Extension of Fermat's theorem?

[ACG]

Hi! I assume you all know Fermat's last theorem. Well, has anyone considered the following extension to it? Assuming we're just using integers:

We know that x1^n + x2^n = y^n has no solution for n > 2. However, what about this?

For which values of k does x1^n+x2^n+...+xk^n = z^n have solutions for n > k?

We know it's true for 1 ((-1)^2 = 1^2) and not true for 2. What about other numbers?

Thanks in advance,

ACG

 

[matt grime]

Every number is the sum of 4 squares. That is one known result.

 

[ACG]

Hi!

That's cool -- I'd wondered about that myself :)

However, that won't help here: the sum of four objects case would have to be a^5+b^5+c^5+d^5 = e^5.

ACG

 

[AlphaNumeric]

Have a read of : http://mathworld.wolfram.com/DiophantineEquation5thPowers.html and click around some of the links at the bottom of the article here : http://mathworld.wolfram.com/DiophantineEquation.html which are similar to the first link.

It's hard to give specifics, many are unproven either way, but a^5+b^5+c^5+d^5 = e^5 has solutions as given in the first link.

 

[barakman]

Hi! I assume you all know Fermat's last theorem. Well, has anyone considered the following extension to it? Assuming we're just using integers:

We know that x1^n + x2^n = y^n has no solution for n > 2. However, what about this?

For which values of k does x1^n+x2^n+...+xk^n = z^n have solutions for n > k?

We know it's true for 1 ((-1)^2 = 1^2) and not true for 2. What about other numbers?

Thanks in advance,

ACG

Yes, I have (considered that, and also wrote a short program to try and find a counter-example).
Have you been able to find anymore details about this one (like, has it been proved, disproved, etc.)?
Thanks.

 

[uart]

like, has it been proved, disproved, etc.

Alphanumeric has already posted links that contain counter examples. For example they show several solutions to a^5 + b^5 + c^5 + d^5 = e^5

 

[barakman]

Alphanumeric has already posted links that contain counter examples. For example they show several solutions to a^5 + b^5 + c^5 + d^5 = e^5

thanks

 

[DeadWolfe]

http://en.wikipedia.org/wiki/Euler's_sum_of_powers_conjecture