Solve equations $x^2+x+1equiv 0(mod 7)$ and $2x-4equiv 0(mod 6)$











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Solve equations $x^2+x+1equiv 0(mod 7)$ and $2x-4equiv 0(mod 6)$



I try from to find x using other equation since $2x equiv 4 (mod 6)$ so $x=6k+2$ or $x=6k+5$, where $kin mathbb N$ if $x=6k+2$ then if we put this in first equation i get $36k^2+30k+7$ so if we want nature number then $7|36k^2+30k$ if I put $k=7$ then I get one solution $x=44$. If I think that $knot =7$ then $k=7m+l, lin {1,2,3,4,5,6,}$, $min mathbb N$ then it show that $7|36k^2+30k$ if $k=7m+5$ so then $x=6(7m+5)+2=42m+32$ then if I put in other equation and first it is true that $x=42m+32$, so $xin {74,116,...}$.



In second option that $x=6k+5$. And if I put in first equation I get $36k^2+60k+25+6k+5+1=36k^2+66k+31$ so this we can write as $35k^2+k^2+63k+3k+28+3$ so I need to show that $7|k^2+3k+3$ from here we can see that one solution $k=1$, then $x=11$ for $k=2$ this is not solution, for $k=3$ this is solution then $x=23$ if we want to find another number then we can write $k=3r+1$ or $k=3r+2$,$rin mathbb N$ it show that $knot =3r+1$and $knot =3r+2$.



So solution $xin {11,23,42m+32}$, $min mathbb N$.



Is this ok? Or you know some better way?






discrete-mathematics modular-arithmetic divisibility







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    up vote
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    down vote

    favorite












    Solve equations $x^2+x+1equiv 0(mod 7)$ and $2x-4equiv 0(mod 6)$



    I try from to find x using other equation since $2x equiv 4 (mod 6)$ so $x=6k+2$ or $x=6k+5$, where $kin mathbb N$ if $x=6k+2$ then if we put this in first equation i get $36k^2+30k+7$ so if we want nature number then $7|36k^2+30k$ if I put $k=7$ then I get one solution $x=44$. If I think that $knot =7$ then $k=7m+l, lin {1,2,3,4,5,6,}$, $min mathbb N$ then it show that $7|36k^2+30k$ if $k=7m+5$ so then $x=6(7m+5)+2=42m+32$ then if I put in other equation and first it is true that $x=42m+32$, so $xin {74,116,...}$.



    In second option that $x=6k+5$. And if I put in first equation I get $36k^2+60k+25+6k+5+1=36k^2+66k+31$ so this we can write as $35k^2+k^2+63k+3k+28+3$ so I need to show that $7|k^2+3k+3$ from here we can see that one solution $k=1$, then $x=11$ for $k=2$ this is not solution, for $k=3$ this is solution then $x=23$ if we want to find another number then we can write $k=3r+1$ or $k=3r+2$,$rin mathbb N$ it show that $knot =3r+1$and $knot =3r+2$.



    So solution $xin {11,23,42m+32}$, $min mathbb N$.



    Is this ok? Or you know some better way?






    discrete-mathematics modular-arithmetic divisibility







    share|cite|improve this question


























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Solve equations $x^2+x+1equiv 0(mod 7)$ and $2x-4equiv 0(mod 6)$



      I try from to find x using other equation since $2x equiv 4 (mod 6)$ so $x=6k+2$ or $x=6k+5$, where $kin mathbb N$ if $x=6k+2$ then if we put this in first equation i get $36k^2+30k+7$ so if we want nature number then $7|36k^2+30k$ if I put $k=7$ then I get one solution $x=44$. If I think that $knot =7$ then $k=7m+l, lin {1,2,3,4,5,6,}$, $min mathbb N$ then it show that $7|36k^2+30k$ if $k=7m+5$ so then $x=6(7m+5)+2=42m+32$ then if I put in other equation and first it is true that $x=42m+32$, so $xin {74,116,...}$.



      In second option that $x=6k+5$. And if I put in first equation I get $36k^2+60k+25+6k+5+1=36k^2+66k+31$ so this we can write as $35k^2+k^2+63k+3k+28+3$ so I need to show that $7|k^2+3k+3$ from here we can see that one solution $k=1$, then $x=11$ for $k=2$ this is not solution, for $k=3$ this is solution then $x=23$ if we want to find another number then we can write $k=3r+1$ or $k=3r+2$,$rin mathbb N$ it show that $knot =3r+1$and $knot =3r+2$.



