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Question on Fermat's Last Theorem

  1. Aug 17, 2012 #1
    According to Fermat's last theorem, a 6th power plus a 6th power cannot equal a 6th power, but a square plus a square can equal a square. But can't 6th powers be written as squares of 3rd powers? Also, can't any even powers be written as squares?
  2. jcsd
  3. Aug 17, 2012 #2
    Yes, they can be written as such. Not really sure what you're getting at...
  4. Aug 17, 2012 #3
    Is this a way to know what numbers cannot be pythagorean triples? for example:
    a2 + b2 = c2
    cannot have integer solutions if a,b, and c are perfect squares, cubes, fourth powers, etc?
  5. Aug 17, 2012 #4


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  6. Aug 19, 2012 #5
    Well yes even powers can be written as squares, but only certain pairs of squares sum to a square. FLT which has been proven states that x^y + y^n = z^n has no solution in integers. Thus, for n > 2 including n equal to an even number greater than 2 there is no solution.

    P.S. it has been shown that all solutions of the form x^2 + y^2 = z^2 are such that the pair x and y are of the form a^2 - b^2 and 2ab for some a and b, but a^2 - b^2 can never equal a cube except in the case |a| = 3 and |b| = 1. However, in that case 2ab = +/- 6 which is not a cube. Thus 8^2 + 6^2 = 10^2 has the property that 8 is cube. Although all even cubes are of the form 2ab, no two cubes sum to a square except in the trival cases where the square is 0 or one of the cubes is 0, i.e. a=b=k^3 or either of (a,b) = 0 in which case the common requirement that a is coprime to b is missing.
    Last edited: Aug 20, 2012
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