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question on Fermat's Last Theorem |
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| Aug17-12, 08:56 PM | #1 |
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question on Fermat's Last Theorem
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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| Aug17-12, 09:01 PM | #2 |
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| Aug17-12, 10:21 PM | #3 |
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a2 + b2 = c2 cannot have integer solutions if a,b, and c are perfect squares, cubes, fourth powers, etc? |
| Aug17-12, 10:59 PM | #4 |
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question on Fermat's Last Theorem |
| Aug19-12, 09:32 PM | #5 |
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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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