Solution to $x^2 equiv 1$ (mod $pq$), $p,q geq 3$ primes.












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Found the following piece online and got stuck:



Let $p,q geq 3$ be different primes. Show that there is an integer $x$ such that $x^2 equiv 1$ (mod pq) with $x$ neither congruent with $1$ or $−1$ (mod pq).



Would appreciate some help.






number-theory modular-arithmetic







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    0












    $begingroup$


    Found the following piece online and got stuck:



    Let $p,q geq 3$ be different primes. Show that there is an integer $x$ such that $x^2 equiv 1$ (mod pq) with $x$ neither congruent with $1$ or $−1$ (mod pq).



    Would appreciate some help.






    number-theory modular-arithmetic







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Found the following piece online and got stuck:



      Let $p,q geq 3$ be different primes. Show that there is an integer $x$ such that $x^2 equiv 1$ (mod pq) with $x$ neither congruent with $1$ or $−1$ (mod pq).



      Would appreciate some help.






      number-theory modular-arithmetic







      share|cite|improve this question











      $endgroup$




      Found the following piece online and got stuck:



      Let $p,q geq 3$ be different primes. Show that there is an integer $x$ such that $x^2 equiv 1$ (mod pq) with $x$ neither congruent with $1$ or $−1$ (mod pq).



      Would appreciate some help.






      number-theory modular-arithmetic




      number-theory modular-arithmetic






      share|cite|improve this question















      share|cite|improve this question













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      edited Jan 15 at 0:52







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      asked Jan 15 at 0:45









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          3 Answers
          3






          active

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          3












          $begingroup$

          By Bezout we have $Ap+Bq=1$. Now consider $x=Bq-Ap$, modulo $p$ and $q$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            $x$ should be $neq $1 or -1 (mod $ pq $ )
            $endgroup$
            – Sota Antonino
            Jan 15 at 0:56












          • $begingroup$
            @SotaAntonino With $p=3$ and $q=5$, You get $x=11$ ...
            $endgroup$
            – Donald Splutterwit
            Jan 15 at 0:59










          • $begingroup$
            $11 neq pm 1 pmod{15}$.
            $endgroup$
            – Donald Splutterwit
            Jan 15 at 1:07





















          1












          $begingroup$

          By the Chinese Remainder Theorem, there exists $x$ such that $x equiv 1 pmod{p}$ and $x equiv -1 pmod{q}$. Then $x^2 equiv 1 pmod{p}$ and $x^2 equiv 1 pmod{q}$, so again by (the uniqueness part of) the Chinese Remainder Theorem it follows that $x^2 equiv 1 pmod{pq}$. However, we cannot have $x equiv 1 pmod{pq}$, or this would imply $x equiv 1 pmod{q}$, so $1 equiv -1 pmod{q} Rightarrow q mid 2$ contradicting the assumption that $q ge 3$. Similarly, from $p ge 3$ we get $x notequiv -1 pmod{pq}$.






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            I prefer somehow to have a "structural" explanation. By the CRT, we have an isomorphism of rings $mathbf Z/pqcong mathbf Z/p times mathbf Z/q$. For $x in mathbf Z$, let us adopt the obvious notation $x_{pq}=(x_p , x_q)$ in the respective quotient rings. We must solve ${x_{pq}}^2=1_{pq}$ , or equivalently $(x_{pq}-1_{pq})(x_{pq}+1_{pq})=0_{pq}$, but we cannot conclude directly because $mathbf Z/pq$ admits divisors of zero, i.e. this ring is not a domain. Working in $mathbf Z/p times mathbf Z/q$ instead, we get immediately the necessary and sufficient condition $x_p=pm 1_p$ and $x_q=pm 1_q$ because $mathbf Z/p$ and $mathbf Z/q$ are fields. Moreover, $x_{pq}=1_{pq}$ iff $x_p=1_p$ and $x_q=1_q$. This means that, if $p$ and $q$ are odd, the non trivial solutions (viewed in $mathbf Z/pqcong mathbf Z/p times mathbf Z/q$ ) are $(1_p , {-1}_q), ({-1}_p , 1_q)$ and $({-1}_p , {-1}_q)$ .






