4 possible choices of sign) to $4$ solution modulo $n=pq$.



                So e.g. solve the systems $xequiv -y pmod p$, $x equiv y pmod q$ using the CRT formula (e.g. see wiki, constructive proof) and
                $xequiv y pmod p$, $x equiv -y pmod q$ to get the two extra solutions besides the already known solutions $y$ and $-y pmod{n} equiv n-y$.






                share|cite|improve this answer











                $endgroup$













                • $begingroup$
                  Thanks for your input, I am aware that I have to use CRT in the end but my issue is how to get rid of the squared. Unfortunately I didn't understand your first part very well and I also think I didn't explain my question well enough as well. Basically my equation is similar to the answer in the following link math.stackexchange.com/a/335181/628342. I am stuck at this part: i.imgur.com/D7fENFp.png. I have no idea how he got rid of the squared and how the right hand side number became +-3. That's what I'm trying to understand.
                  $endgroup$
                  – Mulishia
                  Dec 22 '18 at 13:11












                • $begingroup$
                  @Mulishia The difference is that you are already given a solution, and the quoted question started with no info. You already know that $pm y pmod{p}$ are the solutions to $x^2 equiv r^2 pmod{p}$ while otherwise you have to solve an equation $x^2 equiv a pmod{p}$ from scratch. A solution existence can be checked with Lagrange symbols, a concrete solution from trial and error or knowing a generator for $mathbb{Z}_p$ etc.
                  $endgroup$
                  – Henno Brandsma
                  Dec 22 '18 at 22:50












                • $begingroup$
                  @Mulishia some general theory on this can be found here e.g. But note that you don't need it, really, as you're given $y$ already.
                  $endgroup$
                  – Henno Brandsma
                  Dec 22 '18 at 22:54










                • $begingroup$
                  @Mulishia You can continue with this answer where I completely solve an analogous example.
                  $endgroup$
                  – Bill Dubuque
                  Dec 25 '18 at 20:15


















                1












                $begingroup$

                You know $r^2$ modulo $p$ and $q$ (the prime factors). There we have exactly two solutions: $y$ and $-y$ modulo $p$ resp. $q$. (we have a field modulo a prime so no more then $2$).



                Now the CRT now allows us to combine the $4$ pairs of solutions (corresponding to the 4 possible choices of sign) to $4$ solution modulo $n=pq$.



                So e.g. solve the systems $xequiv -y pmod p$, $x equiv y pmod q$ using the CRT formula (e.g. see wiki, constructive proof) and
                $xequiv y pmod p$, $x equiv -y pmod q$ to get the two extra solutions besides the already known solutions $y$ and $-y pmod{n} equiv n-y$.






                share|cite|improve this answer











                $endgroup$













                • $begingroup$
                  Thanks for your input, I am aware that I have to use CRT in the end but my issue is how to get rid of the squared. Unfortunately I didn't understand your first part very well and I also think I didn't explain my question well enough as well. Basically my equation is similar to the answer in the following link math.stackexchange.com/a/335181/628342. I am stuck at this part: i.imgur.com/D7fENFp.png. I have no idea how he got rid of the squared and how the right hand side number became +-3. That's what I'm trying to understand.
                  $endgroup$
                  – Mulishia
                  Dec 22 '18 at 13:11












                • $begingroup$
                  @Mulishia The difference is that you are already given a solution, and the quoted question started with no info. You already know that $pm y pmod{p}$ are the solutions to $x^2 equiv r^2 pmod{p}$ while otherwise you have to solve an equation $x^2 equiv a pmod{p}$ from scratch. A solution existence can be checked with Lagrange symbols, a concrete solution from trial and error or knowing a generator for $mathbb{Z}_p$ etc.
                  $endgroup$
                  – Henno Brandsma
                  Dec 22 '18 at 22:50












                • $begingroup$
                  @Mulishia some general theory on this can be found here e.g. But note that you don't need it, really, as you're given $y$ already.
                  $endgroup$
                  – Henno Brandsma
                  Dec 22 '18 at 22:54










                • $begingroup$
                  @Mulishia You can continue with this answer where I completely solve an analogous example.
                  $endgroup$
                  – Bill Dubuque
                  Dec 25 '18 at 20:15
















                1












                1








                1





                $begingroup$

                You know $r^2$ modulo $p$ and $q$ (the prime factors). There we have exactly two solutions: $y$ and $-y$ modulo $p$ resp. $q$. (we have a field modulo a prime so no more then $2$).



