Differences between numbers of the form abcdef and acdef which are perfect square.












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$541456-51456=700^2$ is the example I found. Are there other examples of numbers of the form abcdef such that abcdef-acdef is a perfect square?










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    $541456-51456=700^2$ is the example I found. Are there other examples of numbers of the form abcdef such that abcdef-acdef is a perfect square?










    share|cite|improve this question

























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      2








      2







      $541456-51456=700^2$ is the example I found. Are there other examples of numbers of the form abcdef such that abcdef-acdef is a perfect square?










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      $541456-51456=700^2$ is the example I found. Are there other examples of numbers of the form abcdef such that abcdef-acdef is a perfect square?







      number-theory






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      asked Nov 12 at 10:43







      user613967





























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          $$abcdef-acdef=10^4b+10^5a-(10^4a)=10^4(9a+b)$$



          So, all we need is $9a+b$ to be perfect square






          share|cite|improve this answer





























            1














            Of course,
            $$
            begin{aligned}
            overline{abcdef} - overline{acdef}
            &=
            overline{ab0000} - overline{a0000}
            \
            &=10000cdot(overline{ab} - overline{a})
            \
            &=100^2cdot(9a+b) .
            end{aligned}
            $$

            So we can choose $c,d,e,f$ arbitrarily, and for $a,b$ there are the following chances:



            for a in [1..9]:
            for b in [0..9]:
            ab_a = 9*a + b
            if ab_a.is_square():
            print ( "a=%s b=%s %s-%s = %s^2"
            % (a, b, 10*a+b, a, sqrt(ab_a)) )


            And the results are:



            a=1 b=0 10-1 = 3^2
            a=1 b=7 17-1 = 4^2
            a=2 b=7 27-2 = 5^2
            a=3 b=9 39-3 = 6^2
            a=4 b=0 40-4 = 6^2
            a=5 b=4 54-5 = 7^2
            a=7 b=1 71-7 = 8^2
            a=8 b=9 89-8 = 9^2
            a=9 b=0 90-9 = 9^2


            Here i was using sage.






            share|cite|improve this answer





















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              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

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              active

              oldest

              votes






              active

              oldest

              votes









              3














              $$abcdef-acdef=10^4b+10^5a-(10^4a)=10^4(9a+b)$$



              So, all we need is $9a+b$ to be perfect square






              share|cite|improve this answer


























                3














                $$abcdef-acdef=10^4b+10^5a-(10^4a)=10^4(9a+b)$$



                So, all we need is $9a+b$ to be perfect square






                share|cite|improve this answer
























                  3












                  3








                  3






                  $$abcdef-acdef=10^4b+10^5a-(10^4a)=10^4(9a+b)$$



                  So, all we need is $9a+b$ to be perfect square






                  share|cite|improve this answer












                  $$abcdef-acdef=10^4b+10^5a-(10^4a)=10^4(9a+b)$$



                  So, all we need is $9a+b$ to be perfect square







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 12 at 10:46









                  lab bhattacharjee

                  222k15155273




                  222k15155273























                      1














                      Of course,
                      $$
                      begin{aligned}
                      overline{abcdef} - overline{acdef}
                      &=
                      overline{ab0000} - overline{a0000}
                      \
                      &=10000cdot(overline{ab} - overline{a})
                      \
                      &=100^2cdot(9a+b) .
                      end{aligned}
                      $$

                      So we can choose $c,d,e,f$ arbitrarily, and for $a,b$ there are the following chances:



                      for a in [1..9]:
                      for b in [0..9]:
                      ab_a = 9*a + b
                      if ab_a.is_square():
                      print ( "a=%s b=%s %s-%s = %s^2"
                      % (a, b, 10*a+b, a, sqrt(ab_a)) )


                      And the results are:



                      a=1 b=0 10-1 = 3^2
                      a=1 b=7 17-1 = 4^2
                      a=2 b=7 27-2 = 5^2
                      a=3 b=9 39-3 = 6^2
                      a=4 b=0 40-4 = 6^2
                      a=5 b=4 54-5 = 7^2
                      a=7 b=1 71-7 = 8^2
                      a=8 b=9 89-8 = 9^2
                      a=9 b=0 90-9 = 9^2


