0
$\begingroup$

I'm working on the following Fermat little theorem exercise:

Prove that if $p$ is prime and $a^p+b^p = c^p$ then $a+b-c = 0 \mod p$

Also I find a relation with Fermat last theorem which says that no three positive integers $a, b$, and $c$ satisfy the equation $a^n$ + $b^n$ $=$ $c^n$ for any integer value of $n$ greater than $2$.

So is there a solution or a way to solve the problem based on the last theorem? How should I go ahead on this exercise? Any hint or help will be really appreciated.

$\endgroup$
  • $\begingroup$ Fermat's Last Theorem is a distraction here - it says, amongst other things, that there are no solutions to the original for $p\gt 2$. A false statement implies anything, so the implication is then trivially true for $p\gt 2$ and easy for $p=2$. But really the statement here is an early and elementary observation in the journey towards proving the last theorem. $\endgroup$ – Mark Bennet Jan 29 at 7:43
5
$\begingroup$

Write $$a^p+b^p-c^p=(a^p-a)+(b^p-b)-(c^p-c)+a+b-c.$$

$\endgroup$
3
$\begingroup$

By Fermat's little theorem, $0\equiv a^p+b^p-c^p\equiv a+b-c\pmod p$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.