      So solution $xin {11,23,42m+32}$, $min mathbb N$.



      Is this ok? Or you know some better way?






      discrete-mathematics modular-arithmetic divisibility







      share|cite|improve this question















      Solve equations $x^2+x+1equiv 0(mod 7)$ and $2x-4equiv 0(mod 6)$



      I try from to find x using other equation since $2x equiv 4 (mod 6)$ so $x=6k+2$ or $x=6k+5$, where $kin mathbb N$ if $x=6k+2$ then if we put this in first equation i get $36k^2+30k+7$ so if we want nature number then $7|36k^2+30k$ if I put $k=7$ then I get one solution $x=44$. If I think that $knot =7$ then $k=7m+l, lin {1,2,3,4,5,6,}$, $min mathbb N$ then it show that $7|36k^2+30k$ if $k=7m+5$ so then $x=6(7m+5)+2=42m+32$ then if I put in other equation and first it is true that $x=42m+32$, so $xin {74,116,...}$.



      In second option that $x=6k+5$. And if I put in first equation I get $36k^2+60k+25+6k+5+1=36k^2+66k+31$ so this we can write as $35k^2+k^2+63k+3k+28+3$ so I need to show that $7|k^2+3k+3$ from here we can see that one solution $k=1$, then $x=11$ for $k=2$ this is not solution, for $k=3$ this is solution then $x=23$ if we want to find another number then we can write $k=3r+1$ or $k=3r+2$,$rin mathbb N$ it show that $knot =3r+1$and $knot =3r+2$.



      So solution $xin {11,23,42m+32}$, $min mathbb N$.



      Is this ok? Or you know some better way?






      discrete-mathematics modular-arithmetic divisibility




      discrete-mathematics modular-arithmetic divisibility






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      edited Nov 23 at 10:01









      Glorfindel

      3,38481730




      3,38481730










      asked Nov 23 at 8:24









      Marko Škorić

      70310




      70310






















          2 Answers
          2






          active

          oldest

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          up vote
          1
          down vote



          accepted










          We can take $x=3k+2$. So $$(3k+2)^2+3k+2+1 equiv 0 mod 7$$
          begin{aligned}
          begin{align}
          9k^2+12k equiv 0 mod 7&iff k(k+4)equiv 0 mod 7 \
          &iff k=7m text{or} k=7n-4
          end{align}
          end{aligned}
          Hence, all solution set:
          $${21m+2,21n-10:m,nin Bbb Z}$$






          share|cite|improve this answer




























            up vote
            1
            down vote













            As $(7,4)=1$



            $$x^2+x+1equiv0pmod7iff(2x+1)^2equiv-3equiv4$$



            $implies$



            either $2x+1equiv2iff2xequiv1equiv8iff xequiv4pmod7 (1)$



            or $2x+1equiv-2iff2xequiv-3equiv4iff xequiv2pmod7 (2)$



            and we have $2xequiv4pmod6iff xequiv2pmod3 (3)$



            Apply CRT on $(1),(3)$



            and $(2,3)implies$lcm$(3,7)mid(x-2)$






            share|cite|improve this answer





















            • can you write solution of CRT just to check did I get right solution?
              – Marko Škorić
              Nov 23 at 8:41











            Your Answer





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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            1
            down vote



            accepted










            We can take $x=3k+2$. So $$(3k+2)^2+3k+2+1 equiv 0 mod 7$$
            begin{aligned}
            begin{align}
            9k^2+12k equiv 0 mod 7&iff k(k+4)equiv 0 mod 7 \
            &iff k=7m text{or} k=7n-4
            end{align}
            end{aligned}
            Hence, all solution set:
            $${21m+2,21n-10:m,nin Bbb Z}$$






            share|cite|improve this answer

























              up vote
              1
              down vote



              accepted










              We can take $x=3k+2$. So $$(3k+2)^2+3k+2+1 equiv 0 mod 7$$
              begin{aligned}
              begin{align}
              9k^2+12k equiv 0 mod 7&iff k(k+4)equiv 0 mod 7 \
              &iff k=7m text{or} k=7n-4
              end{align}
              end{aligned}
              Hence, all solution set:
              $${21m+2,21n-10:m,nin Bbb Z}$$






