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A Small Algebraic Diversion

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  • Shlomi Fish
    It all started when I noticed the following: 4*6 = 24 5*5 = 25 6*4 = 24 And that 24 was very close to 25. I wondered whether if I have a square a*a I can
    Message 1 of 1 , Dec 10, 2006
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      It all started when I noticed the following:

      4*6 = 24
      5*5 = 25
      6*4 = 24

      And that 24 was very close to 25. I wondered whether if I have a square a*a I
      can subtract 1 from one multiplicand and add 1 to the other and get the same
      result. Let's see:

      (a-1)*(a+1) = a^2-a+a-1=a^2-1

      Nope! However, now that I tought about it I realised that it will hold for
      any "a" no matter how large or how small, which is nice. So 999*1001 =
      1000^2-1.

      Now let's generalise our requirements a bit: what if we have a product of "a"
      times "b", and we want to do the same:

      (a-1)*(b+1) = ab
      ab-b+a-1 = ab
      a - b = 1

      So if we increase a number by one and decrease the other by one, we'll get the
      same product only if their difference is 1. I.e: we switched them.

      But what if we decrease and increase by a \delta (LaTeX notation) that is not
      necessarily 1?

      (a-\delta)*(b+\delta) = ab
      ab-\delta(b-a) + \delta^2 = ab
      \delta^2 = \delta(a-b)
      a-b = \delta

      So again there's no solution except the trivial one.

      I wonder if there are non-trivial solutions for (a-1)b(c+1) or (a-1)(b+1)(c-1)
      (d+1) or any other products like that, but that would probably be somewhat
      hairier algebra.

      Regards,

      Shlomi Fish

      ---------------------------------------------------------------------
      Shlomi Fish shlomif@...
      Homepage: http://www.shlomifish.org/

      Chuck Norris wrote a complete Perl 6 implementation in a day but then
      destroyed all evidence with his bare hands, so no one will know his secrets.
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