5
$\begingroup$

Can somebody help me with this. I am trying to prove something from Fermat's equation.

Fermat's Equation $x^n + y^n = z^n$, where $x,y,z$ and $n$ are positive integers. His last theorem states that this equation has no solution if $n \geq3$. I want to prove that if the equation has no solution if $n$ is prime or $n = 4$, then it must be true that it has no solution if $n \geq3$.

$\endgroup$
  • $\begingroup$ Thank you for your editing. I am not good at it. I appreciate that. $\endgroup$ – KwameNkrumah Apr 22 '14 at 3:49
10
$\begingroup$

Hint: If you had a solution to $x^n+y^n=z^n$, could you write it as a solution to either $a^p+b^p=c^p$ for some prime $p$ or as a solution to $a^4+b^4=c^4$? (You will only need the latter equation for a very special case. The first will suffice for most.)

Double hint: Factor $n$.

$\endgroup$
  • $\begingroup$ Can you please solve this for us because I and my group members have tried it some many times in different ways but we could not answer. $\endgroup$ – KwameNkrumah Apr 22 '14 at 4:06
  • 5
    $\begingroup$ @Kwame I'll do a specific case, and hopefully that will give you an idea how you can do it in general. Let's say you have a nontrivial solution to $x^6+y^6=z^6$. Well, we know that $6 = 2 \cdot 3$, so this is also a solution to $(x^2)^3+(y^2)^3=(z^2)^3$. But since $3$ is prime, and you know there are no nontrivial solutions to $a^3+b^3=c^3$, this is impossible. $\endgroup$ – user98602 Apr 22 '14 at 4:10
  • $\begingroup$ we could not understand it because we are concerned with n being prime or n=4 but 6 is not prime neither equals 4 $\endgroup$ – KwameNkrumah Apr 26 '14 at 19:28

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.