              share|cite|improve this answer























                up vote
                1
                down vote



                accepted







                up vote
                1
                down vote



                accepted






                We can take $x=3k+2$. So $$(3k+2)^2+3k+2+1 equiv 0 mod 7$$
                begin{aligned}
                begin{align}
                9k^2+12k equiv 0 mod 7&iff k(k+4)equiv 0 mod 7 \
                &iff k=7m text{or} k=7n-4
                end{align}
                end{aligned}
                Hence, all solution set:
                $${21m+2,21n-10:m,nin Bbb Z}$$






                share|cite|improve this answer












                We can take $x=3k+2$. So $$(3k+2)^2+3k+2+1 equiv 0 mod 7$$
                begin{aligned}
                begin{align}
                9k^2+12k equiv 0 mod 7&iff k(k+4)equiv 0 mod 7 \
                &iff k=7m text{or} k=7n-4
                end{align}
                end{aligned}
                Hence, all solution set:
                $${21m+2,21n-10:m,nin Bbb Z}$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 23 at 10:59









                1Spectre1

                1018




                1018






















                    up vote
                    1
                    down vote













                    As $(7,4)=1$



                    $$x^2+x+1equiv0pmod7iff(2x+1)^2equiv-3equiv4$$



                    $implies$



                    either $2x+1equiv2iff2xequiv1equiv8iff xequiv4pmod7 (1)$



                    or $2x+1equiv-2iff2xequiv-3equiv4iff xequiv2pmod7 (2)$



                    and we have $2xequiv4pmod6iff xequiv2pmod3 (3)$



                    Apply CRT on $(1),(3)$



                    and $(2,3)implies$lcm$(3,7)mid(x-2)$






                    share|cite|improve this answer





















                    • can you write solution of CRT just to check did I get right solution?
                      – Marko Škorić
                      Nov 23 at 8:41















                    up vote
                    1
                    down vote













                    As $(7,4)=1$



                    $$x^2+x+1equiv0pmod7iff(2x+1)^2equiv-3equiv4$$



                    $implies$



                    either $2x+1equiv2iff2xequiv1equiv8iff xequiv4pmod7 (1)$



                    or $2x+1equiv-2iff2xequiv-3equiv4iff xequiv2pmod7 (2)$



                    and we have $2xequiv4pmod6iff xequiv2pmod3 (3)$



                    Apply CRT on $(1),(3)$



                    and $(2,3)implies$lcm$(3,7)mid(x-2)$






                    share|cite|improve this answer





















                    • can you write solution of CRT just to check did I get right solution?
                      – Marko Škorić
                      Nov 23 at 8:41













                    up vote
                    1
                    down vote










                    up vote
                    1
                    down vote









                    As $(7,4)=1$



                    $$x^2+x+1equiv0pmod7iff(2x+1)^2equiv-3equiv4$$



                    $implies$



                    either $2x+1equiv2iff2xequiv1equiv8iff xequiv4pmod7 (1)$



                    or $2x+1equiv-2iff2xequiv-3equiv4iff xequiv2pmod7 (2)$



                    and we have $2xequiv4pmod6iff xequiv2pmod3 (3)$



                    Apply CRT on $(1),(3)$



                    and $(2,3)implies$lcm$(3,7)mid(x-2)$






                    share|cite|improve this answer












                    As $(7,4)=1$



                    $$x^2+x+1equiv0pmod7iff(2x+1)^2equiv-3equiv4$$



                    $implies$



                    either $2x+1equiv2iff2xequiv1equiv8iff xequiv4pmod7 (1)$



                    or $2x+1equiv-2iff2xequiv-3equiv4iff xequiv2pmod7 (2)$



                    and we have $2xequiv4pmod6iff xequiv2pmod3 (3)$



                    Apply CRT on $(1),(3)$



                    and $(2,3)implies$lcm$(3,7)mid(x-2)$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 23 at 8:29









                    lab bhattacharjee

                    222k15155273




                    222k15155273












                    • can you write solution of CRT just to check did I get right solution?
                      – Marko Škorić
                      Nov 23 at 8:41


















                    • can you write solution of CRT just to check did I get right solution?
                      – Marko Škorić
                      Nov 23 at 8:41
















                    can you write solution of CRT just to check did I get right solution?
                    – Marko Škorić
                    Nov 23 at 8:41




                    can you write solution of CRT just to check did I get right solution?
                    – Marko Škorić
                    Nov 23 at 8:41


















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