Prove that a number raised to a prime is equal to itself and a multiple of the prime?

Question

How do I prove using induction that any number a that is greater than or equal to 1, raised to a power k(which is prime) will be equal to a + kq, where q is some integer?

Answer 1

I like the group theory proof: $\mathbb (Z/kZ)^*$ is a group, with order $k-1$. By Lagrange's theorem, the order of every element divides the order of the group. Therefore $a^{k-1}\cong1\pmod k$. Now multiply through by $a$.

But, to do it by induction, use the "freshman's dream ", which is true in characteristic k: $(x+y)^k=x^k+y^k$.

With that, the inductive step is easy: assume $a^k=a$. Then $(a+1)^k=a^k+1^k=a+1$.