Question on Fermat's Last Theorem

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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?
 

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  • #2
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But can't 6th powers be written as squares of 3rd powers? Also, can't any even powers be written as squares?
Yes, they can be written as such. Not really sure what you're getting at...
 
  • #3
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Yes, they can be written as such. Not really sure what you're getting at...
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
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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?
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.
 
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