            share|cite|improve this answer









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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              3












              $begingroup$

              By Bezout we have $Ap+Bq=1$. Now consider $x=Bq-Ap$, modulo $p$ and $q$.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                $x$ should be $neq $1 or -1 (mod $ pq $ )
                $endgroup$
                – Sota Antonino
                Jan 15 at 0:56












              • $begingroup$
                @SotaAntonino With $p=3$ and $q=5$, You get $x=11$ ...
                $endgroup$
                – Donald Splutterwit
                Jan 15 at 0:59










              • $begingroup$
                $11 neq pm 1 pmod{15}$.
                $endgroup$
                – Donald Splutterwit
                Jan 15 at 1:07


















              3












              $begingroup$

              By Bezout we have $Ap+Bq=1$. Now consider $x=Bq-Ap$, modulo $p$ and $q$.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                $x$ should be $neq $1 or -1 (mod $ pq $ )
                $endgroup$
                – Sota Antonino
                Jan 15 at 0:56












              • $begingroup$
                @SotaAntonino With $p=3$ and $q=5$, You get $x=11$ ...
                $endgroup$
                – Donald Splutterwit
                Jan 15 at 0:59










              • $begingroup$
                $11 neq pm 1 pmod{15}$.
                $endgroup$
                – Donald Splutterwit
                Jan 15 at 1:07
















              3












              3








              3





              $begingroup$

              By Bezout we have $Ap+Bq=1$. Now consider $x=Bq-Ap$, modulo $p$ and $q$.






              share|cite|improve this answer









              $endgroup$



              By Bezout we have $Ap+Bq=1$. Now consider $x=Bq-Ap$, modulo $p$ and $q$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Jan 15 at 0:52









              Donald SplutterwitDonald Splutterwit

              23.1k21446




              23.1k21446












              • $begingroup$
                $x$ should be $neq $1 or -1 (mod $ pq $ )
                $endgroup$
                – Sota Antonino
                Jan 15 at 0:56












              • $begingroup$
                @SotaAntonino With $p=3$ and $q=5$, You get $x=11$ ...
                $endgroup$
                – Donald Splutterwit
                Jan 15 at 0:59










              • $begingroup$
                $11 neq pm 1 pmod{15}$.
                $endgroup$
                – Donald Splutterwit
                Jan 15 at 1:07




















              • $begingroup$
                $x$ should be $neq $1 or -1 (mod $ pq $ )
                $endgroup$
                – Sota Antonino
                Jan 15 at 0:56












              • $begingroup$
                @SotaAntonino With $p=3$ and $q=5$, You get $x=11$ ...
                $endgroup$
                – Donald Splutterwit
                Jan 15 at 0:59










              • $begingroup$
                $11 neq pm 1 pmod{15}$.
                $endgroup$
                – Donald Splutterwit
                Jan 15 at 1:07


















              $begingroup$
              $x$ should be $neq $1 or -1 (mod $ pq $ )
              $endgroup$
              – Sota Antonino
              Jan 15 at 0:56






              $begingroup$
              $x$ should be $neq $1 or -1 (mod $ pq $ )
              $endgroup$
              – Sota Antonino
              Jan 15 at 0:56














              $begingroup$
              @SotaAntonino With $p=3$ and $q=5$, You get $x=11$ ...
              $endgroup$
              – Donald Splutterwit
              Jan 15 at 0:59




              $begingroup$
              @SotaAntonino With $p=3$ and $q=5$, You get $x=11$ ...
              $endgroup$
              – Donald Splutterwit
              Jan 15 at 0:59












              $begingroup$
              $11 neq pm 1 pmod{15}$.
              $endgroup$
              – Donald Splutterwit
              Jan 15 at 1:07






              $begingroup$
              $11 neq pm 1 pmod{15}$.
              $endgroup$
              – Donald Splutterwit
              Jan 15 at 1:07













