4
$\begingroup$

I'm trying to solve $8^x \equiv 2 \mod 23$ using Fermat's little theorem.

We have $2^{3x} \equiv 2 \mod 23$, then $3x=23$, but this doesn't work.

Could somebody please help?

$\endgroup$
1
$\begingroup$

As $(2,11)=1$, we have $2^{3x-1}\equiv1\pmod{23}$

Now, $2^2=4\not\equiv1,2^5=32\equiv9,2^{10}\equiv9^2\equiv12,2^{11}\equiv24\equiv1$

So, $3x-1\equiv0\pmod{11}\iff3x\equiv1\pmod{11}\equiv1+11$

$\implies x\equiv4\pmod{11}$ as $(3,11)=1$

$\endgroup$
0
$\begingroup$

You have $2^{22}\equiv 1$ so you should be able to see that there are alternatives to $3x=23$.

$\endgroup$

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.