2 variable induction proof divisibility

Question

Prove that $$a^{2^n}-1$$ is divisible by $4 \times 2^n$ for all odd integers $a$ and for all integers $n$.

I cannot for the life of me figure out how to solve this when theres 2 variables involved in the induction. Can someone please help me out?

Answer 1

Fix any odd $a$, i.e., let $a =2k+1$ for some integer $k$, and induct on $n$.

Answer 2

Hint: $\;a^{2^n}-1=a^{2 \cdot 2^{n-1}}-1=\left(a^{2^{n-1}}\right)^2-1=\left(a^{2^{n-1}}-1\right)\left(a^{2^{n-1}}+1\right)\,$, where the first factor is divisible by $\,4 \cdot 2^{n-1}\,$ by the induction hypothesis on $\,n\,$, and the second one is even.

Answer 3

$$a^{2^n}-1=(a^2-1)(a^2+1)(a^4+1)...(a^{2^{n-1}}+1)$$it is obvious that $a^{2^k}+1$ is divisible by $2$ and there are $n-1$ of them. Also the square of every odd integer is of form $8q+1$ implying that $a^2-1$ is divisible by $8$ therefore whole the multiplication is divisible by $8\cdot 2^{n-1}=4\cdot 2^n$