2-adic valuation of $5^{2^{k-2}}-1$

Question

I have the following exercice :

1. Consider that the 2-adic valuation of $x$ is $k \geq 2$. Prove that the 2-adic valuation of $(1+x)^2- 1$ is $k+1$.

2. Using the previous question, conclude that for $k\geq 2$, the 2-adic valuation of $5^{2^{k-2}}-1$ is $k$.

For the first l question, we know that $(1+x)^2 - 1= x(x+2) $. So, $$v_2((1+x)^2-1) = v_2(x)+\min(v_2(x),v_2(2))= k+1.$$

For the last question, I tried several methods but didn't work. Can someone give me a hint please?

Thanks!

Answer 1

1. If $v_2(x)=k\ge 2$ then $v_2(x+2)=1$ and $v_2( (1+x)^2-1)=v_2 (x(x+2)) =v_2 (x)+v_2(x+2) =k+1$.

2. Let $x_{k+2}=5^{2^k}-1=(1+x_{k+1})^2-1, v_2(x_2)=2$, by induction if $v_2(x_k)=k$ then $v_2(x_{k+1})=k+1$.