Solving "A Little Problem" with a^n+b^n=c^n

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The discussion revolves around the question of finding natural numbers n greater than 2 such that a^n + b^n = c^n for real numbers a, b, and c. It is clarified that while Fermat's Last Theorem states there are no integer solutions for n > 2, there are indeed infinite solutions when a, b, and c are real numbers. Participants debate the interpretation of "real numbers" versus "integers," emphasizing the importance of precise language in mathematical discussions. The conversation highlights the distinction between different number sets and their implications for the problem posed. Ultimately, the focus remains on the nature of solutions available under the specified conditions.
tyutyu fait le train
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Here is the question: "could you fine n>2 (n natural number) so that: a^n+ b^n=c^n with a, b and c real numbers".

I have got an idea but I am not sure if it works.

Thank you very much

tyu
 
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Sure, no problem. Take n = 3 and a = 1, b = 2 and c = 3^(2/3)...
 
http://en.wikipedia.org/wiki/Fermat's_last_theorem

There are none for integers a,b, and c.
 
yep it is fermat's theorem...ok thank you!
 
But what you posted isn't Fermat's Last Theorem. For real a,b and c, there are an infinite number of solutions for every natural number value of n.
 
He just neglected to mention the integer requirement, which I amended.
 
whozum said:
He just neglected to mention the integer requirement, which I amended.

No he didn't neglect anything, he said "with a, b and c real numbers".
 
How is that not neglecting to mention the integer requirement.
 
whozum said:
How is that not neglecting to mention the integer requirement.
The Real Number set is an entirely different number set to the Integers. The poster said Real Numbers.
 
  • #10
The issue here is of word choice. I said "neglecting to mention the integer requirement" is the same as "not mentioning the integer requirement".

He mentioned a problem very similar to FLT in which I referenced him to the correct form. It turns out that is what he is looking for.
 

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