              1












              $begingroup$

              By the Chinese Remainder Theorem, there exists $x$ such that $x equiv 1 pmod{p}$ and $x equiv -1 pmod{q}$. Then $x^2 equiv 1 pmod{p}$ and $x^2 equiv 1 pmod{q}$, so again by (the uniqueness part of) the Chinese Remainder Theorem it follows that $x^2 equiv 1 pmod{pq}$. However, we cannot have $x equiv 1 pmod{pq}$, or this would imply $x equiv 1 pmod{q}$, so $1 equiv -1 pmod{q} Rightarrow q mid 2$ contradicting the assumption that $q ge 3$. Similarly, from $p ge 3$ we get $x notequiv -1 pmod{pq}$.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                By the Chinese Remainder Theorem, there exists $x$ such that $x equiv 1 pmod{p}$ and $x equiv -1 pmod{q}$. Then $x^2 equiv 1 pmod{p}$ and $x^2 equiv 1 pmod{q}$, so again by (the uniqueness part of) the Chinese Remainder Theorem it follows that $x^2 equiv 1 pmod{pq}$. However, we cannot have $x equiv 1 pmod{pq}$, or this would imply $x equiv 1 pmod{q}$, so $1 equiv -1 pmod{q} Rightarrow q mid 2$ contradicting the assumption that $q ge 3$. Similarly, from $p ge 3$ we get $x notequiv -1 pmod{pq}$.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  By the Chinese Remainder Theorem, there exists $x$ such that $x equiv 1 pmod{p}$ and $x equiv -1 pmod{q}$. Then $x^2 equiv 1 pmod{p}$ and $x^2 equiv 1 pmod{q}$, so again by (the uniqueness part of) the Chinese Remainder Theorem it follows that $x^2 equiv 1 pmod{pq}$. However, we cannot have $x equiv 1 pmod{pq}$, or this would imply $x equiv 1 pmod{q}$, so $1 equiv -1 pmod{q} Rightarrow q mid 2$ contradicting the assumption that $q ge 3$. Similarly, from $p ge 3$ we get $x notequiv -1 pmod{pq}$.






                  share|cite|improve this answer









                  $endgroup$



                  By the Chinese Remainder Theorem, there exists $x$ such that $x equiv 1 pmod{p}$ and $x equiv -1 pmod{q}$. Then $x^2 equiv 1 pmod{p}$ and $x^2 equiv 1 pmod{q}$, so again by (the uniqueness part of) the Chinese Remainder Theorem it follows that $x^2 equiv 1 pmod{pq}$. However, we cannot have $x equiv 1 pmod{pq}$, or this would imply $x equiv 1 pmod{q}$, so $1 equiv -1 pmod{q} Rightarrow q mid 2$ contradicting the assumption that $q ge 3$. Similarly, from $p ge 3$ we get $x notequiv -1 pmod{pq}$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 15 at 1:28









                  Daniel ScheplerDaniel Schepler

                  9,4291821




                  9,4291821























                      0












                      $begingroup$

                      I prefer somehow to have a "structural" explanation. By the CRT, we have an isomorphism of rings $mathbf Z/pqcong mathbf Z/p times mathbf Z/q$. For $x in mathbf Z$, let us adopt the obvious notation $x_{pq}=(x_p , x_q)$ in the respective quotient rings. We must solve ${x_{pq}}^2=1_{pq}$ , or equivalently $(x_{pq}-1_{pq})(x_{pq}+1_{pq})=0_{pq}$, but we cannot conclude directly because $mathbf Z/pq$ admits divisors of zero, i.e. this ring is not a domain. Working in $mathbf Z/p times mathbf Z/q$ instead, we get immediately the necessary and sufficient condition $x_p=pm 1_p$ and $x_q=pm 1_q$ because $mathbf Z/p$ and $mathbf Z/q$ are fields. Moreover, $x_{pq}=1_{pq}$ iff $x_p=1_p$ and $x_q=1_q$. This means that, if $p$ and $q$ are odd, the non trivial solutions (viewed in $mathbf Z/pqcong mathbf Z/p times mathbf Z/q$ ) are $(1_p , {-1}_q), ({-1}_p , 1_q)$ and $({-1}_p , {-1}_q)$ .






