For which integers $k \geq 0$ is the number $2^{k}+3^{k}$ divisible by 7?
We recall that for any integer $a, a^{0}=1$. It is appropriate to study all the values that $2^{k}+3^{k}(\bmod 7)$ can take. $2^{0} \equiv 1,2^{1} \equiv 2,2^{2} \equiv 4,2^{3} \equiv 1$ and since $2^{3} \equiv 1,2^{k+3} \equiv 2^{k}$ which means that the sequence of $2^{k}(\bmod 9)$ is periodic with a period of 3. It can be demonstrated that the same is true for the sequence of powers of any integer modulo any other. For example, for $3,3^{k} \equiv 1,3,2,6,4,5,1, \cdots(\bmod 7)$ when $k=0,1,2,3,4,5,6, \cdots$. The sequence is periodic with a period of 6. We write this in a table, and find that the sequence of $2^{k}+3^{k}$ is periodic with a period of 6, and is modulo $7: 2,5,6,0,6,2$. It is therefore for $k \equiv 3(\bmod 6)$ that $2^{k}+3^{k}$ is divisible by 7.