                      Here i was using sage.






                      share|cite|improve this answer


























                        1














                        Of course,
                        $$
                        begin{aligned}
                        overline{abcdef} - overline{acdef}
                        &=
                        overline{ab0000} - overline{a0000}
                        \
                        &=10000cdot(overline{ab} - overline{a})
                        \
                        &=100^2cdot(9a+b) .
                        end{aligned}
                        $$

                        So we can choose $c,d,e,f$ arbitrarily, and for $a,b$ there are the following chances:



                        for a in [1..9]:
                        for b in [0..9]:
                        ab_a = 9*a + b
                        if ab_a.is_square():
                        print ( "a=%s b=%s %s-%s = %s^2"
                        % (a, b, 10*a+b, a, sqrt(ab_a)) )


                        And the results are:



                        a=1 b=0 10-1 = 3^2
                        a=1 b=7 17-1 = 4^2
                        a=2 b=7 27-2 = 5^2
                        a=3 b=9 39-3 = 6^2
                        a=4 b=0 40-4 = 6^2
                        a=5 b=4 54-5 = 7^2
                        a=7 b=1 71-7 = 8^2
                        a=8 b=9 89-8 = 9^2
                        a=9 b=0 90-9 = 9^2


                        Here i was using sage.






                        share|cite|improve this answer
























                          1












                          1








                          1






                          Of course,
                          $$
                          begin{aligned}
                          overline{abcdef} - overline{acdef}
                          &=
                          overline{ab0000} - overline{a0000}
                          \
                          &=10000cdot(overline{ab} - overline{a})
                          \
                          &=100^2cdot(9a+b) .
                          end{aligned}
                          $$

                          So we can choose $c,d,e,f$ arbitrarily, and for $a,b$ there are the following chances:



                          for a in [1..9]:
                          for b in [0..9]:
                          ab_a = 9*a + b
                          if ab_a.is_square():
                          print ( "a=%s b=%s %s-%s = %s^2"
                          % (a, b, 10*a+b, a, sqrt(ab_a)) )


                          And the results are:



                          a=1 b=0 10-1 = 3^2
                          a=1 b=7 17-1 = 4^2
                          a=2 b=7 27-2 = 5^2
                          a=3 b=9 39-3 = 6^2
                          a=4 b=0 40-4 = 6^2
                          a=5 b=4 54-5 = 7^2
                          a=7 b=1 71-7 = 8^2
                          a=8 b=9 89-8 = 9^2
                          a=9 b=0 90-9 = 9^2


                          Here i was using sage.






                          share|cite|improve this answer












                          Of course,
                          $$
                          begin{aligned}
                          overline{abcdef} - overline{acdef}
                          &=
                          overline{ab0000} - overline{a0000}
                          \
                          &=10000cdot(overline{ab} - overline{a})
                          \
                          &=100^2cdot(9a+b) .
                          end{aligned}
                          $$

                          So we can choose $c,d,e,f$ arbitrarily, and for $a,b$ there are the following chances:



                          for a in [1..9]:
                          for b in [0..9]:
                          ab_a = 9*a + b
                          if ab_a.is_square():
                          print ( "a=%s b=%s %s-%s = %s^2"
                          % (a, b, 10*a+b, a, sqrt(ab_a)) )


                          And the results are:



                          a=1 b=0 10-1 = 3^2
                          a=1 b=7 17-1 = 4^2
                          a=2 b=7 27-2 = 5^2
                          a=3 b=9 39-3 = 6^2
                          a=4 b=0 40-4 = 6^2
                          a=5 b=4 54-5 = 7^2
                          a=7 b=1 71-7 = 8^2
                          a=8 b=9 89-8 = 9^2
                          a=9 b=0 90-9 = 9^2


                          Here i was using sage.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 12 at 10:56









                          dan_fulea

                          6,2301312




                          6,2301312






























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