                Now the CRT now allows us to combine the $4$ pairs of solutions (corresponding to the 4 possible choices of sign) to $4$ solution modulo $n=pq$.



                So e.g. solve the systems $xequiv -y pmod p$, $x equiv y pmod q$ using the CRT formula (e.g. see wiki, constructive proof) and
                $xequiv y pmod p$, $x equiv -y pmod q$ to get the two extra solutions besides the already known solutions $y$ and $-y pmod{n} equiv n-y$.






                share|cite|improve this answer











                $endgroup$



                You know $r^2$ modulo $p$ and $q$ (the prime factors). There we have exactly two solutions: $y$ and $-y$ modulo $p$ resp. $q$. (we have a field modulo a prime so no more then $2$).



                Now the CRT now allows us to combine the $4$ pairs of solutions (corresponding to the 4 possible choices of sign) to $4$ solution modulo $n=pq$.



                So e.g. solve the systems $xequiv -y pmod p$, $x equiv y pmod q$ using the CRT formula (e.g. see wiki, constructive proof) and
                $xequiv y pmod p$, $x equiv -y pmod q$ to get the two extra solutions besides the already known solutions $y$ and $-y pmod{n} equiv n-y$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 22 '18 at 12:46

























                answered Dec 22 '18 at 12:38









                Henno BrandsmaHenno Brandsma

                111k348119




                111k348119












                • $begingroup$
                  Thanks for your input, I am aware that I have to use CRT in the end but my issue is how to get rid of the squared. Unfortunately I didn't understand your first part very well and I also think I didn't explain my question well enough as well. Basically my equation is similar to the answer in the following link math.stackexchange.com/a/335181/628342. I am stuck at this part: i.imgur.com/D7fENFp.png. I have no idea how he got rid of the squared and how the right hand side number became +-3. That's what I'm trying to understand.
                  $endgroup$
                  – Mulishia
                  Dec 22 '18 at 13:11












                • $begingroup$
                  @Mulishia The difference is that you are already given a solution, and the quoted question started with no info. You already know that $pm y pmod{p}$ are the solutions to $x^2 equiv r^2 pmod{p}$ while otherwise you have to solve an equation $x^2 equiv a pmod{p}$ from scratch. A solution existence can be checked with Lagrange symbols, a concrete solution from trial and error or knowing a generator for $mathbb{Z}_p$ etc.
                  $endgroup$
                  – Henno Brandsma
                  Dec 22 '18 at 22:50












                • $begingroup$
                  @Mulishia some general theory on this can be found here e.g. But note that you don't need it, really, as you're given $y$ already.
                  $endgroup$
                  – Henno Brandsma
                  Dec 22 '18 at 22:54










                • $begingroup$
                  @Mulishia You can continue with this answer where I completely solve an analogous example.
                  $endgroup$
                  – Bill Dubuque
                  Dec 25 '18 at 20:15




















                • $begingroup$
                  Thanks for your input, I am aware that I have to use CRT in the end but my issue is how to get rid of the squared. Unfortunately I didn't understand your first part very well and I also think I didn't explain my question well enough as well. Basically my equation is similar to the answer in the following link math.stackexchange.com/a/335181/628342. I am stuck at this part: i.imgur.com/D7fENFp.png. I have no idea how he got rid of the squared and how the right hand side number became +-3. That's what I'm trying to understand.
                  $endgroup$
                  – Mulishia
                  Dec 22 '18 at 13:11












                • $begingroup$
                  @Mulishia The difference is that you are already given a solution, and the quoted question started with no info. You already know that $pm y pmod{p}$ are the solutions to $x^2 equiv r^2 pmod{p}$ while otherwise you have to solve an equation $x^2 equiv a pmod{p}$ from scratch. A solution existence can be checked with Lagrange symbols, a concrete solution from trial and error or knowing a generator for $mathbb{Z}_p$ etc.
                  $endgroup$
                  – Henno Brandsma
                  Dec 22 '18 at 22:50