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        I prefer somehow to have a "structural" explanation. By the CRT, we have an isomorphism of rings $mathbf Z/pqcong mathbf Z/p times mathbf Z/q$. For $x in mathbf Z$, let us adopt the obvious notation $x_{pq}=(x_p , x_q)$ in the respective quotient rings. We must solve ${x_{pq}}^2=1_{pq}$ , or equivalently $(x_{pq}-1_{pq})(x_{pq}+1_{pq})=0_{pq}$, but we cannot conclude directly because $mathbf Z/pq$ admits divisors of zero, i.e. this ring is not a domain. Working in $mathbf Z/p times mathbf Z/q$ instead, we get immediately the necessary and sufficient condition $x_p=pm 1_p$ and $x_q=pm 1_q$ because $mathbf Z/p$ and $mathbf Z/q$ are fields. Moreover, $x_{pq}=1_{pq}$ iff $x_p=1_p$ and $x_q=1_q$. This means that, if $p$ and $q$ are odd, the non trivial solutions (viewed in $mathbf Z/pqcong mathbf Z/p times mathbf Z/q$ ) are $(1_p , {-1}_q), ({-1}_p , 1_q)$ and $({-1}_p , {-1}_q)$ .






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          I prefer somehow to have a "structural" explanation. By the CRT, we have an isomorphism of rings $mathbf Z/pqcong mathbf Z/p times mathbf Z/q$. For $x in mathbf Z$, let us adopt the obvious notation $x_{pq}=(x_p , x_q)$ in the respective quotient rings. We must solve ${x_{pq}}^2=1_{pq}$ , or equivalently $(x_{pq}-1_{pq})(x_{pq}+1_{pq})=0_{pq}$, but we cannot conclude directly because $mathbf Z/pq$ admits divisors of zero, i.e. this ring is not a domain. Working in $mathbf Z/p times mathbf Z/q$ instead, we get immediately the necessary and sufficient condition $x_p=pm 1_p$ and $x_q=pm 1_q$ because $mathbf Z/p$ and $mathbf Z/q$ are fields. Moreover, $x_{pq}=1_{pq}$ iff $x_p=1_p$ and $x_q=1_q$. This means that, if $p$ and $q$ are odd, the non trivial solutions (viewed in $mathbf Z/pqcong mathbf Z/p times mathbf Z/q$ ) are $(1_p , {-1}_q), ({-1}_p , 1_q)$ and $({-1}_p , {-1}_q)$ .






                          share|cite|improve this answer









                          $endgroup$



                          I prefer somehow to have a "structural" explanation. By the CRT, we have an isomorphism of rings $mathbf Z/pqcong mathbf Z/p times mathbf Z/q$. For $x in mathbf Z$, let us adopt the obvious notation $x_{pq}=(x_p , x_q)$ in the respective quotient rings. We must solve ${x_{pq}}^2=1_{pq}$ , or equivalently $(x_{pq}-1_{pq})(x_{pq}+1_{pq})=0_{pq}$, but we cannot conclude directly because $mathbf Z/pq$ admits divisors of zero, i.e. this ring is not a domain. Working in $mathbf Z/p times mathbf Z/q$ instead, we get immediately the necessary and sufficient condition $x_p=pm 1_p$ and $x_q=pm 1_q$ because $mathbf Z/p$ and $mathbf Z/q$ are fields. Moreover, $x_{pq}=1_{pq}$ iff $x_p=1_p$ and $x_q=1_q$. This means that, if $p$ and $q$ are odd, the non trivial solutions (viewed in $mathbf Z/pqcong mathbf Z/p times mathbf Z/q$ ) are $(1_p , {-1}_q), ({-1}_p , 1_q)$ and $({-1}_p , {-1}_q)$ .







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 15 at 21:03









                          nguyen quang donguyen quang do

                          9,2141724




                          9,2141724






























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