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Posers and Puzzles

Posers and Puzzles

  1. Standard member genius
    Wayward Soul
    16 Nov '06 12:32
    if a^x+b^x=c^x, x>=3 has no solutions for integers a,b,c, prove that c must be irrational if a and b are rational, or find a valid example where a,b and c are all rational. (e.g. 3^2+4^2=5^2 for x=2, but x>=3...)

    the proof is actually surprisingly easy...either that, or i got it wrong
  2. 16 Nov '06 14:06 / 2 edits
    a and b are rational. Therefore a can be written a1/a2, b as b1/b2 (a1,a2,b1,b2 all integers).

    Assume that c is rational too. So c can be written as c1/c2 (c1 and c2 integers).

    (a1/a2)^n + (b1/b2)^n = (c1/c2)^n

    Multiply both sides by (a2*b2*c2)^n

    This gives (a1*b2*c2)^n + (b1*a2*c2)^n = (c1*a2*b2)^n

    Clearly a1*b2*c2, b1*a2*c2 and c1*a2*b2 are all integers and satisfy A^n + B^n = C^n, which we know has no integer solutions.

    Therefore our assumption that c is rational must be false.
  3. Standard member genius
    Wayward Soul
    16 Nov '06 14:08 / 2 edits
    Originally posted by Fat Lady
    a and b are rational. Therefore a can be written a1/a2, b as b1/b2 (a1,a2,b1,b2 all integers).

    Assume that c is rational too. So c can be written as c1/c2 (c1 and c2 integers).

    (a1/a2)^n + (b1/b2)^n = (c1/c2)^n

    Multiply both sides by (a2*b2*c2)^n

    This gives (a1*b2*c2)^n + (b1*a2*c2)^n = (c1*a2*b2)^n

    Clearly a1*b2*c2, b1*a2*c2 and c1*a2*b2 are ...[text shortened]... we know has no integer solutions.

    Therefore our assumption that c is rational must be false.
    see-simple

    EDIT: i'm bored now. therefore, next question,

    can 1/2+1/3+...+1/n=A where A is an integer?

    this one is -erm- slightly tougher...