                • $begingroup$
                  @Mulishia some general theory on this can be found here e.g. But note that you don't need it, really, as you're given $y$ already.
                  $endgroup$
                  – Henno Brandsma
                  Dec 22 '18 at 22:54










                • $begingroup$
                  @Mulishia You can continue with this answer where I completely solve an analogous example.
                  $endgroup$
                  – Bill Dubuque
                  Dec 25 '18 at 20:15


















                $begingroup$
                Thanks for your input, I am aware that I have to use CRT in the end but my issue is how to get rid of the squared. Unfortunately I didn't understand your first part very well and I also think I didn't explain my question well enough as well. Basically my equation is similar to the answer in the following link math.stackexchange.com/a/335181/628342. I am stuck at this part: i.imgur.com/D7fENFp.png. I have no idea how he got rid of the squared and how the right hand side number became +-3. That's what I'm trying to understand.
                $endgroup$
                – Mulishia
                Dec 22 '18 at 13:11






                $begingroup$
                Thanks for your input, I am aware that I have to use CRT in the end but my issue is how to get rid of the squared. Unfortunately I didn't understand your first part very well and I also think I didn't explain my question well enough as well. Basically my equation is similar to the answer in the following link math.stackexchange.com/a/335181/628342. I am stuck at this part: i.imgur.com/D7fENFp.png. I have no idea how he got rid of the squared and how the right hand side number became +-3. That's what I'm trying to understand.
                $endgroup$
                – Mulishia
                Dec 22 '18 at 13:11














                $begingroup$
                @Mulishia The difference is that you are already given a solution, and the quoted question started with no info. You already know that $pm y pmod{p}$ are the solutions to $x^2 equiv r^2 pmod{p}$ while otherwise you have to solve an equation $x^2 equiv a pmod{p}$ from scratch. A solution existence can be checked with Lagrange symbols, a concrete solution from trial and error or knowing a generator for $mathbb{Z}_p$ etc.
                $endgroup$
                – Henno Brandsma
                Dec 22 '18 at 22:50






                $begingroup$
                @Mulishia The difference is that you are already given a solution, and the quoted question started with no info. You already know that $pm y pmod{p}$ are the solutions to $x^2 equiv r^2 pmod{p}$ while otherwise you have to solve an equation $x^2 equiv a pmod{p}$ from scratch. A solution existence can be checked with Lagrange symbols, a concrete solution from trial and error or knowing a generator for $mathbb{Z}_p$ etc.
                $endgroup$
                – Henno Brandsma
                Dec 22 '18 at 22:50














                $begingroup$
                @Mulishia some general theory on this can be found here e.g. But note that you don't need it, really, as you're given $y$ already.
                $endgroup$
                – Henno Brandsma
                Dec 22 '18 at 22:54




                $begingroup$
                @Mulishia some general theory on this can be found here e.g. But note that you don't need it, really, as you're given $y$ already.
                $endgroup$
                – Henno Brandsma
                Dec 22 '18 at 22:54












                $begingroup$
                @Mulishia You can continue with this answer where I completely solve an analogous example.
                $endgroup$
                – Bill Dubuque
                Dec 25 '18 at 20:15






                $begingroup$
                @Mulishia You can continue with this answer where I completely solve an analogous example.
                $endgroup$
                – Bill Dubuque
                Dec 25 '18 at 20:15













                1












                $begingroup$

                Similar to 'random' comment we have:



                $(y-r)(y+r)≡0mod n$



                $n=79061=173times457$



                Following cases can be considered:



                a: $y-r=173$ ⇒ $r=y-173=12654-173=12421$



                b: $y+r=173$ ⇒ $r=173-12654=-12421$



                And we have:



                $ 12654^2-(±12421)^2=55times 79061$



                c: $y-r=457$⇒ $r=12654+457=13111$



                d: $y+r=457$⇒ $r=-13111$



                And we have:



                $12654^2-13111^2=148.93...times79061$



                So $r= ±12421$ is acceptable.






