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Is it true...

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  • Jon Perry
    that there exists x,y, a,b 1 s.t. x^a - y^b = n, for all n integers? Jon Perry perry@globalnet.co.uk http://www.users.globalnet.co.uk/~perry/maths/
    Message 1 of 7 , Apr 1, 2003
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      that there exists x,y, a,b>1 s.t. x^a - y^b = n, for all n integers?

      Jon Perry
      perry@...
      http://www.users.globalnet.co.uk/~perry/maths/
      http://www.users.globalnet.co.uk/~perry/DIVMenu/
      BrainBench MVP for HTML and JavaScript
      http://www.brainbench.com
    • Jon Perry
      6,14,21,29,32,34,42,46,51,58,59,62,66,69,70,75,77,78,84,85,86,88,90,93,96,. .. (Any pattern there, I wonder?) I asked for a,b 1, hence every odd is accounted
      Message 2 of 7 , Apr 2, 2003
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        '6,14,21,29,32,34,42,46,51,58,59,62,66,69,70,75,77,78,84,85,86,88,90,93,96,.
        ..
        (Any pattern there, I wonder?)'

        I asked for a,b>1, hence every odd is accounted for, e.g. 6^2-5^2=11,
        7^2-6^2=13, so your list becomes;

        6,14,32,34,42,46,62,66,70,78,84,86,88,90,96

        Is 32 the only 2^k present?

        Jon Perry
        perry@...
        http://www.users.globalnet.co.uk/~perry/maths/
        http://www.users.globalnet.co.uk/~perry/DIVMenu/
        BrainBench MVP for HTML and JavaScript
        http://www.brainbench.com
      • Jack Brennen
        ... Probably not. I don t think any solution is known for n=6, and I suspect that no solution exists for n=6. For n=6, I think that it can be shown easily
        Message 3 of 7 , Apr 2, 2003
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          > Is it true...
          > that there exists x,y, a,b>1 s.t. x^a - y^b = n, for all n integers?

          Probably not. I don't think any solution is known for n=6, and I
          suspect that no solution exists for n=6. For n=6, I think that it
          can be shown easily that if a solution exists, one of the exponents
          must be >= 5, and the heuristics (and the scarcity of 5th-and-above
          powers) would lead one to strongly suspect that no solution exists.
        • mikeoakes2@aol.com
          In a message dated 03/04/03 02:34:44 GMT Daylight Time, jack@brennen.net ... I agree that heuristics based on density of powers would indicate probably not .
          Message 4 of 7 , Apr 3, 2003
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            In a message dated 03/04/03 02:34:44 GMT Daylight Time, jack@...
            writes:


            > > Is it true...
            > > that there exists x,y, a,b>1 s.t. x^a - y^b = n, for all n integers?
            >
            > Probably not. I don't think any solution is known for n=6, and I
            > suspect that no solution exists for n=6. For n=6, I think that it
            > can be shown easily that if a solution exists, one of the exponents
            > must be >= 5, and the heuristics (and the scarcity of 5th-and-above
            > powers) would lead one to strongly suspect that no solution exists.
            >

            I agree that heuristics based on density of powers would indicate "probably
            not".

            Since only 0 and 1 are quadratic residues mod 4, a = b = 2 is no good for n =
            2 mod 4.

            The next best bet, in terms of density of powers, is a=2, b=3. A couple of
            months ago, as a by-product of investigating the "Fermat-Catalan conjecture",
            I established that for (x^2) and (y^3) both <= 10^10, no less than 736754 of
            the numbers <= 1 million could /not/ be represented as abs(x^2-y^3), the list
            starting as follows:-
            6,14,21,29,32,34,42,46,51,58,59,62,66,69,70,75,77,78,84,85,86,88,90,93,96,...
            (Any pattern there, I wonder?)

            Mike Oakes


            [Non-text portions of this message have been removed]
          • mikeoakes2@aol.com
            In a message dated 03/04/03 12:33:28 GMT Daylight Time, perry@globalnet.co.uk ... No. 2048 and 131072 are also in the list. Mike [Non-text portions of this
            Message 5 of 7 , Apr 3, 2003
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              In a message dated 03/04/03 12:33:28 GMT Daylight Time, perry@...
              writes:


              > Is 32 the only 2^k present?
              >
              >
              No.
              2048 and 131072 are also in the list.

              Mike


              [Non-text portions of this message have been removed]
            • mikeoakes2@aol.com
              In a message dated 03/04/03 12:33:28 GMT Daylight Time, perry@globalnet.co.uk ... For (x^2) and (y^2) both
              Message 6 of 7 , Apr 3, 2003
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                In a message dated 03/04/03 12:33:28 GMT Daylight Time, perry@...
                writes:


                > I asked for a,b>1, hence every odd is accounted for, e.g. 6^2-5^2=11,
                > 7^2-6^2=13, so your list becomes;
                >
                > 6,14,32,34,42,46,62,66,70,78,84,86,88,90,96
                >

                For (x^2) and (y^2) both <= 10^10, 19023 (even) integers <= 100000 can't be
                expressed as abs(x^2-y^2), the list starting:-
                6,14,22,30,38,42,46,54,62,66,70,78,86,94

                So 32,34,84,88,90 can be removed from your list, leaving:-
                6,14,42,46,62,66,70,78,86,96
                needing higher powers.

                Except for 96, all these are = 2 mod 4.

                Mike Oakes


                [Non-text portions of this message have been removed]
              • mikeoakes2@aol.com
                Apologies: I should have written:- ... Of course, 4n = (2n+1)^2 - (2n-1)^2, so only the case n = 2 mod 4 poses any problems. Mike [Non-text portions of this
                Message 7 of 7 , Apr 3, 2003
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                  Apologies: I should have written:-

                  > So 32,34,84,88,90,96 can be removed from your list, leaving:-
                  > 6,14,42,46,62,66,70,78,86
                  > needing higher powers.

                  Of course, 4n = (2n+1)^2 - (2n-1)^2, so only the case n = 2 mod 4 poses any
                  problems.

                  Mike


                  [Non-text portions of this message have been removed]
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