A Congruence Problem for RMO basics

Problem:

Find the solution of the congruence $x^{24}+7x \equiv 2 \pmod {13}$ .

Solution:

It is clear that $x \equiv 0 \pmod {13}$ is not a solution. So, let $1 \leq x \leq 12$ . Then, from Fermat’s Little Theorem, we have that $x^{12} \equiv 1 \pmod{13}$ and this is why $x^{24}= 1 \pmod{13}$ . The congruence to be solved can therefore be reduced to $7x \equiv 1 \pmod{13}$ which leads to solution $x \equiv 2 \pmod{13}$ .