Solving higher congruence

Question

How do you go about solving $x^{11} \equiv 5 \pmod{41}$. I'm very new to this topic and have watched numerous YouTube tutorials with not much luck so far.

Thanks in advance.

Answer 1

Expanding on @DanielFischer 's comment. Since $41$ is prime, one knows that $x^{41}\equiv x\pmod{41}$ via Fermat's little theorem. Now, you know that $5^{41}\equiv 5\mod 41$.

Our job is now to rewrite $5^{41}$ as $x^{11}$. Since $5^{40}\equiv 1\pmod{41}$ by Fermat's little theorem, we wish to find a $k$ so that $(5^{40})^k5^{41}$ is of the form $x^{11}$. In other words, you want to find a $k$ so that the exponent of $5^{40k+41}$ is a multiple of $11$.

Therefore, you will solve $40k+41\equiv 0\pmod{11}$. Once you have that $k$, you know that $40k+41=11r$ and your answer is $5^r$.