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 $n \geq3$.

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

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 $n$.

  • $\begingroup$ 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
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    $\begingroup$ @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

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