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  Similar to 'random' comment we have:



                  $(y-r)(y+r)≡0mod n$



                  $n=79061=173times457$



                  Following cases can be considered:



                  a: $y-r=173$ ⇒ $r=y-173=12654-173=12421$



                  b: $y+r=173$ ⇒ $r=173-12654=-12421$



                  And we have:



                  $ 12654^2-(±12421)^2=55times 79061$



                  c: $y-r=457$⇒ $r=12654+457=13111$



                  d: $y+r=457$⇒ $r=-13111$



                  And we have:



                  $12654^2-13111^2=148.93...times79061$



                  So $r= ±12421$ is acceptable.






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    Similar to 'random' comment we have:



                    $(y-r)(y+r)≡0mod n$



                    $n=79061=173times457$



                    Following cases can be considered:



                    a: $y-r=173$ ⇒ $r=y-173=12654-173=12421$



                    b: $y+r=173$ ⇒ $r=173-12654=-12421$



                    And we have:



                    $ 12654^2-(±12421)^2=55times 79061$



                    c: $y-r=457$⇒ $r=12654+457=13111$



                    d: $y+r=457$⇒ $r=-13111$



                    And we have:



                    $12654^2-13111^2=148.93...times79061$



                    So $r= ±12421$ is acceptable.






                    share|cite|improve this answer









                    $endgroup$



                    Similar to 'random' comment we have:



                    $(y-r)(y+r)≡0mod n$



                    $n=79061=173times457$



                    Following cases can be considered:



                    a: $y-r=173$ ⇒ $r=y-173=12654-173=12421$



                    b: $y+r=173$ ⇒ $r=173-12654=-12421$



                    And we have:



                    $ 12654^2-(±12421)^2=55times 79061$



                    c: $y-r=457$⇒ $r=12654+457=13111$



                    d: $y+r=457$⇒ $r=-13111$



                    And we have:



                    $12654^2-13111^2=148.93...times79061$



                    So $r= ±12421$ is acceptable.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 22 '18 at 13:20









                    siroussirous

                    1,6871514




                    1,6871514























                        0












                        $begingroup$

                        The equation can be rewritten as $(y-r)(y+r)=y^2 -r^2equiv 0 pmod n$. Now each of the two factors of $n$ must divide at least one of the factors of $y^2 -r^2$.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          The equation can be rewritten as $(y-r)(y+r)=y^2 -r^2equiv 0 pmod n$. Now each of the two factors of $n$ must divide at least one of the factors of $y^2 -r^2$.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            The equation can be rewritten as $(y-r)(y+r)=y^2 -r^2equiv 0 pmod n$. Now each of the two factors of $n$ must divide at least one of the factors of $y^2 -r^2$.






                            share|cite|improve this answer









                            $endgroup$



                            The equation can be rewritten as $(y-r)(y+r)=y^2 -r^2equiv 0 pmod n$. Now each of the two factors of $n$ must divide at least one of the factors of $y^2 -r^2$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 22 '18 at 12:29









                            randomrandom

                            57626




                            57626























                                0












                                $begingroup$

                                $$ 12654^2 - (91542 - 79061 m)^2 = 79061 (-79061 m^2 + 183084 m - 103968) $$
                                $$ r=91542 - 79061 m $$






                                share|cite|improve this answer









                                $endgroup$


















                                  0












                                  $begingroup$

                                  $$ 12654^2 - (91542 - 79061 m)^2 = 79061 (-79061 m^2 + 183084 m - 103968) $$
                                  $$ r=91542 - 79061 m $$






                                  share|cite|improve this answer









                                  $endgroup$
















                                    0












                                    0








                                    0





                                    $begingroup$

                                    $$ 12654^2 - (91542 - 79061 m)^2 = 79061 (-79061 m^2 + 183084 m - 103968) $$
                                    $$ r=91542 - 79061 m $$






                                    share|cite|improve this answer









                                    $endgroup$



                                    $$ 12654^2 - (91542 - 79061 m)^2 = 79061 (-79061 m^2 + 183084 m - 103968) $$
                                    $$ r=91542 - 79061 m $$







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Dec 22 '18 at 18:17









                                    S. I.S. I.

                                    112




                                    112






























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