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It was mathematically proven impossible for integer values for a, b, c to satisfy aⁿ+bⁿ=cⁿ for any integer n greater than 2! What did I violate here? >.> -1³ is -1, 1³ is 1, 0³ is 0, and -1 plus 1 equals zero, so the calculation is correct. I'm sure I didn't violate Fermat's Last Theorem... only that I didn't remember one condition... are negative values for a, b, and c not allowed?
MathematicallyMindedFractal MathematicallyMindedFractal 16-17, F 4 Answers May 31 in Hobbies

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"For any integer value of n greater than two."

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Yes, I chose n=3.

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Oh, duh, I copied the wrong part. "No three positive integers a, b, and c can satisfy the equation" for yada yada yada.

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It isn't the condition. It's what's being tested. It was not proven for hundreds of years. Fermat died before he could write his proof down. It is why it is his last theorem. I'm wondering what part of the proof I violated.

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I dunno.....but reading this just made my head split open. hehehe :p

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I tried to put it simply xD

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ehhh, that wasn't simple enough. LOL :p

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How about this: You basically cannot do the Pythagorean Theorem for any power greater than two where a, b, and c are all without decimals.

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(Thar's the theorem; it wasn't proven for hundreds of years)

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ah. well, you're the math genius, so I should probably agree with you. haha :)

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so a^n+b^n=c^n is impossible? SO... the Pythagorum Theorum doesn't work... lol I can't spell tonight...

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a^2+b^2=c^ (3+4=5 at right angles) (9+16=25)

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Not quite. It is syntaxically similar to the Pythagorean theorem, but should not hold for any n greater than 2.

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oh... n greater than 2... ok! got you ! :D

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Any n >2 and if a is not equal to b...

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Any n>2 and if a is not equal to b, and a and b are > or <0.

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Basically, a³+b³≠c³, a⁴+b⁴≠c⁴ and so on.

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Unless a is 1 b is -1 n is odd... lol

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yup >.>

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Irrational, my dear Watson.....

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What do you mean?

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In electronics, we call it "j". I think mathematicians call it "i".

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That's an imaginary number, not an irrational number. Imaginary numbers don't apply here, though >.>

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My bad.....

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xD

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