# Extension of Fermat's theorem?

• ACG
In summary, the conversation is about an extension to Fermat's last theorem where the question is posed for which values of k does the equation x1^n + x2^n + ... +xk^n = z^n have solutions for n > k. Alphanumeric provides links to counterexamples and mentions that it is hard to give specifics as many cases are still unproven. The conversation also briefly touches on the sum of four squares, with a link provided for more information.

#### 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?

ACG

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

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

ACG said:
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?

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.

like, has it been proved, disproved, etc.
Alphanumeric has already posted links that contain counter examples. For example they show several solutions to $a^5 + b^5 + c^5 + d^5 = e^5$

uart said:
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