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?










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    homunculus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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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      homunculus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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 14 hours ago









      homunculus

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          2 Answers
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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






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            Of course,
            $$
            beginaligned
            overlineabcdef - overlineacdef
            &=
            overlineab0000 - overlinea0000
            \
            &=10000cdot(overlineab - overlinea)
            \
            &=100^2cdot(9a+b) .
            endaligned
            $$

            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.






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              Your Answer





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              2 Answers
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              2 Answers
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              up vote
              4
              down vote













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



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






              share|cite|improve this answer
























                up vote
                4
                down vote













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



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






                share|cite|improve this answer






















                  up vote
                  4
                  down vote










                  up vote
                  4
                  down vote









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



                  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 14 hours ago









                  lab bhattacharjee

                  219k14153268




                  219k14153268




















                      up vote
                      1
                      down vote













                      Of course,
                      $$
                      beginaligned
                      overlineabcdef - overlineacdef
                      &=
                      overlineab0000 - overlinea0000
                      \
                      &=10000cdot(overlineab - overlinea)
                      \
                      &=100^2cdot(9a+b) .
                      endaligned
                      $$

                      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
























                        up vote
                        1
                        down vote













                        Of course,
                        $$
                        beginaligned
                        overlineabcdef - overlineacdef
                        &=
                        overlineab0000 - overlinea0000
                        \
                        &=10000cdot(overlineab - overlinea)
                        \
                        &=100^2cdot(9a+b) .
                        endaligned
                        $$

                        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






















                          up vote
                          1
                          down vote










                          up vote
                          1
                          down vote









                          Of course,
                          $$
                          beginaligned
                          overlineabcdef - overlineacdef
                          &=
                          overlineab0000 - overlinea0000
                          \
                          &=10000cdot(overlineab - overlinea)
                          \
                          &=100^2cdot(9a+b) .
                          endaligned
                          $$

                          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,
                          $$
                          beginaligned
                          overlineabcdef - overlineacdef
                          &=
                          overlineab0000 - overlinea0000
                          \
                          &=10000cdot(overlineab - overlinea)
                          \
                          &=100^2cdot(9a+b) .
                          endaligned
                          $$

                          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 14 hours ago









                          dan_fulea

                          5,7251312




                          5,7251312




















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