Prove that $5^{2^n} - 1$ is divisible by $2^{n+1}$ for all $n ≥ 1$

Question

I made the following using induction:

If $n=1$ then the proposition is true: $5^{2^1} - 1=24$ is divisible by $2^{1+1} = 4$

Now I suppose that for a natural number $k$, $5^{2^k} - 1$ is divisible by $2^{k+1}$ is true. And I want to prove (using this) that the proposition is true for $n=k+1$ but I don't know how to do this.

I appreciate the help you give me.

Answer 1

Step $n+1$:

\begin{align} 5^{2^{n+1}}-1 &= 5^{2^n \times 2}-1 \\ &= (5^{2^n})^2-1 \\ &= (5^{2^n}-1)(5^{2^n}+1) \\ &=(k2^{n+1})(5^{2^n}+1) \end{align}

For the first factor above the hypothesis (step $n$) is used,

the second factor is even, say $2p$, since a power of $5$ is odd.

Combining:

$$(k2^{n+1})2p= (kp)2^{n+2}$$

Answer 2

$$5^{2^{k+1}}=(5^{2^k})^2=(a2^{k+1}+1)^2=1+2^{k+2}a(?)$$

Answer 3

We have:

$$5^{2^{k+1}} - 1 = \left(5^{2^k} - 1\right)\left(5^{2^k}+1\right)$$

By induction the first factor is divisible by $2^{k+1}$, while the second one is obviously even and hence $5^{2^{k+1}} - 1$ is divisible by $2^{k+2}$

Answer 4

Consider $n=k+1$:

$$5^{2^{k+1}}-1=\left(5^{2^k}\right)^2-1=(5^{2^k}-1)(5^{2^k}+1)=\cdots$$

Answer 5

First assume that $5^{2^k} - 1$ is divisible by $2^{k + 1}$ so $2^{k + 1}\,|\,5^{2^k} - 1$. Now consider $$5^{2^{k + 1}} - 1 = 5^{2\cdot 2^k} - 1 = (5^{2^k})^2 - 1 = (5^{2^k} - 1)(5^{2^k} + 1)$$

Now we need to show that $2^{(k + 1) + 1} = 2^{k + 2}$ divides $(5^{2^k} - 1)(5^{2^k} + 1)$. Our inductive assumption allows to know that $2^{k + 1}\,|\, 5^{2^k} - 1$. Now we just need to show that $2 \,|\, 5^{2^k} + 1$. Well $5^{2^k}$ will always be an odd number, so adding one makes it even and hence $2 \,|\, 5^{2^k} + 1$.

Since $2$ and $2^{k + 1}$ divide $5^{2^k} + 1$ and $5^{2^k} - 1$, then $2\cdot 2^{k + 1} = 2^{k + 2} \,| 5^{2^{k + 1}} - 1\,$. By induction then $2^{n+ 1}\,|\, 5^{2^n} - 1$ for all $n \in \mathbb{N}$.