Show that if x,y,z are not divisible by 53, then $x^{26}+4y^{26} \neq\ z^{26}$ Show that if x,y,z are not divisible by 53, then $x^{26}+4y^{26} \neq\ z^{26}$
I've got that $x,y,z$ to the 52nd power are congruent to 1 modulo 53 from Fermat's.  How is it continued? Help would be appreciated.

I think you have meant $x^{26}+4y^{26}=z^{26}$
Now using Fermat's Litlle theorem, $u^{52}\equiv1\pmod{53}$ for $53\nmid u\iff(u,53)=1$
$\implies 53|(u^{26}-1)(u^{26}+1)\implies u^{26}\equiv\pm1\pmod{53}$
Observe that $\pm1+4(\pm1)\not\equiv\pm1\pmod{53}$
