# A little problem

1. May 25, 2005

### tyutyu fait le train

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

2. May 25, 2005

### Muzza

Sure, no problem. Take n = 3 and a = 1, b = 2 and c = 3^(2/3)...

3. May 25, 2005

### whozum

http://en.wikipedia.org/wiki/Fermat's_last_theorem

There are none for integers a,b, and c.

4. May 25, 2005

### tyutyu fait le train

yep it is fermat's theorem...ok thank you!

5. May 25, 2005

### Curious3141

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.

6. May 25, 2005

### whozum

He just neglected to mention the integer requirement, which I amended.

7. May 25, 2005

### uart

No he didn't neglect anything, he said "with a, b and c real numbers".

8. May 25, 2005

### whozum

How is that not neglecting to mention the integer requirement.

9. May 25, 2005

### Zurtex

The Real Number set is an entirely different number set to the Integers. The poster said Real Numbers.

10. May 25, 2005

### whozum

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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