I have seen the proof that Fermat gives for $$x^4 +y^4 \neq z^2$$ which we know also works for $z^4$.

BUT I am wondering if the same basic argument can be used for the power of $2^n$.

Thinks 8,16,32

Can we write a proof saying $$x^8+y^8 \neq z^8$$ More generally written as: $$x^{2^n} +y^{2^n} \neq z^{2^n}$$

Anyone have a proper way to go about this?? Maybe some helpful links?


We don't need to write a separate argument. If $x^8+y^8=z^8$ then $(x^2)^4+(y^2)^4=(z^2)^4$. But if $a^4+b^4=c^4$, then one of $a$ or $b$ is $0$.

More generally, the result about $x^4+y^4=z^2$ shows immediately that there are no non-trivial solutions for $x^{4a}+y^{4b}=z^{2c}$.

  • $\begingroup$ Is it that easy?? After I prove it for n=4 I can say that it proves the same for the form $x^{4a} + y^{4b} =x^{2c}$?? $\endgroup$ – user148072 May 8 '14 at 23:08
  • 1
    $\begingroup$ Yes, because if the above equation holds, then $u=x^a$, $v=y^b$, $w=z^c$ satisfies the equation $u^4+v^4=w^2$. But Fermat proved this equation has non solution in positive integers. Similarly, to prove Fermat's Last Theorem for all $n$, it is enough to show $x^4+y^4=z^4$ has no positive solutions, and $x^p+y^p=z^p$ has no positive solutions for all primes $p$. $\endgroup$ – André Nicolas May 9 '14 at 0:46

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