Someone posted an interesting number theory question earlier. I think I'm making a mistake with Fermat's Little Theorem, but I"m not sure how.
Original problem, find $x$, $y$, $z$ natural numbers so that $7^x+13^y=2^{3z}$.
One obvious solution is $x=1$, $y=0$, and $z=1$.
Re-expressing the equation modulo 7 implies $y$ is even.
Given $y$ is even then the equation in modulo 8 implies $x$ is odd.
Modulo 13, 2 and 7 are multiplicative inverses.
So $7^{x+3z}=1 \pmod{13}$
By Fermat's Little Theorem, it follows that $x+3z=0 \pmod{12}$
But this congruence is not satisfied if x=1 and z=1, a known solution to the problem.
Am I missing something?
Best Answer
You assume that $13^y$ is divisible by $13$ which is wrong for $y=0$. So your claim $7^{x+3z}\equiv 1\pmod{13}$ is incorrect.