Can Fermat's Equation Be Solved with Non-Positive Integers?

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The discussion confirms that Fermat's Equation, represented as an + bn = cn for n > 2, cannot be solved using non-positive integers. The participants clarify that even with negative integers, the equation can be rearranged to yield positive integer solutions. Specifically, for even n, the equation holds true as negative values can be converted to their positive counterparts. For odd n, the same principle applies, demonstrating that any solution involving negative integers can be transformed into a solution involving positive integers.

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avery
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hi,

is there integers a, b, c that satisfy the equation an + bn = cn for n>2
(I don't mean positive integers)

thx.
 
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I assume you mean an + bn = cn. Even if a, b, or c is negative, it won't make any difference. If n is even, it is obvious, while if n is odd you can replace the negative by its positive counterpart and rearrange terms so the equation is in its original form.
 
There are many solutions where one of the numbers is 0: two examples are 64+04=64 and 53+(-5)3=03.

There can be no solution with non-zero numbers, and this can be inferred from non-existence of positive solutions.
For even n, this is trivial: since an=(-a)n, a solution with negatives is also a solution with positive integers.
For odd n, (-a)n=-an, and you can always rearrange the terms to have a solution in positive numbers. This is a bit different for each assignment of signs to a, b, and c, but as an example, if a<0,b>0,c>0 and an+bn=cn, then cn+(-a)n=bn is a solution in positive numbers.
 

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