Prove $\frac{2^{ab}-1}{2^a-1} $ is composite number let $b>a>1$ be postive integers,show that
$$\dfrac{2^{ab}-1}{2^a-1} $$ is composite number
I try use $$(x^n-1)=(x-1)(x^{n-1}+x^{n-2}+\cdots+x+1)$$
so $x-1|x^n-1$.then let $x=2^a,n=b$ we have
$$\dfrac{2^{ab}-1}{2^a-1}=2^{(b-1)a}+2^{(b-2)a}+\cdots+2^a+1$$
but How to prove $2^{(b-1)a}+2^{(b-2)a}+\cdots+2^a+1$ is composite ？

We use this $fact*$ that:
if m is a natural number and $a>1$ we have:
$$\big(\frac{a^m-1}{a-1}, a-1\big)= (a-1, m)$$
Proof:
Suppose $\big(\frac{a^m-1}{a-1}, a-1\big)=d$
Using following identity:
$\frac{a^m-1}{a-1}=(x^{m-1}-1)+(x^{m-2}-1)+(x^{m-3}-1)+\cdot\cdot\cdot+(a-1)+m$ $\space\space\space\space\space (1)$
and the fact that ($a^k-1$) is divisible by $(a-1)$ we conclude that m is divisible by d. If numbers $(a-1)$ and $m$ have a common divisor like ($\sigma>d$), considering relation (1) we conclude that ($\frac{a^m-1}{a-1}$) is divisible by ($\sigma$) and numbers ($\frac{a^m-1}{a-1}$) and ($a-1$) have a common divisor $(\sigma>d)$., which results in $(a-1, m)=d$
Now we have :
$A=\frac {2^{ab}-1}{2^a-1}=\big(\frac{(2^a)^b-1}{2^a-1}\big)$ which has a common divisor like d with $(m=b)$ that means $A$ is divisible by d, so it is a composite number.

*

*This fact and its proof is from a number theory book by Sierpinski.

